x_{0}&\dots&x_{i-1}&(x_{i}&x_{i})&x_{i+2}&\dots&x_{n}\end{pmatrix}$

indicates that $x_{i}$ is inserted in $f_{0},\dots,f_{n}$ and then also into
$f_{0}^{\prime},\dots,f_{n}^{\prime}$.

Extending this to zeros of multiplicity $m$ for $\mathcal{C}^{m-1}$-functions
is straight forward and we leave it to the reader to write down the formulas.
Similar to (36) we write for any $a\leq x_{0}\leq x_{1}\leq\dots\leq x_{n}\leq
b$ the matrix as

$\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}^{*}$ (37)

when $f_{0},\dots,f_{n}$ are sufficiently differentiable.

We often want to express polynomials $f\in\mathrm{lin}\,\mathcal{F}$ as
determinants (31) only by knowing their zeros $x_{1},\dots,x_{k}$. If
arbitrary multiplicities appear we only have $x_{1}\leq x_{2}\leq\dots\leq
x_{n}$ where we include zeros multiple times according to their algebraic
multiplicities. Hence, for

$x_{0}=\dots=x_{i_{1}}\;<\;x_{i_{1}+1}=\dots=x_{i_{2}}\;<\;\dots\;<\;x_{i_{k}+1}=\dots=x_{n}$

we introduce a simpler notation to write down (36):

$\left(\begin{array}[]{c|cccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{n}\\\
x&\,x_{1}&x_{2}&\dots&x_{n}\end{array}\right):=\begin{pmatrix}f_{0}&f_{1}\;\dots\;f_{i_{1}}&f_{i_{1}+1}\;\dots\;f_{i_{2}}&\dots&f_{i_{k}+1}\;\dots\;f_{i_{k}+1}\\\
x&(x_{1}\;\dots\;x_{i_{1}})&(x_{i_{1}+1}\;\dots\;x_{i_{2}})&\dots&(x_{i_{k}+1}\;\dots\;\;\;x_{n})\;\,\end{pmatrix}.$
(38)

Clearly $(\ref{eq:doublezeroDfn2})\in\mathrm{lin}\,\mathcal{F}$. For (38) to
be well-defined we need $\mathcal{F}\subseteq\mathcal{C}^{m-1}$ where $m$ is
the largest multiplicity of any zero.

We see here why we require in 1
$\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}\subseteq\mathcal{C}^{n}([a,b],\mathds{R})$.
In the case of $x_{0}=x_{1}=\dots=x_{n}$ the functions $f_{i}$ need to be
$\mathcal{C}^{n}([a,b],\mathds{R})$, not just
$\mathcal{C}^{n-1}([a,b],\mathds{R})$.

Similar to 5 we have the following.

######  Theorem 3 (Kre (51) or e.g. (KN, 77, p. 37, P.1.1)).

Let $n\in\mathds{N}$ and
$\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}\subseteq\mathcal{C}^{n}([a,b],\mathds{R})$
with $a<b$. Then the following are equivalent:

  1. (i)

$\mathcal{F}$ is an ET-system.

  2. (ii)

We have

$\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}^{*}\neq 0$

for every $a\leq x_{0}\leq x_{1}\leq\dots\leq x_{n}\leq b$.

######  Proof 27.1.

Let $x_{0},\dots,x_{n}\in[a,b]$ with

$a\;\leq\;x_{0}=\dots=x_{i_{1}}\;<\;x_{i_{1}+1}=\dots=x_{i_{2}}\;<\;\dots\;<\;x_{i_{k}}=\dots=x_{n}\;\leq\;b$

be the zeros of some $f=\sum_{i=0}^{n}a_{n}f_{i}\in\mathrm{lin}\,\mathcal{F}$.
We get the coefficients $a_{0},\dots,a_{n}$ from the system

$0=\begin{pmatrix}f(x_{0})\\\ f^{\prime}(x_{0})\\\ \vdots\\\
f^{(i_{1})}(x_{0})\\\ f(x_{i_{1}+1})\\\ \vdots\\\
f^{(n-i_{k})}(x_{i_{k}})\end{pmatrix}=\underbrace{\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}^{*}}_{=:M}\cdot\begin{pmatrix}a_{0}\\\
a_{1}\\\ \vdots\\\ a_{n}\end{pmatrix}.$ (39)

Hence, since $x_{0},\dots,x_{n}$ are arbitrary we have (i) $\mathcal{F}$ is an
ET-systems $\Leftrightarrow a_{0}=\dots=a_{n}=0\Leftrightarrow$ (39) has only
the trivial solution $\Leftrightarrow$ $M$ has full rank $\Leftrightarrow$
(ii).

######  Remark 4.

Similar to 7 for T-systems we can assume after a sign change in $f_{n}$ that
for every ET-system $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ on $[a,b]$ we have
that

$\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}^{*}>0$

holds for all $a\leq x_{0}\leq x_{1}\leq\dots\leq x_{n}\leq b$ since
$\mathcal{F}\subseteq\mathcal{C}^{n}([a,b],\mathds{R})$. $\circ$

An even more special case of ET-systems and therefore T-systems are the ECT-
systems which we define now.

######  Definition 5.

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ with $a<b$. The family
$\mathcal{F}=\\{f_{0}\\}_{i=0}^{n}$ is called an _extended complete
Tchebycheff system_ (short _ECT-system_) _on $[a,b]$_ if
$\\{f_{i}\\}_{i=0}^{k}$ is an ET-system on $[a,b]$ for all $k=0,\dots,n$.

####  28 Wronskian Determinant

To handle and work with ECT-systems it is useful to introduce the following
determinant.

######  Definition 6.

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ be with $a<b$. For
each $k=0,\dots,n$ we define the _Wronskian determinant_ (short _Wronskian_)
$\mathcal{W}(f_{0},\dots,f_{k})$ of $f_{0},\dots,f_{k}$ to be

$\mathcal{W}(f_{0},f_{1},\dots,f_{k}):=\det\begin{pmatrix}f_{0}&f_{0}^{\prime}&\dots&f_{0}^{(k)}\\\
f_{1}&f_{1}^{\prime}&\dots&f_{1}^{(k)}\\\ \vdots&\vdots&&\vdots\\\
f_{k}&f_{k}^{\prime}&\dots&f_{k}^{(k)}\end{pmatrix}.$ (40)

The Wronskian is a common tool in the theory of ordinary differential
equations.

In the previous definition (40) we could also shortly write

$\mathcal{W}(f_{0},\dots,f_{k})(x):=\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{k}\\\
x&x&\dots&x\end{pmatrix}^{*}$

for all $x\in[a,b]$.

Let $m_{1},\dots,m_{k}\in\mathds{N}$ with $m_{1}+\dots+m_{k}=n+1$ and
$x_{1}<\dots<x_{k}$. Then the first $m_{j}$ columns of
$\mathcal{W}(f_{0},\dots,f_{n})$ are the $m_{j}$ columns in

$\begin{pmatrix}f_{0}&\dots&f_{m_{1}-1}&f_{m_{1}}&\dots&f_{m_{1}+m_{2}-1}&f_{m_{1}+m_{2}}&\dots&f_{n}\\\
x_{1}&\dots&x_{1}&x_{2}&\dots&x_{2}&x_{3}&\dots&x_{k}\end{pmatrix}^{*}$

involving $x_{j}$.

######  Lemma 7.

Let $n\in\mathds{N}_{0}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system on $[a,b]$ with $a<b$, and let $g\in\mathcal{C}^{n}([a,b],\mathds{R})$
with $g>0$. Then

$\mathcal{G}:=\\{g_{i}\\}_{i=0}^{n}\quad\text{with}\quad g_{i}:=g\cdot f_{i}$

is an ET-system and we have

$\mathcal{W}(g_{0},\dots,g_{n})=g^{n+1}\cdot\mathcal{W}(f_{0},\dots,f_{n}).$

######  Proof 28.1.

See Problem 21.

######  Lemma 8.

Let $n\in\mathds{N}_{0}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system on $[c,d]$, and $g\in\mathcal{C}^{n}([a,b],[c,d])$ with $g^{\prime}>0$
on $[a,b]$. Then

$\mathcal{G}:=\\{f_{i}\circ g\\}_{i=0}^{n}\quad\text{with}\quad
g_{i}:=f_{i}\circ g$

is an ET-system on $[a,b]$ with

$\mathcal{W}(g_{0},\dots,g_{n})=(g^{\prime})^{\frac{n(n+1)}{2}}\cdot\mathcal{W}(f_{0},\dots,f_{n})\circ
g.$

######  Proof 28.2.

See Problem 22.

For the Wronskian the following reduction property holds.

######  Lemma 9 (see e.g. (KS, 66, p. 377)).

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ be with $a<b$ and
$f_{0}>0$. Then for the _reduced system_
$g_{0},\dots,g_{n-1}\in\mathcal{C}^{n-1}([a,b],\mathds{R})$ defined by

$g_{i}:=\left(\frac{f_{i+1}}{f_{0}}\right)^{\prime}$ (41)

for all $i=0,\dots,n-1$ we have

$\mathcal{W}(f_{0},\dots,f_{n})=f_{0}^{n+1}\cdot\mathcal{W}(g_{0},\dots,g_{n-1}).$
(42)

######  Proof 28.3.

See Problem 23.

######  Remark 10.

Since $f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ we have that
$\mathcal{W}(f_{0},\dots,f_{k})(x)$ is continuous in $x\in[a,b]$ and hence
after adjusting the signs of $f_{0},\dots,f_{n}$ we have that (40) being non-
zero on $[a,b]$ is equivalent to $\mathcal{W}(f_{0},\dots,f_{k})>0$ on $[a,b]$
for all $k=0,\dots,n$, see also 7 and 4. $\circ$

######  Lemma 11 (see e.g. (KS, 66, pp. 242–245, Lem. 5.1 - 5.3)).

Let $n\in\mathds{N}_{0}$ and let
$f_{1},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ be such that

$\mathcal{W}(f_{0})>0,\quad\dots,\quad\mathcal{W}(f_{0},\dots,f_{n})>0$

on $[a,b]$. Define functions $g_{0},\dots,g_{n}:[a,b]\to\mathds{R}$ by

$\displaystyle g_{0}$ $\displaystyle:=f_{0}$ $\displaystyle g_{1}$
$\displaystyle:=D_{0}f_{1}$ $\displaystyle g_{2}$
$\displaystyle:=D_{1}D_{0}f_{2}$ $\displaystyle\ \;\;\vdots$ $\displaystyle
g_{n}$ $\displaystyle:=D_{n-1}\dots D_{1}D_{0}f_{n}$

with

$D_{j}f:=\left(\frac{f}{g_{j}}\right)^{\prime},\quad i.e.,\quad
D_{j}=\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{g_{j}}.$ (43)

Then

  1. (i)

$g_{i}\in\mathcal{C}^{n-i}([a,b],\mathds{R})$ are well defined with

$g_{1}=\frac{\mathcal{W}(f_{0},f_{1})}{f_{0}^{2}}\qquad\text{and}\qquad
g_{i}=\frac{\mathcal{W}(f_{0},\dots,f_{i})\cdot\mathcal{W}(f_{0},\dots,f_{i-2})}{\mathcal{W}(f_{0},\dots,f_{i-1})^{2}}$

for all $i=2,\dots,n$,

  2. (ii)

$g_{i}>0$ on $[a,b]$ for all $i=0,\dots,n$,

  3. (iii)

for any $g_{n+1}\in\mathcal{C}([a,b],\mathds{R})$ with $g_{n+1}>0$ on $[a,b]$
we define

$f_{n+1}(x):=g_{0}(x)\int_{a}^{x}g_{1}(y_{1})\int_{a}^{y_{1}}g_{2}(y_{2})\dots\int_{a}^{y_{n}}g_{n+1}(y_{n+1})~{}\mathrm{d}y_{n+1}\dots\mathrm{d}y_{1}$

and we get

$g_{n+1}=D_{n}\dots D_{1}D_{0}f_{n+1},$

  4. (iv)

for all $k=0,\dots,n+1$ we have

$\mathcal{W}(f_{0},\dots,f_{k})=g_{0}^{k+1}g_{1}^{k}\cdots g_{k}$

with $g_{n+1}$ and $f_{n+1}$ from (iii),

  5. (v)

there exists a $f_{n+1}\in\mathcal{C}^{n+1}([a,b],\mathds{R})$ such that

$\mathcal{W}(f_{0},\dots,f_{n},f_{n+1})>0$

on $[a,b]$, and

  6. (vi)

for all $k=0,\dots,n+1$ the families $\\{f_{i}\\}_{i=0}^{k}$ are T-systems on
$[a,b]$.

######  Proof 28.4.

(i) and (ii): Since $\mathcal{W}(f_{0})>0$ we have $f_{0}>0$ and hence
$g_{1}=(f_{1}/f_{0})^{\prime}$ is well-defined and we have

$\frac{\mathcal{W}(f_{0},f_{1})}{f_{0}^{2}}=f_{0}^{-2}\cdot\det\begin{pmatrix}f_{0}&f_{0}^{\prime}\\\
f_{1}&f_{1}^{\prime}\end{pmatrix}=\frac{f_{0}f_{1}^{\prime}-f_{1}f_{0}^{\prime}}{f_{0}^{2}}=\left(\frac{f_{1}}{f_{0}}\right)^{\prime}=g_{1},$

i.e., $g_{1}>0$ on $[a,b]$. The relations for $g_{i}$ for all $i=2,\dots,n$
follow by induction from Sylvester’s identity Syl (51); AAM (96).

(iii): From the definition of $f_{n+1}$ we get immediately $g_{n+1}=D_{n}\dots
D_{1}D_{0}f_{n+1}$.

(iv): Follows immediately from (i).

(v): Take the $f_{n+1}$ from (iii).

(vi): For $k=0$ it is clear that $\\{f_{i}\\}_{i=0}^{0}$ is a T-system since
$f_{0}>0$ on $[a,b]$. So assume that for any $f_{0},\dots,f_{n-1}$ with
$\mathcal{W}(f_{0},\dots,f_{k})>0$ on $[a,b]$ for all $k=0,\dots,n-1$ we have
that all $\\{f_{i}\\}_{i=0}^{k}$ with $k=0,\dots,n-1$ are T-systems. We show
that $\\{f_{i}\\}_{i=0}^{n}$ is also a T-system. So let
$x_{0},\dots,x_{n}\in[a,b]$ with $x_{0}<x_{1}<\dots<x_{n}$. We then have

$\displaystyle\det\begin{pmatrix}f_{0}&\dots&f_{n}\\\
x_{0}&\dots&x_{n}\end{pmatrix}=\det(f_{i}(x_{j}))_{i,j=0}^{n}$ and factoring
out $f_{0}(x_{j})>0$ in each column gives
$\displaystyle=\prod_{j=0}^{n}f_{0}(x_{j})\cdot\det\left(\tilde{f}_{i}(x_{j})\right)_{i,j=0}^{n}$
with $\tilde{f}_{i}:=f_{i}/f_{0}$ for all $i=0,\dots,n$ and substracting from
each row its predecessor (the row above) gives
$\displaystyle=\prod_{j=0}^{n}f_{0}(x_{j})\cdot\det\left(\delta_{0,j},\tilde{f}_{1}(x_{j})-\tilde{f}_{1}(x_{j-1}),\dots,\tilde{f}_{n}(x_{j})-\tilde{f}_{n}(x_{j-1})\right)_{j=0}^{n}.$
Expanding along the first column and applying the theorem of the mean gives
$\displaystyle=\prod_{j=0}^{n}f_{0}(x_{j})\cdot\prod_{i=0}^{n-1}(x_{i+1}-x_{i})\cdot\det\left(\hat{f}_{i}(y_{j})\right)_{i,j=0}^{n-1}$

for some $y_{0},\dots,y_{n-1}$ with
$x_{0}<y_{0}<x_{1}<y_{1}<\dots<y_{n-1}<x_{n}$ and
$\hat{f}_{i}:=(f_{i+1}/f_{0})^{\prime}$ for all $i=0,\dots,n-1$. The family
$\\{\hat{f}_{i}\\}_{i=0}^{n-1}$ is the reduced system from 9 and hence by (42)
we have

$\mathcal{W}(\hat{f}_{0},\dots,\hat{f}_{k-1})=\frac{\mathcal{W}(f_{0},\dots,f_{k})}{f_{0}^{k+1}}>0$

on $[a,b]$ for all $k=1,\dots,n$. By the induction hypothesis we have that
$\\{\hat{f}_{i}\\}_{i=0}^{n-1}$ is a T-system, i.e.,

$\det\left(\hat{f}_{i}(y_{j})\right)_{i,j=0}^{n-1}\neq
0\quad\Rightarrow\quad\det\begin{pmatrix}f_{0}&\dots&f_{n}\\\
x_{0}&\dots&x_{n}\end{pmatrix}\neq 0$

and $\\{f_{i}\\}_{i=0}^{n}$ is a T-system which ends the proof.

The previous lemma is used to characterize all ECT-systems.

####  29 Characterizations of ECT-Systems

We have the following characterization of ECT-systems.

######  Theorem 12 (see e.g. (KS, 66, p. 376, Thm. 1.1)).

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ be with $a<b$. The
following are equivalent:

  1. (i)

$\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ is an ECT-system.

  2. (ii)

For all $k=0,\dots,n$ we have that $\mathcal{W}(f_{0},\dots,f_{k})\neq 0$ on
$[a,b]$.

After adjusting the signs of $f_{0},\dots,f_{n}$ by 10 we can in 12 (ii) also
assume that $\mathcal{W}(f_{0},\dots,f_{k})>0$ on $[a,b]$ for all
$k=0,\dots,n$.

The following proof is adapted from (KS, 66, pp. 376–379).

######  Proof 29.1.

(i) $\Rightarrow$ (ii): Since every ECT-system is also an ET-system the
statement is 3 (i) $\Rightarrow$ (ii) because

$\mathcal{W}(f_{0},\dots,f_{k})(x)=\begin{pmatrix}f_{0}&f_{1}&\dots&f_{k}\\\
x&x&\dots&x\end{pmatrix}^{*}$

for all $x\in[a,b]$.

(ii) $\Rightarrow$ (i): To show that $\mathcal{F}$ is an ECT-system we have to
show that $\\{f_{i}\\}_{i=0}^{k}$ is an ET-system for all $k=0,\dots,n$. And
to show that $\\{f_{i}\\}_{i=0}^{k}$ is an ET-system it is by 3 sufficient to
show

$\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{k}\\\
x_{0}&x_{1}&\dots&x_{k}\end{pmatrix}^{*}\neq 0$

for every $a\leq x_{0}\leq x_{1}\leq\dots\leq x_{k}\leq b$. We make two case
distinctions:

  1. Case I:

All $x_{0},\dots,x_{k}$ are pairwise distinct: $x_{0}<x_{1}<\dots<x_{n}$.

  2. Case II:

At least once we have $x_{j}=x_{j+1}$ for some $j=0,\dots,n-1$.

After renaming $x_{0},\dots,x_{k}$ we can assume $a\leq
x_{1}<x_{2}<\dots<x_{l}\leq b$ and $m_{1},\dots,m_{l}\in\mathds{N}$ are the
algebraic multiplicities with $m_{1}+\dots+m_{l}=n+1$ for some
$l\in\mathds{N}_{0}$.

Case I: We have $m_{0}=\dots=m_{k}=1$ and that is 11 (vi).

Case II: We assume $m_{j}\geq 2$ for some $j$. We show that we can reduce the
system.

We show this reduction by induction over $n$.

Induction beginning ($n=0$): Since $\mathcal{W}(f_{0})(x)\neq 0$ it is an ET-
and an ECT-system. We can assume by changing the sign of $f_{0}$ that
$f_{0}>0$ on $[a,b]$.

Induction step ($n-1\to n$): By the induction beginning ($n=0$) we can assume
$f_{0}>0$ on $[a,b]$. Then we have to show that

$\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{m_{1}-1}&f_{m_{1}}&\dots&f_{n}\\\
x_{1}&x_{1}&\dots&x_{1}&x_{2}&\dots&x_{l}\end{pmatrix}^{*}$ (44)

is non-zero. To show this we factor $f_{0}(x_{j})>0$ out of the $m_{j}$ rows
containing $x_{j}$ in (44) for each $j=0,\dots,l$ to get

$\det\begin{pmatrix}1&\frac{f_{0}^{\prime}}{f_{0}}(x_{1})&\dots&\frac{f_{0}^{(m_{1}-1)}}{f_{0}}(x_{1})&1&\dots&\frac{f_{0}^{(m_{l}-1)}}{f_{0}}(x_{l})\\\
\frac{f_{1}}{f_{0}}(x_{1})&\frac{f_{1}^{\prime}}{f_{0}}(x_{1})&\dots&\frac{f_{1}^{(m_{1}-1)}}{f_{0}}(x_{1})&\frac{f_{1}}{f_{0}}(x_{2})&\dots&\frac{f_{1}^{(m_{l}-1)}}{f_{0}}(x_{l})\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\\
\frac{f_{n}}{f_{0}}(x_{1})&\frac{f_{n}^{\prime}}{f_{0}}(x_{1})&\dots&\frac{f_{n}^{(m_{1}-1)}}{f_{0}}(x_{1})&\frac{f_{n}}{f_{0}}(x_{2})&\dots&\frac{f_{n}^{(m_{l}-1)}}{f_{0}}(x_{l})\end{pmatrix}.$

Then subtract from each of the columns containing $x_{j}$ a linear combination
of its predecessors to obtain for these $m_{j}$ columns the first $m_{j}$
columns of $\mathcal{W}(1,f_{1}/f_{0},\dots,f_{n}/f_{0})$ evaluated at
$x_{j}$:

$\det\begin{pmatrix}1&\left(\frac{f_{0}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{0}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&1&\dots&\left(\frac{f_{0}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\\\
\frac{f_{1}}{f_{0}}(x_{1})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\frac{f_{1}}{f_{0}}(x_{2})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\\
\frac{f_{n}}{f_{0}}(x_{1})&\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\frac{f_{n}}{f_{0}}(x_{2})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\end{pmatrix}\\\
=\det\begin{pmatrix}1&0&\dots&0&1&\dots&0\\\
\frac{f_{1}}{f_{0}}(x_{1})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\frac{f_{1}}{f_{0}}(x_{2})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\\
\frac{f_{n}}{f_{0}}(x_{1})&\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\frac{f_{n}}{f_{0}}(x_{2})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\end{pmatrix}.$

The Leibniz rule on differentiation, here for us explicitly

$\left(\frac{f_{i}}{f_{0}}\right)^{(k)}=\sum_{j=0}^{k}\binom{k}{j}\cdot
f_{i}^{(k-j)}\cdot\left(\frac{1}{f_{0}}\right)^{(j)},$

ensures that this is always possible.

We then subtract from each column which starts with a $1$ its predecessor
which also starts with a $1$ and apply the mean value theorem to get apart
from the positive factor $(x_{j+1}-x_{j})$

$\det\begin{pmatrix}1&0&\dots&0&0&0&\dots&0\\\
\frac{f_{1}}{f_{0}}(x_{1})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(y_{2})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{2})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\\
\frac{f_{n}}{f_{0}}(x_{1})&\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(y_{2})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{2})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\end{pmatrix}$

with $x_{1}<y_{2}<x_{2}<\dots<x_{l}$ and expanding by the first row gives

$\det\begin{pmatrix}\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(y_{2})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{2})&\dots&\left(\frac{f_{1}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\\\
\vdots&&\vdots&\vdots&&\vdots\\\
\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(x_{1})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{1}-1)}(x_{1})&\left(\frac{f_{n}}{f_{0}}\right)^{\prime}(y_{2})&\left(\frac{f_{1}}{f_{0}}\right)^{\prime}(x_{2})&\dots&\left(\frac{f_{n}}{f_{0}}\right)^{(m_{l}-1)}(x_{l})\end{pmatrix}.$
(45)

In (45) we now have the reduced system $g_{i}:=(f_{i+1}/f_{0})^{\prime}$ with
$i=0,\dots,n-1$ from (41) in 9. By (42) in 9 and since the reduced systems is
of dimension $n-1$ where the inductions hypotheses holds we have that (45) is
non-zero and hence also (44) is non-zero which we wanted to prove.

######  Remark 13 (see e.g. (KS, 66, p. 379, Rem. 1.2)).

We find the following complete characterization of ECT-systems which requires
the additional property (46). Fortunately, this seemingly additional property
can always be generated by a change of basis vectors, i.e., for any vector
space spanned by an ECT-system a suitable basis with (46) can be found.
$\circ$

######  Theorem 14 (see e.g. (KS, 66, p. 379, Thm. 1.2)).

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n}\in\mathcal{C}^{n}([a,b],\mathds{R})$ be such that

$f_{j}^{(k)}(a)=0$ (46)

holds for all $k=0,\dots,j-1$ and $j=1,\dots,n$. After suitable sign changes
in $f_{0},\dots,f_{n}$ the following are equivalent:

  1. (i)

There exist $g_{0},\dots,g_{n}$ with
$g_{i}\in\mathcal{C}^{n-i}([a,b],\mathds{R})$ and $g_{i}>0$ on $[a,b]$ for all
$i=0,\dots,n$ such that

$\displaystyle f_{0}(x)$ $\displaystyle=g_{0}(x)$ $\displaystyle f_{1}(x)$
$\displaystyle=g_{0}(x)\cdot\int_{a}^{x}g_{1}(y_{1})~{}\mathrm{d}y_{1}$
$\displaystyle f_{2}(x)$
$\displaystyle=g_{0}(x)\cdot\int_{a}^{x}g_{1}(y_{1})\cdot\int_{a}^{y_{1}}g_{2}(y_{2})~{}\mathrm{d}y_{2}~{}\mathrm{d}y_{1}$
$\displaystyle\;\,\vdots$ $\displaystyle f_{n}(x)$
$\displaystyle=g_{0}(x)\cdot\int_{a}^{x}g_{1}(y_{1})\cdot\int_{a}^{y_{1}}g_{2}(y_{2})~{}{\dots}\int_{a}^{y_{n-1}}g_{n}(y_{n})~{}\mathrm{d}y_{n}~{}\dots~{}\mathrm{d}y_{2}~{}\mathrm{d}y_{1}.$

  2. (ii)

$\\{f_{i}\\}_{i=0}^{n}$ is an ECT-system on $[a,b]$.

  3. (iii)

$\mathcal{W}(f_{0},\dots,f_{k})>0$ on $[a,b]$ for all $k=0,\dots,n$.

If one and therefore all of the equivalent conditions (i) – (iii) hold then
the $g_{i}$ in (i) are given by

$g_{0}:=f_{0}\quad\text{and}\quad g_{i}:=D_{i-1}\dots
D_{1}D_{0}f_{i}\quad\text{with}\quad
D_{i}:=\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{f_{0}}$

for all $i=1,\dots,n$ or equivalently by

$g_{0}:=f_{0},\quad
g_{1}:=\frac{\mathcal{W}(f_{0},f_{1})}{f_{0}^{2}},\quad\text{and}\quad
g_{i}:=\frac{\mathcal{W}(f_{0},\dots,f_{i})\cdot\mathcal{W}(f_{0},\dots,f_{i-2})}{\mathcal{W}(f_{0},\dots,f_{i-1})^{2}}$

for all $i=2,\dots,n$.

######  Proof 29.2.

“(ii) $\Leftrightarrow$ (iii)” is 12, “(iii) $\Rightarrow$ (i)” is 11 (i) –
(iii), and “(i) $\Rightarrow$ (iii)” is 11 (iv).

Condition (ii) in 14 is of course to be understood after suitable sign changes
in $f_{0},\dots,f_{n}$.

The partial statement 14 (i) $\Rightarrow$ (ii) can be found e.g. in (KS, 66,
p. 19, Exm. 12) and (KN, 77, pp. 39–40, P.2.4).

####  30 Examples of ET- and ECT-Systems

An equivalent result as 3 for T-systems, i.e., restricting the domain
$\mathcal{X}$ of a T-system leads again to a T-system, also holds for ET- and
ECT-systems. We leave that to the reader, see Problem 24. Hence, it is
sufficient to give (examples of) ET- and ECT-systems with the largest possible
domain $\mathcal{X}\subseteq\mathds{R}$.

While the condition of being an ET-system or being even an ECT-system seems
very restrictive, several examples are known.

######  Example 15.

Let $n\in\mathds{N}_{0}$ and $\mathcal{F}=\\{x^{i}\\}_{i=0}^{n}$. Then
$\mathcal{F}$ on $\mathds{R}$ is an ECT-system. $\circ$

######  Proof 30.1.

Clearly, $\mathcal{F}\subset\mathcal{C}^{\infty}(\mathds{R},\mathds{R})$ and
every non-trivial $f\in\mathrm{lin}\,\mathcal{F}=\mathds{R}[x]_{\leq n}$ has
at most $n$ real zeros counting multiplicities by the fundamental theorem of
algebra, i.e., $\mathcal{F}$ is an ET-systems. Besides that we have that

$\mathcal{W}(1,x,x^{2},\dots,x^{k})(x)=\det\begin{pmatrix}1&0&0&\dots&0\\\
x&1&0&\dots&0\\\ x^{2}&2x&2&\dots&0\\\ \vdots&\vdots&\vdots&&\vdots\\\
x^{k}&kx^{k}&k(k-1)x^{k-1}&\dots&k!\end{pmatrix}\geq 1$

holds for all $x\in\mathds{R}$ and $k=0,\dots,n$ which shows that
$\mathcal{F}$ is also an ECT-system.

######  Example 16.

Let $\mathcal{F}=\\{1,x,x^{3}\\}$ on $[0,b]$ with $b>0$. Then $\mathcal{F}$ is
a T-system (see 15) but not an ET-system. To see this let
$x_{0}=x_{1}=x_{2}=0$, then

$\begin{pmatrix}f_{0}&f_{1}&f_{2}\\\
0&0&0\end{pmatrix}^{*}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&0\end{pmatrix}.$

This shows that $\mathcal{F}$ is not an ET-system. $\circ$

In the previous example the position $x=0$ prevents the T-system to be an ET-
system. If $x=0$ is removed then it is even an ECT-system.

######  Example 17.

Let $\alpha_{0},\dots,\alpha_{n}\in\mathds{N}_{0}$ with
$\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$. Then
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$ on $(0,\infty)$ is an ECT-system.
For $n=2m$ and $0<x_{1}<x_{2}<\dots<x_{m}$ we often encounter a specific
polynomial structure and hence we write it down explicitly once:

$\displaystyle\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}$
$\displaystyle=\lim_{\varepsilon\to
0}\varepsilon^{-m}\cdot\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&x_{1}&x_{1}+\varepsilon&\dots&x_{m}&x_{m}+\varepsilon\end{pmatrix}$
$\displaystyle=\lim_{\varepsilon\to
0}\left[\prod_{i=1}^{m}(x_{i}-x)(x_{i}+\varepsilon-x)\right]\cdot\left[\prod_{1\leq
i<j\leq m}(x_{j}-x_{i})^{2}(x_{j}-x_{i}-\varepsilon)(x_{j}+\varepsilon-
x_{i})\right]$ (47) $\displaystyle\qquad\times
s_{\alpha}(x,x_{1},x_{1}+\varepsilon,\dots,x_{m},x_{m}+\varepsilon)$
$\displaystyle=\prod_{i=1}^{m}(x_{i}-x)^{2}\cdot\prod_{1\leq i<j\leq
m}(x_{j}-x_{i})^{4}\cdot s_{\alpha}(x,x_{1},x_{1},\dots,x_{m},x_{m})$

where $s_{\alpha}$ is the Schur polynomial of
$\alpha=(\alpha_{0},\dots,\alpha_{n})$ Mac (95). Hence,

$s_{\alpha}(x,x_{1},x_{1},\dots,x_{m},x_{m})$

is not divisible by any $(x_{i}-x)$. $\circ$

######  Proof 30.2.

Combine the induction

$f^{(m+1)}(x)=\lim_{h\to 0}\frac{f^{(m)}(x+h)-f^{(m)}(x)}{h}$

and

$\det\begin{pmatrix}x^{\alpha_{0}}&\dots&x^{\alpha_{n}}\\\
x_{0}&\dots&x_{n}\end{pmatrix}=\prod_{0\leq i<j\leq n}(x_{j}-x_{i})\cdot
s_{\alpha}(x_{0},\dots,x_{n})$

where $s_{\alpha}$ is the Schur polynomial of
$\alpha=(\alpha_{0},\dots,\alpha_{n})$.

With 14 the previous example can be generalized.

######  Examples 18 (Examples 16 and 17 continued).

Let $n\in\mathds{N}_{0}$ and let

$-\infty<\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}<\infty$

be reals. Then

  1. (a)

$\mathcal{F}=\\{x^{\alpha_{0}},\dots,x^{\alpha_{n}}\\}$ on
$\mathcal{X}=(0,\infty)$ (16) and

  2. (b)

$\mathcal{G}=\\{e^{\alpha_{0}x},\dots,e^{\alpha_{n}x}\\}$ on
$\mathcal{Y}=\mathds{R}$ (17)

are ECT-systems. $\circ$

######  Proof 30.3.

See Problem 26.

In Problem 25 we will see that also 19 are ET- and ECT-systems.

####  31 Representation as a Determinant, Zeros, and Non-Negativity

Similar to 20 we have the following for ET-systems, i.e., knowing $n$ zeros of
a polynomial $f$ counting multiplicities determines $f$ uniquely up to a
scalar.

######  Theorem 19.

Let $n\in\mathds{N}_{0}$ and let
$\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}\subseteq\mathcal{C}^{n}([a,b],\mathds{R})$
be an ET-system. Let $x_{1},\dots,x_{n}\in[a,b]$ with

$x_{1}=\dots=x_{i_{1}}\;<\;x_{i_{1}+1}=\dots=x_{i_{1}+i_{2}}\;<\;\dots\;<\;x_{i_{1}+\dots+i_{k-1}+1}=\dots=x_{i_{1}+\dots+i_{k}=n}$

for some $k,i_{1},\dots,i_{k}\in\mathds{N}$ and let
$f\in\mathrm{lin}\,\mathcal{F}$. The following are equivalent:

  1. (i)

$f^{(l)}(x_{j})=0$ for all $j=1,\dots,k$ and $l=0,\dots,i_{j}-1$.

  2. (ii)

There exists a constant $c\in\mathds{R}$ such that

$f(x)=c\cdot\det\left(\begin{array}[]{c|cccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{n}\\\
x&\,x_{1}&x_{2}&\dots&x_{n}\end{array}\right).$

######  Proof 31.1.

(ii) $\Rightarrow$ (i): Clear.

(i) $\Rightarrow$ (ii): If $f=0$ then $c=0$ so the assertion holds. If $f\neq
0$ then there exists a point
$x_{0}\in\mathcal{X}\setminus\\{x_{1},\dots,x_{n}\\}$ such that $f(x_{0})\neq
0$ since $\mathcal{F}$ is an ET-system. Then also the determinant in (ii) is
non-zero and we can choose $c$ such that both $f$ and the scaled determinant
coincide also in $x_{0}$. Since $\mathcal{F}$ is an ET-system we have by 3
that

$\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}^{*}$

has full rank, i.e., the coefficients of $f$ and

$c\cdot\det\left(\begin{array}[]{c|cccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{n}\\\
x&\,x_{1}&x_{2}&\dots&x_{n}\end{array}\right)$

coincide.

The following result is a strengthened version of 26. It is a small extension
of e.g. (KS, 66, p. 28, Thm. 5.1) with explicit multiplicities of the zeros of
a non-negative polynomial.

######  Theorem 20.

Let $n\in\mathds{N}_{0}$ and let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system on $[a,b]$ with $a<b$. Let $x_{1}<\dots<x_{k}$ in $[a,b]$ and let
$m_{1},\dots,m_{k}\in\mathds{N}$ for some $k\in\mathds{N}$. The following
hold:

  1. (a)

If $m_{1}+\dots+m_{k}\leq n$ and $m_{i}\in 2\mathds{N}$ for all
$x_{i}\in(a,b)$ then there exists a $f\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (i)

$f\geq 0$ on $[a,b]$,

    2. (ii)

$f$ has precisely the zeros $x_{1},\dots,x_{k}$,

    3. (iii)

the zeros $x_{i}\in(a,b)$ of $f$ have multiplicity $m_{i}$,

    4. (iv)

if $x_{1}=a$ then $x_{1}=a$ has multiplicity $m_{1}$ or $m_{1}+1$, and

    5. (v)

if $x_{k}=b$ then $x_{k}=b$ has multiplicity $m_{k}$ or $m_{k}+1$.

  2. (b)

If $\mathcal{F}$ is an ECT-system or $m_{1}+\dots+m_{k}=n$ then there exists a
$f\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (i)

$f\geq 0$ on $[a,b]$,

    2. (ii)

$f$ has precisely the zeros $x_{1},\dots,x_{k}$, and

    3. (iii)

the zeros $x_{i}$ of $f$ have multiplicity exactly $m_{i}$.

######  Proof 31.2.

(a): Set $m:=m_{1}+\dots+m_{k}$. If all $x_{1},\dots,x_{k}\in(a,b)$ and
$n=m+p$ for some $p\in\mathds{N}_{0}$ then the polynomial

$\displaystyle f(x)$
$\displaystyle=(-1)^{p}\cdot\det\left(\begin{array}[]{c|cccccccc}f_{0}&\,f_{1}&\dots&f_{p}&f_{p+1}&\dots&f_{p+m_{1}}&\dots&f_{n}\\\
x&\,(a&\dots&a)&(x_{1}&\dots&x_{1})&\dots&x_{k})\end{array}\right)$
$\displaystyle\quad+\det\left(\begin{array}[]{c|cccccccc}f_{0}&\,f_{1}&\dots&f_{m_{1}}&\dots&f_{m}&f_{m+1}&\dots&f_{n}\\\
x&\,(x_{1}&\dots&x_{1})&\dots&x_{k})&(b&\dots&b)\end{array}\right)$

fulfills the requirements. If $x_{1}=a$ and/or $x_{k}=b$ then include
$x_{1}=a$ with multiplicity $m_{1}$ or $m_{1}+1$ and $x_{k}=b$ with
multiplicity $m_{k}$ or $m_{k}+1$. Use the choice $m_{1}$ or $m_{1}+1$ resp.
$m_{k}$ or $m_{k}+1$ to let $p\in 2\mathds{N}_{0}$ and add $y$ and $z$ with
$x_{k-1}<y<z<x_{k}$. Once construct a polynomial with the zeros
$x_{1},\dots,x_{k},y$ with the corresponding multiplicities and add another
polynomial with the zeros $x_{1},\dots,x_{k},z$ with the corresponding
multiplicities to it as above.

(b): Use $\\{f_{i}\\}_{i=0}^{m}$ as the ET-system in (a).

#### Problems

######  Problem 21.

Prove 7.

######  Problem 22.

Prove 8.

######  Problem 23.

Prove 9.

######  Problem 24.

  1. (a)

Let $n\in\mathds{N}_{0}$ and let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system on $[a,b]$ for some $a<b$. Show that $\mathcal{F}$ on
$[a^{\prime},b^{\prime}]$ with $a<a^{\prime}<b^{\prime}<b$ is also an ET-
system.

  1. (b)

Show (a) for ECT-systems.

######  Problem 25.

Prove that 19 is an ECT-system.

######  Problem 26.

Prove that the 18 are ECT-systems.

######  Problem 27.

Let

$\mathcal{F}:=\\{1,x^{2},x^{3},x^{5},x^{8},x^{11},x^{13},x^{42}\\}$

on $[0,\infty)$. Give an algebraic polynomial $f\in\mathrm{lin}\,\mathcal{F}$
such that

  1. (a)

$f$ is non-negative on $[0,\infty)$,

  2. (b)

$f$ has $x_{1}=1$ as a zero with multiplicity $m_{1}=2$,

  3. (c)

$f$ has $x_{2}=3$ as a zero with multiplicity $m_{2}=4$, and

  4. (d)

$f$ has no zeros in $[0,\infty)$ other than $x_{1}$ and $x_{2}$.

Life is a short affair;
we should try to make it smooth, and free from strife.
Euripides: The Suppliant Women (Eur, 13, p. 175)

###  Chapter 6 Generating ET-Systems from T-Systems by Using Kernels

We have seen that ET- and especially ECT-systems have much nicer properties
than T-systems. Therefore, especially for technical reasons, it is desirable
to smoothen a T-system into an ET-system. Usually, a function is smoothed by
convolution with e.g. the Gaussian kernel. This procedure is also used for
T-systems.

####  32 Kernels

Let $\mathcal{X}$ and $\mathcal{Y}$ be sets and

$K:\mathcal{X}\times\mathcal{Y}\to\mathds{R}$

be a bivariate function, also called _kernel_. A family
$\\{f_{i}\\}_{i=0}^{n}$ on $\mathcal{Y}$ can then be seen as a special case of
$K$ with $\mathcal{X}=\\{0,1,\dots,n\\}$, i.e., $f_{i}=K(i,\,\cdot\,)$ for all
$i\in\mathcal{X}$. For a kernel $K$ we define the short hand notation

$K\\!\begin{pmatrix}x_{0}&x_{1}&\dots&x_{n}\\\
y_{0}&y_{1}&\dots&y_{n}\end{pmatrix}:=\det(K(x_{i},y_{j}))_{i,j=0}^{n}.$ (48)

######  Definition 1.

Let $k\in\mathds{N}_{0}$, $\mathcal{X}$ and $\mathcal{Y}$ be ordered sets, and
$K:\mathcal{X}\times\mathcal{Y}\to\mathds{R}$ be a kernel. The kernel $K$ is
called _totally positive_ (_of order $k$_), short (TPk) property, if for all
$i=0,1,\dots,k$ we have

$K\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{i}\\\
y_{1}&y_{2}&\dots&y_{i}\end{pmatrix}\geq 0$

for all $x_{1}<x_{2}<\dots<x_{i}$, $y_{1}<y_{2}<\dots<y_{i}$, and
$(x_{l},y_{m})\in\mathcal{X}\times\mathcal{Y}$ for all $l,m=1,\dots,i$. The
kernel $K$ is called _strictly totally positive_ (_of order $k$_), short
(STPk), if we always have

$K\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{i}\\\
y_{1}&y_{2}&\dots&y_{i}\end{pmatrix}>0.$

For more on sign regular kernels see e.g. Kar (68) and GM (96).

######  Corollary 2 (see e.g. (KS, 66, p. 10, Exm. 3)).

Let $n\in\mathds{N}_{0}$, let $K$ be a STPn+1 kernel with $\mathcal{X}=[a,b]$,
$\mathcal{Y}=[c,d]$ and $K(x,\,\cdot\,)\in\mathcal{C}([c,d],\mathds{R})$ for
all $x\in\mathcal{X}$, and let $x_{0}<x_{1}<\dots<x_{n}$ in $\mathcal{X}$.

Then $\\{K(x_{i},\,\cdot\,)\\}_{i=0}^{k}$ is a continuous T-system on
$\mathcal{Y}=[c,d]$ for all $k=0,\dots,n$.

######  Proof 32.1.

Follows immediately from 5.

######  Definition 3.

Let $k\in\mathds{N}$, $\mathcal{X}=[a,b]$, $\mathcal{Y}=[c,d]$, and
$K:\mathcal{X}\times\mathcal{Y}\to\mathds{R}$ be a kernel such that
$K(x,\,\cdot\,)\in\mathcal{C}^{k}(\mathcal{Y},\mathds{R})$ for all
$x\in\mathcal{X}$. We define

$K^{*}\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{k}\\\
y_{1}&y_{2}&\dots&y_{k}\end{pmatrix}:=\det\begin{pmatrix}K(x_{1},\,\cdot\,)&K(x_{2},\,\cdot\,)&\dots&K(x_{k},\,\cdot\,)\\\
y_{1}&y_{2}&\dots&y_{k}\end{pmatrix}^{*}$ (49)

for all $x_{1}<x_{2}<\dots<x_{k}$ in $\mathcal{X}$ and $y_{1}\leq
y_{2}\leq\dots\leq y_{k}$ in $\mathcal{Y}$.

We say $K$ is _extended totally positive_ (_of order $k$_), short ETPk, if for
all $i=1,2,\dots,k$ we have

$K^{*}\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{i}\\\
y_{1}&y_{2}&\dots&y_{i}\end{pmatrix}>0$

for all $x_{1}<x_{2}<\dots<x_{i}$ in $\mathcal{X}$ and $y_{1}\leq
y_{2}\leq\dots\leq y_{i}$ in $\mathcal{Y}$.

######  Corollary 4 (see e.g. (KS, 66, p. 10, Exm. 3)).

Let $n\in\mathds{N}_{0}$, let $K$ be an ETPn+1 kernel with
$\mathcal{X}=[a,b]$, $\mathcal{Y}=[c,d]$ and
$K(x,\,\cdot\,)\in\mathcal{C}^{n}([c,d],\mathds{R})$ for all
$x\in\mathcal{X}$, and let $x_{0}<x_{1}<\dots<x_{n}$ in $\mathcal{X}$.

Then $\\{K(x_{i},\,\cdot\,)\\}_{i=0}^{n}$ is an ECT-system on
$\mathcal{Y}=[c,d]$.

######  Proof 32.2.

Follows immediately from 3.

######  Example 5.

Let $\mathcal{X}=\mathds{R}$, $\mathcal{Y}=[a,b]\subset(0,\infty)$, and
$K(x,y)=y^{x}$. Then $K$ is ETPk for all $k\in\mathds{N}_{0}$. $\circ$

######  Proof 32.3.

Follows immediately from 18.

######  Example 6 (see e.g. (KS, 66, p. 11, Exm. 5)).

For any $\sigma>0$ the Gaussian kernel

$K_{\sigma}(x,y):=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{1}{2}\left(\frac{x-y}{\sigma}\right)^{2}\right)\qquad\text{on\;
$\mathcal{X}\times\mathcal{Y}=\mathds{R}^{2}$}$ (50)

is ETPk for any $k\in\mathds{N}$.

The proof is adapted from (KS, 66, p. 11).

######  Proof 32.4.

It is sufficient to show that $K(x,y)=e^{-(x-y)^{2}}$ is ETPk for all
$k\in\mathds{N}_{0}$.

In Example 18 (b) we have seen that $\\{e^{\alpha_{i}x}\\}_{i=0}^{n}$ is an
ECT-system on $\mathds{R}$ for all $n\in\mathds{N}_{0}$ and all
$\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$ in $\mathds{R}$. Hence, by writing

$f_{n}(x):=\sum_{i=0}^{n}a_{n}\cdot e^{-(x_{i}-x)^{2}}\qquad\text{as}\qquad
f_{n}(x)=e^{-x^{2}}\cdot\sum_{i=0}^{n}a_{i}\cdot e^{-x_{i}^{2}}\cdot
e^{2x_{i}x}$

we see that $f_{n}$ has at most $n$ zeros (counting multiplicities) in
$\mathds{R}$ if $a_{0},\dots,a_{n}\in\mathds{R}$ with
$a_{0}^{2}+\dots+a_{n}^{2}>0$.

####  33 The Basic Composition Formulas

The following equations (51) and (53) are the _basic composition formulas_.

######  Lemma 7 (see e.g. (KS, 66, pp. 13–14, Exm. 8)).

Let $K:[a,b]\times[c,d]\to\mathds{R}$ and $L:[c,d]\times[e,f]\to\mathds{R}$ be
kernels. Let $\mu$ be a $\sigma$-finite measure such that $M(x,z)$ defined by

$M:[a,b]\times[e,f]\to\mathds{R},\quad M(x,z):=\int_{c}^{d}K(x,y)\cdot
L(y,z)~{}\mathrm{d}\mu(y)$

exists for all $(x,z)\in[a,b]\times[e,f]$. The following hold:

  1. (i)

$M$ is a kernel.

  2. (ii)

For all $k\in\mathds{N}$, $x_{1}<\dots<x_{k}$ in $[a,b]$, and
$z_{1}<\dots<z_{k}$ in $[e,f]$ we have

$M\\!\begin{pmatrix}x_{1}&\dots&x_{k}\\\ z_{1}&\dots&z_{k}\end{pmatrix}=\\\
\underset{c\leq y_{1}<\dots<y_{k}\leq
d}{\int\;\,\dots\;\,\int}K\\!\begin{pmatrix}x_{1}&\dots&x_{k}\\\
y_{1}&\dots&y_{k}\end{pmatrix}\cdot L\\!\begin{pmatrix}y_{1}&\dots&y_{k}\\\
z_{1}&\dots&z_{k}\end{pmatrix}~{}\mathrm{d}\mu(y_{1})\dots\mathrm{d}\mu(y_{k}).$
(51)

  3. (iii)

If $L(y,\,\cdot\,)\in\mathcal{C}^{k-1}([e,f],\mathds{R})$ for some
$k\in\mathds{N}$ and

$\partial_{z}^{i}M(x,z):=\int_{c}^{d}K(x,y)\cdot\partial_{z}^{i}L(y,z)~{}\mathrm{d}\mu(y)$
(52)

holds for all $i=0,\dots,k-1$ then

$M^{*}\\!\begin{pmatrix}x_{1}&\dots&x_{k}\\\
z_{1}&\dots&z_{k}\end{pmatrix}=\\\ \underset{c\leq y_{1}<\dots<y_{k}\leq
d}{\int\;\,\dots\;\,\int}K\\!\begin{pmatrix}x_{1}&\dots&x_{k}\\\
y_{1}&\dots&y_{k}\end{pmatrix}\cdot
L^{*}\\!\begin{pmatrix}y_{1}&\dots&y_{k}\\\
z_{1}&\dots&z_{k}\end{pmatrix}~{}\mathrm{d}\mu(y_{1})\dots\mathrm{d}\mu(y_{k})$
(53)

for all $x_{1}<\dots<x_{k}$ in $[a,b]$, and $z_{1}\leq\dots\leq z_{k}$ in
$[e,f]$.

######  Proof 33.1.

(i) is clear, (ii) follows by straight forward calculations, see e.g. (PS, 70,
p. 48, No. 68), and (iii) follows from (ii) with (52).

####  34 Smoothing T-Systems into ET-Systems

With the Gaussian kernel from 6 we get from 7 the following smoothing result.

######  Corollary 8 (see e.g. (KS, 66, p. 15)).

Let $n\in\mathds{N}_{0}$ and $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a
continuous T-system on $[a,b]$. For any $\sigma>0$ let

$K_{\sigma}(x):=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{1}{2}\left(\frac{x}{\sigma}\right)^{2}\right)\qquad\text{on\;
$\mathcal{X}=\mathds{R}$}$

be the Gaussian kernel and define $f_{i,\sigma}:=f_{i}*K_{\sigma}$ for all
$i=0,\dots,n$. Then $\mathcal{F}_{\sigma}:=\\{f_{i,\sigma}\\}_{i=0}^{n}$ is an
ET-system.

######  Proof 34.1.

See Problem 10.

If $\mathcal{F}$ is a continuous T-system on $[a,b]$ then

$\lim_{\sigma\searrow 0}f_{i,\sigma}(x)=f_{i}(x)$

for all $x\in(a,b)$ and $i=0,\dots,n$.

######  Corollary 9.

If $\\{f_{i}\\}_{i=0}^{k}$ in 8 is a T-system for all $k=0,\dots,n$ then
$\mathcal{F}_{\sigma}$ is an ECT-system.

######  Proof 34.2.

Apply 8 for every $k=0,1,\dots,n$.

Approximating a T-system by ET-systems with the Gaussian kernel is often used
GK (02); Sch (53); Kar (68), see also (KS, 66, p. 16). We will need it in the
proof of 1.

#### Problems

######  Problem 10.

Prove 8 from 7.

##  Part III Karlin’s Positivstellensätze and Nichtnegativstellensätze

Beauty is the first test: there is no permanent place
in this world for ugly mathematics.
Godfrey Harold Hardy (Har, 69, §10, p. 85)

###  Chapter 7 Karlin’s Positivstellensatz and Nichtnegativstellensatz on
$[a,b]$

We now come to the main result (1) and its variations: 3 for T-systems on
$[a,b]$ and 6 for ET-systems on $[a,b]$. Earlier versions were already
developed in KS (53). Both results are used in the following chapters to prove
1 for T-systems on $[0,\infty)$, 3 for ET-systems on $[0,\infty)$, 4 for
T-systems on $\mathds{R}$, and finally 5 for ET-systems on $\mathds{R}$.

The main applications and examples will be the various sparse algebraic
Positivstellensätze and sparse algebraic Nichtnegativstellensätze in Part IV.

####  35 Karlin’s Positivstellensatz for T-Systems on $[a,b]$

For the following main result we remind the reader what it means that a set
has an index, see 24: If $x\in(a,b)$ then its index is $2$ and if $x=a$ or $b$
then its index is $1$. The following result is due to Karlin and we name it
therefore after him.

######  Karlin’s Theorem 1 (for $f>0$ on $[a,b]$; (Kar, 63, Thm. 1) or e.g.
(KS, 66, p. 66, Thm. 10.1)).

Let $n\in\mathds{N}_{0}$, $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a continuous
T-system of order $n$ on $[a,b]$ with $a<b$, and let
$f\in\mathcal{C}([a,b],\mathds{R})$ with $f>0$ on $[a,b]$ be a strictly
positive continuous function. The following hold:

  1. (i)

There exists a unique polynomial $f_{*}\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (a)

$f(x)\geq f_{*}(x)\geq 0$ for all $x\in[a,b]$,

    2. (b)

$f_{*}$ vanishes on a set with index $n$,

    3. (c)

the function $f-f_{*}$ vanishes at least once between each pair of adjacent
zeros of $f_{*}$,

    4. (d)

the function $f-f_{*}$ vanishes at least once between the larges zero of
$f_{*}$ and the end point $b$, and

    5. (e)

$f_{*}(b)>0$.

  2. (ii)

There exists a unique polynomial $f^{*}\in\mathrm{lin}\,\mathcal{F}$ which
satisfies the conditions (a) to (d) of (i) and

    1. (e’)

$f^{*}(b)=0$.

Examples of $f_{*}$ and $f^{*}$ are depicted in Figure 1 for an odd and an
even $n$.

(a) $n=5$

(b) $n=6$

Figure 1: The functions $f\in\mathcal{C}([a,b],\mathds{R})$ with $f>0$
(black), $f_{*}\in\mathrm{lin}\,\mathcal{F}$ (red), and
$f^{*}\in\mathrm{lin}\,\mathcal{F}$ (blue) from the 3 with $n=5$ and $n=6$.

The proof is taken from (KS, 66, pp. 68–71). The proof constructs the
polynomials $f_{*}$ and $f^{*}$ by using the Fixed Point Theorem of Brouwer
(Bro, 11, Satz 4), see also e.g. (Zei, 86, Prop. 2.6).101010Note that in Zei
(86) the work Über Abbildungen von Mannigfaltigkeiten Bro (11) is incorrectly
dated in the references and Proposition 2.6 on p. 52 to the year 1912 while
the paper actually appeared in 1911 in the Mathematische Annalen. However, we
also want to point out that Zeidler gives three proofs of the Fixed Point
Theorem of Brouwer, including a constructive one in (Zei, 86, pp. 254–255,
Problem 6.7e).

######  Proof 35.1.

We distinguish three different cases.

Case 1: Let $n=2m$ and let $\mathcal{F}$ be an ET-system. We construct $f_{*}$
in (i) as follows. For each point $\xi=(\xi_{0},\dots,\xi_{m})$ in the
$m$-dimensional simplex

$\Xi^{m}:=\left\\{(\xi_{0},\dots,\xi_{m})\in\mathds{R}^{m+1}\,\middle|\,\xi_{i}\geq
0,\ i=0,1,\dots,m,\ \sum_{i=0}^{m}\xi_{i}=b-a\right\\}$ (54)

set

$x_{i}:=a+\sum_{k=0}^{i-1}\xi_{k}$

for all $i=0,\dots,m$ and define

$f_{\xi}(x):=c_{\xi}\cdot\det\left(\begin{array}[]{c|ccccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{n-1}&f_{n}\\\
x&\,x_{1}&x_{1}&\dots&x_{m}&x_{m}\end{array}\right)$ (55)

with $c_{\xi}\in\mathds{R}$ such that $f_{\xi}=\sum_{i=0}^{n}a_{i}f_{i}\geq 0$
on $[a,b]$ with $a_{0}^{2}+\dots+a_{n}^{2}=1$. If $p$ of the points $x_{i}$
coincide, this common point is to have multiplicity $2p$.

Define

$\delta_{i}(\xi):=\min\\{\delta\geq 0\,|\,\delta\cdot f\geq u_{\xi}\
\text{on}\ [x_{i},x_{i+1}]\\}$ (56)

for all $i=0,\dots,m$ with $x_{0}=a$ and $x_{m+1}=b$. The coefficients
$a_{i}(\xi)$ are continuous in $\xi$ and hence the functions $\delta_{i}(\xi)$
are continuous in $\xi$.

Next, define

$F_{i}(\xi):=\delta_{i}(\xi)-\min_{k}\delta_{k}(\xi)$ (57)

for all $i=0,\dots,m$ and set $F_{m+1}(\xi):=F_{0}(\xi)$. If there does not
exist a point $\xi$ such that $F_{i}(\xi)=0$ for all $i=0,\dots,m$, then
$\sum_{i=0}^{m}F_{i}(\xi)>0$ for all $\xi\in\Xi^{m}$. In this event the
continuous mapping

$\cdot\,^{\prime}:\Xi^{m}\to\Xi^{m},\
\xi\mapsto\xi^{\prime}\quad\text{with}\quad\xi_{i}^{\prime}:=\frac{F_{i+1}(\xi)}{\sum_{k=0}^{m}F_{k}(\xi)}\cdot(b-a)$

for all $i=0,\dots,m$ is well-defined. The Fixed Point Theorem of Brouwer
affirms the existence of a point $\xi^{*}\in\Xi^{m}$ for which

$\xi_{i}^{*}:=\frac{F_{i+1}(\xi^{*})}{\sum_{k=0}^{m}F_{k}(\xi^{*})}\cdot(b-a)$
(58)

for all $i=0,\dots,m$. By (57) we have that for any $\xi\in\Xi^{m}$ we have
$F_{i}(\xi)=0$ for some $i$. Suppose $F_{j}(\xi^{*})=0$ for some fixed
$j=0,\dots,m$. Then (58) implies $\xi_{j-1}^{*}=0$. By (56) and (57) imply
$F_{j-1}(\xi^{*})=0$. Continuing in this way we get $F_{i}(\xi^{*})=0$ for all
$i=0,\dots,j$ and since $F_{m+1}(\xi)=F_{0}(\xi)$ we have $F_{i}(\xi^{*})=0$
for all $i=0,\dots,m$. But this contradicts our assumption
$\sum_{i=0}^{m}F_{i}(\xi^{*})>0$. Therefore, there exists at least one point
$\xi^{*}\in\Xi^{m}$ such that $\delta_{i}(\xi^{*})=\delta$ for all
$i=0,\dots,m$. Since $f_{\xi}\neq 0$ it follows that $\delta>0$ and hence all
$x_{i}$ are distinct, i.e.,

$a=x_{0}<x_{1}<\dots<x_{m}=b.$

Hence, $f_{*}:=\delta^{-1}\cdot f_{\xi^{*}}$ by the nature of its construction
fulfills the requirements (a) – (e) of (i).

For $f^{*}$ we let $x_{0}=a$ and $x_{m}=b$ and we define similar to (55) the
polynomial

$g_{\xi}(x):=d_{\xi}\cdot\det\left(\begin{array}[]{c|ccccccc}f_{0}&\,f_{1}&f_{2}&f_{3}&\dots&f_{n-2}&f_{n-1}&f_{n}\\\
x&\,a&x_{1}&x_{1}&\dots&x_{m-1}&x_{m-1}&b\end{array}\right).$

Repeating the arguments from above we get $f^{*}$ which fulfills (a) – (d) and
(e’) in (ii).

Case 2: Let $n=2m+1$ and let $\mathcal{F}$ be an ET-system. Similar to case 1,
we define the polynomials

$f_{\xi}(x):=d_{\xi}\cdot\det\left(\begin{array}[]{c|cccccc}f_{0}&\,f_{1}&f_{2}&f_{3}&\dots&f_{n-1}&f_{n}\\\
x&\,a&x_{1}&x_{1}&\dots&x_{m}&x_{m}\end{array}\right).$

and

$g_{\xi}(x):=d_{\xi}\cdot\det\left(\begin{array}[]{c|cccccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{n-2}&f_{n-1}&f_{n}\\\
x&\,x_{1}&x_{1}&\dots&x_{m}&x_{m}&b\end{array}\right).$

Repeating the procedure of case 1 gives the statement.

Case 3: Let $n=2m$ and $\mathcal{F}$ be a T-systems. Then we consider the
functions

$f_{i}(x;\sigma):=\int_{a}^{b}K_{\sigma}(x,y)\cdot f_{i}(y)~{}\mathrm{d}y$

where

$K_{\sigma}(x,y):=\frac{1}{\sigma\cdot\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{x-y}{\sigma}\right)^{2}\right]$

with $\sigma>0$, see Chapter 6. By 8 we have that
$\mathcal{F}_{\sigma}:=\\{f_{i}(\,\cdot\,;\sigma)\\}_{i=0}^{n}$ is an ET-
system on $[a,b]$ and hence also on any subinterval $[a^{\prime},b^{\prime}]$
with $a<a^{\prime}<b^{\prime}<b$. The need to restrict the system
$\mathcal{F}_{\sigma}$ to the proper interval $[a^{\prime},b^{\prime}]$ is due
to the annoyance that at the end points $x=a$ and $x=b$ we have

$\lim_{\sigma\searrow 0}f_{i}(x;\sigma)=\frac{1}{2}f_{i}(x)$

for all $i=0,\dots,n$ while for $x\in(a,b)$ we have

$\lim_{\sigma\searrow 0}f_{i}(x;\sigma)=f_{i}(x).$

From the cases 1 and 2 we find that for any $\sigma>0$ we have a polynomial
$f_{*,\sigma}$ satisfying conditions (a) – (e) of (i) on the interval
$[a^{\prime},b^{\prime}]$. If

$f_{*,\sigma}=\sum_{i=0}^{n}a_{i}(\sigma)\cdot f_{i}(\,\cdot\,,\sigma)$

we can chose a sequence $\sigma_{k}\searrow 0$ and let
$x_{1}^{(k)},\dots,x_{m}^{(k)}$ be the zeros of $f_{*,\sigma_{k}}$.
Additionally, let $y_{1}^{(k)},\dots,y_{m+1}^{(k)}$ be the points which
interlace with $\\{x_{i}^{(k)}\\}_{i=0}^{m}$, i.e.,
$a^{\prime}<y_{1}^{(k)}<x_{1}^{(k)}<\dots<x_{m}^{(k)}<y_{m+1}^{(k)}\leq
b^{\prime}$ and satisfying $f(y_{i}^{(k)})=f_{*,\sigma_{k}}(y_{i}^{(k)})$ for
all $i=0,\dots,m+1$.

Since $f(x)\geq f_{*,\sigma}\geq 0$ on $[a^{\prime},b^{\prime}]$ and solving
the system of equations

$f_{*,\sigma}(x_{j})=\sum_{i=0}^{n}a_{i}(\sigma)\cdot f_{i}(x_{j};\sigma)$

for $i=0,\dots,n$ we find that these quantities are uniformly bounded. We now
select a subsequence $\\{\sigma_{k^{\prime}}\\}$ from $\\{\sigma_{k}\\}$ with
the property that as $k^{\prime}\to\infty$ we obtain

$\displaystyle a_{i}(\sigma_{k^{\prime}})$ $\displaystyle\to a_{i}$
$\displaystyle\text{for all}\ i=0,\dots,n,$ $\displaystyle
y_{j}^{(k^{\prime})}$ $\displaystyle\to y_{j}$ $\displaystyle\text{for all}\
j=1,\dots,m+1,$ $\displaystyle x_{l}^{(k^{\prime})}$ $\displaystyle\to x_{l}$
$\displaystyle\text{for all}\ l=1,\dots,m$

and

$a^{\prime}\leq y_{1}\leq x_{1}\leq\dots\leq x_{m}\leq y_{m+1}\leq
b^{\prime}.$

The function $f_{*,a^{\prime},b^{\prime}}:=\sum_{i=0}^{n}a_{i}\cdot f_{i}$
vanishes at all $x_{l}$, $l=1,\dots,m$, and equals $f$ at all $y_{j}$,
$j=1,\dots,m+1$. Therefore, since $f_{*,a^{\prime},b^{\prime}}$ is continuous
we see that

$a^{\prime}\leq y_{1}<x_{1}<\dots<x_{m}<y_{m+1}\leq b^{\prime}.$

Hence, $f_{*,a^{\prime},b^{\prime}}$ satisfies (a) – (e) of (i) on the
interval $[a^{\prime},b^{\prime}]$.

Performing a last limiting procedure letting $a^{\prime}\searrow a$ and
$b^{\prime}\nearrow b$ we obtain a polynomial $f_{*}$ satisfying (a) – (e) in
(i) on the full interval $[a,b]$.

For $f^{*}$ the same procedure gives the desired polynomial satisfying the
conditions (a) – (d) and (e’).

Uniqueness of $f_{*}$ and $f^{*}$: Let $n=2m$. Observe that if another
polynomial $\tilde{f}_{*}$ with properties (a) – (e) exists then it must have
$m$ interior zeros $\tilde{x}_{1},\dots,\tilde{x}_{m}$. Denote by
$x_{1},\dots,x_{m}$ the zeros of $f_{*}$. Without loss of generality we can
assume that either $\tilde{x}_{1}<x_{1}$ or $\tilde{x}_{1}=x_{1}$ and
$f_{*}-\tilde{f}_{*}$ is non-negative in some interval
$(x_{1}-\varepsilon,x_{1})$. Otherwise we interchange the roles of $f_{*}$ and
$\tilde{f}_{*}$. We count the zeros of $g:=f_{*}-\tilde{f}_{*}$. We say $g$
has a zero in the closed interval $[c,d]$ if

  * •

$g(t_{0})=0$ for $t_{0}\in(c,d)$,

  * •

$g(c)=0$ and $g\geq 0$ on $(c,c+\varepsilon)$, or

  * •

$g(d)=0$ and $g\geq 0$ on $(d-\varepsilon,d)$.

Counting zeros in this fashion we see that $g$ has at least two zeros in each
of the intervals $[x_{i-1},x_{i}]$ for $i=1,\dots,m$ where $x_{0}=a$ and at
least one in the interval $[x_{m},b]$. In total $g$ vanishes at least $n+1$
times. Notice, that certain non-nodal zeros of $g$ have been counted twice and
hence by 22 we have $g=0$.

In a similar way we get uniqueness of $f^{*}$ and also in the case $n=2m+1$.

Note, in the previous result we do not need to have
$f\in\mathrm{lin}\,\mathcal{F}$. The function $f$ only needs to be continuous
and strictly positive on $[a,b]$.

An earlier version of (or at least connected to) 1 combined with 22 (which was
used in the proof of 1) is a lemma by Markov Mar (84), see also (ST, 43, p.
80).

######  Lemma 2 (Mar (84), see also (ST, 43, p. 80)).

Let $m\in\mathds{N}$ and let $f\in\mathcal{C}^{n+1}([a,b],\mathds{R})$ be such
that $f>0$ and $f^{(k)}\geq 0$ for all $k=1,\dots,m+1$ in $[a,b]$. Let
$p_{m}\in\mathds{R}[x]_{\leq m}$ and $c\in(a,b)$. Let $m_{1}\in\mathds{N}$ be
the number of zeros in $(a,c)$ of the function $f-p_{m}$ and $m_{2}$ be the
number of zeros of $p_{m}$ in $(c,b)$, both counted with multiplicity. Then
$m_{1}+m_{2}\leq m+1$.

1 is of course much more general. As a consequence of 1 we get Karlin’s
Positivstellensatz for T-systems on $[a,b]$.

######  Karlin’s Positivstellensatz 3 (for T-Systems on $[a,b]$; see (Kar, 63,
Cor. 1) or e.g. (KS, 66, p. 71, Cor. 10.1(a))).

Let $n\in\mathds{N}_{0}$, let $\mathcal{F}$ be a continuous T-system of order
$n$ on $[a,b]$ with $a<b$, and let $f\in\mathrm{lin}\,\mathcal{F}$ with $f>0$
on $[a,b]$. Then there exists a unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that

  1. (i)

$f_{*},f^{*}\geq 0$ on $[a,b]$,

  2. (ii)

the zeros of $f_{*}$ and $f^{*}$ each are sets of index $n$,

  3. (iii)

the zeros of $f_{*}$ and $f^{*}$ strictly interlace,

  4. (iv)

$f_{*}(b)=f(b)>0$, and

  5. (v)

$f^{*}(b)=0$.

######  Proof 35.2.

Let $f_{*}$ be the unique $f_{*}$ from 1(i). Then
$f-f_{*}\in\mathrm{lin}\,\mathcal{F}$ is a polynomial and fulfills (a) – (d),
and (e’) of $f^{*}$ in 1. But since also $f^{*}$ is unique we have
$f-f_{*}=f^{*}$.

####  36 The Snake Theorem: An Interlacing Theorem

In 1 a polynomial $f_{*}\in\mathrm{lin}\,\mathcal{F}$ was found with $0\leq
f_{*}\leq f$ for some given $f\in\mathcal{C}([a,b],\mathds{R})$ with $f>0$ on
$[a,b]$. This can be extended to find a function
$f_{*}\in\mathrm{lin}\,\mathcal{F}$ between some
$g_{1},g_{2}\in\mathcal{C}([a,b],\mathds{R})$ as the following result shows.
In (KS, 66, p. 368, Thm. 6.1) M. G. Krein and A. A. Nudel’man called it the
_Snake Theorem_ which is an accurate description of its graphical
representation, see Figure 2.

######  Snake Theorem 4 ((Kar, 63, Thm. 2) or e.g. (KS, 66, p. 72, Thm. 10.2)
and (KN, 77, p. 368, Thm. 6.1)).

Let $n\in\mathds{N}_{0}$, $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a continuous
T-system of order $n$ on $[a,b]$ with $a<b$, and let
$g_{1},g_{2}\in\mathcal{C}([a,b],\mathds{R})$ be two continuous functions on
$[a,b]$ such that there exists a function $g\in\mathrm{lin}\,\mathcal{F}$ with

$g_{1}<g<g_{2}$

on $[a,b]$. Then the following hold:

  1. (i)

There exists a unique polynomial $f_{*}\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (a)

$g_{1}(x)\leq f_{*}(x)\leq g_{2}(x)$ for all $x\in[a,b]$, and

    2. (b)

there exist $n+1$ points $x_{1}<\dots<x_{n+1}$ in $[a,b]$ such that

$f_{*}(x_{n+1-i})=\begin{cases}g_{1}(x_{n+1-i})&\text{for}\ i=1,3,5,\dots,\\\
g_{2}(x_{n+1-i})&\text{for}\ i=0,2,4,\dots.\end{cases}$

  2. (ii)

There exists a unique polynomial $f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (a’)

$g_{1}(x)\leq f^{*}(x)\leq g_{2}(x)$ for all $x\in[a,b]$, and

    2. (b’)

there exist $n+1$ points $y_{1}<\dots<y_{n+1}$ in $[a,b]$ such that

$f^{*}(y_{n+1-i})=\begin{cases}g_{2}(y_{n+1-i})&\text{for}\ i=1,3,5,\dots,\\\
g_{1}(y_{n+1-i})&\text{for}\ i=0,2,4,\dots.\end{cases}$

The functions $g_{1}$, $g_{2}$, $g$, $f_{*}$, and $f^{*}$ of the 4 are
illustrated in Figure 2.

Figure 2: The functions $g_{1},g_{2}\in\mathcal{C}([a,b],\mathds{R})$ (black,
$g_{1}$ bottom, $g_{2}$ top), $g\in\mathrm{lin}\,\mathcal{F}$ (blue, dashed),
$f_{*}\in\mathrm{lin}\,\mathcal{F}$ (red), and
$f^{*}\in\mathrm{lin}\,\mathcal{F}$ (green) from the 4.

The following proof is taken from (KS, 66, p. 73).

######  Proof 36.1.

Let $n=2m$ and $\mathcal{F}$ be an ET-system. We proceed as in the proof of 1.
For each $\xi=(\xi_{0},\dots,\xi_{n})\in\Xi^{n}$ and
$\sum_{i=0}^{n}\xi_{i}=b-a$ we construct the polynomial

$f_{\xi}(x)=\sum_{i=0}^{n}a_{i}(\xi)\cdot
f_{i}(x)=c_{\xi}\cdot\det\left(\begin{array}[]{c|ccc}f_{0}&\,f_{1}&\dots&f_{n}\\\
x&\,x_{1}&\dots&x_{n}\end{array}\right)$

which vanishes at each of the points

$x_{i}:=a+\sum_{k=0}^{i-1}\xi_{k}$

for all $i=0,\dots,n$ and let $c_{\xi}\in\mathds{R}$ be such that
$a_{0}(\xi)^{2}+\dots+a_{n}(\xi)^{2}=1$ and $f_{\xi}\geq 0$ on
$[x_{i},x_{i+1}]$ if $i$ is even.

For $i=0,2,4,\dots,n$ we define

$\delta_{i}(\xi):=\min\left\\{\delta\geq 0\,\middle|\,\delta\cdot(g_{2}-g)\geq
f_{\xi}\ \text{on}\ [x_{i},x_{i+1}]\right\\},$

where $x_{0}=a$ and $x_{n+1}=b$, while for $i=1,3,\dots,n-1$ we define

$\delta_{i}(\xi):=\min\left\\{\delta\geq
0\,\middle|\,f_{\xi}\geq\delta\cdot(g-g_{1})\ \text{on}\
[x_{i},x_{i+1}]\right\\}.$

As in 1 we define $F_{k}(\xi):=\delta_{k}(\xi)-\min_{i}\delta_{i}(\xi)$. And
as before assuming $\sum_{k=0}^{n}F_{k}(\xi)>0$ for all $\xi\in\Xi^{n}$ leads
to a contradiction. Therefore, there exists a $\xi^{*}\in\Xi^{n}$ for which
$\delta_{i}(\xi^{*})=\delta$ for all $i=0,\dots,n$. It is clear that
$\delta>0$ and that the polynomial $f_{*}:=\delta^{-1}\cdot f_{\xi^{*}}+g$
satisfies the conditions of the theorem.

The polynomial $f^{*}$ is constructed employing the same line of arguments.

The extension encompassing the case where $\mathcal{F}$ is merely a T-system
and the proof of the uniqueness proceed as in the proof of 1.

####  37 Karlin’s Nichtnegativstellensatz for ET-Systems on $[a,b]$

While 1 with $f>0$ can be proved for T-systems, an equivalent version allowing
zeros in $f\in\mathcal{C}([a,b],\mathds{R})$, i.e., $f\geq 0$ but not $f>0$,
needs to assume that $\mathcal{F}$ is an ET-system.

######  Karlin’s Theorem 5 (for $f\geq 0$ on $[a,b]$; (Kar, 63, Thm. 3) or
e.g. (KS, 66, p. 74, Thm. 10.3)).

Let $n\in\mathds{N}_{0}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a
continuous ET-system of order $n$ on $[a,b]$ with $a<b$, and let
$f\in\mathcal{C}^{n}([a,b],\mathds{R})$ be such that $f\geq 0$ on $[a,b]$ and
$f$ has $r<n$ zeros (counting multiplicities). The following hold:

  1. (i)

There exists a unique polynomial $f_{*}\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (a)

$f(x)\geq f_{*}(x)\geq 0$ for all $x\in[a,b]$,

    2. (b)

$f_{*}$ has $n$ zeros counting multiplicities,

    3. (c)

if $x_{1}<\dots<x_{{n-r}}$ in $(a,b)$ are the zeros of $f_{*}$ which remain
after removing the $r$ zeros of $f$ then $f-f_{*}$ vanishes at least twice
more (counting multiplicities) in each open interval $(x_{i},x_{i+1})$,
$i=1,\dots,n-r-1$, and at least once more in each of the intervals $[a,x_{1})$
and $(x_{n-r},b]$,

    4. (d)

the zeros $x_{1},\dots,x_{n-r}$ of (c) are a set of index $n-r$, and

    5. (e)

$x_{n-r}<b$.

  2. (ii)

There exists a unique polynomial $f^{*}\in\mathrm{lin}\,\mathcal{F}$ which
satisfies the conditions (a) to (d) and

    1. (e’)

$x_{n-r}=b$.

The proof is taken from (KS, 66, pp. 74–75).

######  Proof 37.1.

Let $z_{1},\dots,z_{p}$ be the distinct zeros of $f$ with multiplicities
$m_{1},\dots,m_{p}$ where $\sum_{i=1}^{p}m_{i}=r\leq n-1$ and set
$n^{\prime}:=n-r$. The proof is now similar to the proof of 1 where $n$ is
replaced by $n^{\prime}$. Since the odd and the even cases are again somewhat
the same and for the sake of some slight variety we treat now the odd case
$n^{\prime}=2m^{\prime}+1$. The construction of $f_{*}$ in part (i) proceeds
as follows. For each $\xi\in\Xi^{m^{\prime}}$ we construct the polynomial

$f_{\xi}(x):=\sum_{i=0}^{n}a_{i}(\xi)\cdot f_{i}\\\
=c_{\xi}\cdot\det\left(\begin{array}[]{c|cccccccc}f_{0}&\,f_{1}\dots
f_{m_{1}}&\dots&f_{m_{1}+m_{p-1}+1}&\dots
f_{r}&f_{r+1}\;f_{r+2}&\dots&f_{n-2}\;f_{n-1}&f_{n}\\\
x&\,z_{1}\,\dots\,z_{1}&\dots&z_{p}&\dots\,z_{p}&x_{1}\;\;\;x_{1}&\dots&x_{n^{\prime}}\;\;\,x_{n^{\prime}}&a\end{array}\right)$
(59)

where $c_{\xi}\in\mathds{R}$ is chosen such that
$a_{1}(\xi)^{2}+\dots+a_{n}(\xi)^{2}=1$ and

$x_{i}:=a+\sum_{k=0}^{i}\xi_{k}$

for all $i=1,\dots,m^{\prime}$ are the zeros of multiplicity two and $a$ is a
zero of multiplicity one. Now we define

$\delta_{i}(\xi):=\min\left\\{\delta\geq
0\,\middle|\,\delta\geq\frac{f_{\xi}}{f}\ \text{on}\ [x_{i},x_{i+1}]\right\\}$

for $i=1,\dots,m^{\prime}+1$ with $x_{m^{\prime}+2}=b$ where the ratio is
evaluated by l’Hopital’s rule at the zeros $z_{1},\dots,z_{p}$ of $f$.

By examining $\frac{f_{\xi}}{f}$ first in the neighborhood of each of the
points $z_{1},\dots,z_{p}$ and then over the remaining part we find that if
$\xi^{(k)}\to\xi$ then

$\frac{f_{\xi^{(k)}}}{f}\to\frac{f_{\xi}}{f}$

uniformly on $[a,b]$. Consequently, each of th $\delta_{i}$ is continuous in
$\xi$ and $\delta_{i}(\xi)=0$ if and only $\xi_{i}=0$.

The same arguments used in the proof of 1 now show that for some
$\xi^{*}\in\mathrm{int}\,\Xi^{m^{\prime}}$ we have
$\delta_{i}(\xi^{*})=\delta>0$ for all $i=1,\dots,m^{\prime}+1$. It is simple
to see that $f_{*}:=\delta^{-1}\cdot f_{\xi^{*}}$ possesses the properties
(a), (b), (d), and (e) in (i). To show property (c) observe that if
$x_{i}=z_{j}$ for some $j$ then $f_{\xi^{*}}$ has a zero at $z_{j}$ with
multiplicity exceeding that of $f$ so that $\delta$ is strictly greater than
$f_{\xi^{*}}\cdot f^{-1}$ in some neighborhood of $z_{j}$. This implies the
equality $\delta=f_{\xi^{*}}(x)\cdot f(x)^{-1}$ for some $x$ in each of the
open intervals $(x_{1},x_{2})$, …, $(x_{m^{\prime}},x_{m^{\prime}+1})$ and
somewhere in $(x_{m^{\prime}+1},b]$. Thus, in each $(x_{i},x_{i+1})$, either
$f(x)-\delta^{-1}\cdot f_{\xi^{*}}(x)$ vanishes somewhere other than at the
zeros of $f$ or the multiplicity of one of the common zeros of $f$ and
$\delta^{-1}\cdot f_{\xi^{*}}$ is increased by two. In the interval
$(x_{m^{\prime}+1},b]$ the function $f-\delta^{-1}\cdot f_{\xi}$ may vanish at
$b$ with multiplicity only one greater than the zero of $f$ at this point.
This concludes that $f_{*}$ also fulfills (c) in (i).

The polynomial $f^{*}$ when $n^{\prime}=2m^{\prime}+1$ is constructed in the
same manner by replacing $a$ in (59) by $b$.

Uniqueness: Assume another polynomial $g$ satisfies the same properties as
$f_{*}$. Without loss of generality we can assume that the first zero of $f-g$
other than the zeros of $f$ is less than or equal to first zero of $f-f_{*}$.
Define $h:=\frac{f_{*}-g}{f}$. A zero of $h$ occurring at one of the values
$x_{i}$, $i=2,\dots,n^{\prime}+1$ is necessarily at least a double zero. In
this case we assign one zero to each of the intervals $[x_{i-1},x_{i}]$ and
$[x_{i},x_{i+1}]$ with $x_{m^{\prime}+2}=b$. Under this counting procedure,
and taking account of the oscillation properties of $f_{*}$ and $g$, we deduce
that $h$ has at least three zeros in $[a,x_{1}]$, at least two zeros in each
of the intervals $[x_{i},x_{i+1}]$, $i=2,\dots,m^{\prime}$, and at least one
zero in $[x_{m^{\prime}+1},b]$. Clearly, all of these zeros are other than the
$r$ zeros of $f$, so that $f_{*}-g$ has at least $3+2(m^{\prime}-1)+1+r=n+1$
zeros (counting multiplicities). Hence, $h=0$ and $f_{*}=g$.

If $f\in\mathrm{lin}\,\mathcal{F}$ in 5 we get similar to 3 the following
Nichtnegativstellensatz on $[a,b]$ due to Karlin.

######  Karlin’s Nichtnegativstellensatz 6 (for ET-Systems on $[a,b]$; (Kar,
63, p. 603, Cor. after Thm. 3) or e.g. (KS, 66, p. 76, Cor. 10.3)).

Let $n\in\mathds{N}_{0}$, $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-system
of order $n$ on $[a,b]$ with $a<b$, and let $f\in\mathrm{lin}\,\mathcal{F}$ be
such that $f\geq 0$ on $[a,b]$ and $f$ has $r<n$ zeros $a\leq z_{1}\leq
z_{2}\leq\dots\leq z_{r}\leq b$ (counting multiplicities). Then there exists a
unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that

  1. (i)

$f_{*},f^{*}\geq 0$ on $[a,b]$,

  2. (ii)

for $f_{*}$ and $f^{*}$ the sets of zeros counting algebraic multiplicities is
after removing the zeros of $f$ with algebraic multiplicity a set of index
$n-r$ which strictly interlace, and

  3. (iii)

the set of zeros of $f^{*}$ contains after removing the zeros of $f$ with
algebraic multiplicities the point $b$.

######  Proof 37.2.

Let $f_{*}$ be the polynomial from 5 and set $g:=f-f_{*}$. Then $g$ fulfills
the conditions of $f^{*}$ in 5 and by its uniqueness we have $g=f^{*}$ which
proves the statement.

######  Remark 7.

Since 6 (ii) might be a little bit confusing we explain it more detailed.

Let $\mathcal{F}$ be an ET-system of order $n\in\mathds{N}_{0}$ on $[a,b]$
with $a<b$ and let $f\in\mathrm{lin}\,\mathcal{F}$ be such that $f\geq 0$ on
$[a,b]$ and $f$ has the zeros $z_{1},\dots,z_{l}$ with algebraic
multiplicities $m_{1},\dots,m_{l}$, $m_{1}+\dots+m_{l}=:r<n$.

  1. (i)

If $n-r=2m$ is _even_ then the zeros of $f_{*}$ from 6 are $x_{1},\dots,x_{m}$
all with algebraic multiplicity $2$ and the zeros of $f^{*}$ are
$y_{0},y_{1},\dots,y_{m}$ where $y_{0}$ and $y_{m}$ have algebraic
multiplicity $1$ and otherwise the $y_{i}$ have algebraic multiplicity $2$.
They interlace, i.e., we have

$a=y_{0}<x_{1}<y_{1}<\dots<x_{m}<y_{m}=b.$

The $f_{*}$ and $f^{*}$ are then given by

$f_{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|cccccccc}f_{0}&\,f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}&f_{2m+1}&\dots&f_{n}\\\
x&\,x_{1}&x_{1}&\dots&x_{m}&x_{m}&z_{1}&\dots&z_{l}\end{array}\right)$

and

$f^{*}(x)=c^{*}\cdot\det\left(\begin{array}[]{c|ccccccccc}f_{0}&\,f_{1}&f_{2}\;f_{3}\dots&f_{2m-2}&f_{2m-1}&f_{2m}&f_{2m+1}&\dots&f_{n}\\\
x&\,a&y_{1}\;y_{1}\dots&y_{m-1}&y_{m-1}&b&z_{1}&\dots&z_{l}\end{array}\right)$

where $c_{*},c^{*}\in\mathds{R}\setminus\\{0\\}$ and the signs are such that
$f_{*},f^{*}\geq 0$ on $[a,b]$. The zeros $z_{1},\dots,z_{l}$ are included
with their corresponding algebraic multiplicities $m_{1},\dots,m_{l}$, i.e.,
$z_{1}$ is included $m_{1}$-times, …, $z_{l}$ is included $m_{l}$-times.

  2. (ii)

If $n-r=2m+1$ is _odd_ then the zeros of $f_{*}$ from 6 are
$x_{0},\dots,x_{m}$ where $x_{0}$ has algebraic multiplicity $1$ and the other
algebraic multiplicity $2$. For $f^{*}$ we have the zeros $y_{0},\dots,y_{m}$
where $y_{0},\dots,y_{m-1}$ have algebraic multiplicity $2$ and $y_{m}$ has
algebraic multiplicity $1$. They interlace, i.e., we have

$x_{0}=a<y_{0}<\dots<x_{m}<y_{m}=b.$

The $f_{*}$ and $f^{*}$ are then given by

$f_{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|ccccccccc}f_{0}&\,f_{1}&f_{2}\;f_{3}\dots&f_{2m}&f_{2m+1}&f_{2m+2}&\dots&f_{n}\\\
x&\,a&x_{1}\;x_{1}\dots&x_{m}&x_{m}&z_{1}&\dots&z_{l}\end{array}\right)$

and

$f^{*}(x)=c^{*}\cdot\det\left(\begin{array}[]{c|cccccccc}f_{0}&\,f_{1}\;f_{2}&\dots&f_{2m-1}&f_{2m}&f_{2m+1}&f_{2m+2}&\dots&f_{n}\\\
x&\,y_{0}\;y_{0}&\dots&y_{m-1}&y_{m-1}&b&z_{1}&\dots&z_{l}\end{array}\right)$

where $c_{*},c^{*}\in\mathds{R}\setminus\\{0\\}$ and the signs are such that
$f_{*},f^{*}\geq 0$ on $[a,b]$. The zeros $z_{1},\dots,z_{l}$ are included
with their corresponding algebraic multiplicities $m_{1},\dots,m_{l}$, i.e.,
$z_{1}$ is included $m_{1}$-times, …, $z_{l}$ is included $m_{l}$-times.
$\circ$

With the proof of 5 one can prove a similar interlacing theorem as the 4 when
$g_{2}-g_{1}$ has a certain number of zeros, see (KS, 66, p. 76, Rem. 10.1).

We stated here 3 and 6 for functions on $[a,b]$. There are also similar
statements for periodic functions, see (Kar, 63, Thm. 6 and 7). The cases on
$[0,\infty)$ and $\mathds{R}$ are given in the next chapter.

#### Problems

######  Problem 8.

Examine the proof of 5 more closely. In the statement of the theorem it is
required that $\mathcal{F}$ is an ET-system on $[a,b]$. But for a given $f\geq
0$ where does the family $\mathcal{F}$ actually only needs to be an ET-system?

Look at the conclusion! And try to think of a familiar
theorem having the same or a similar conclusion.
George Pólya (Pól, 45, p. 25)

###  Chapter 8 Karlin’s Positivstellensätze and Nichtnegativstellensätze on
$[0,\infty)$ and $\mathds{R}$

In this chapter we extend the results of the previous chapter, i.e., we extend
3 on $[a,b]$ to $[0,\infty)$ in 1 and to $\mathds{R}$ in 4 as well as we
extend 6 on $[a,b]$ to $[0,\infty)$ in 3 and to $\mathds{R}$ in 5.

####  38 Karlin’s Positivstellensatz for T-Systems on $[0,\infty)$

By a transformation of $[a,b]$ to $[0,\infty]$ and then restriction to
$[0,\infty)$ we get from 3 the following.

######  Karlin’s Positivstellensatz 1 (for T-Systems on $[0,\infty)$; see
(Kar, 63, Thm. 9) or e.g. (KS, 66, p. 169, Thm. 8.1)).

Let $n\in\mathds{N}_{0}$ and $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a
continuous T-system of order $n$ on $[0,\infty)$ such that

  1. (a)

there exists a $C>0$ such that $f_{n}(x)>0$ for all $x\geq C$,

  2. (b)

$\displaystyle\lim_{x\to\infty}\frac{f_{i}(x)}{f_{n}(x)}=0$ for all
$i=0,\dots,n-1$, and

  3. (c)

$\\{f_{i}\\}_{i=0}^{n-1}$ is a continuous T-system on $[0,\infty)$.

Then for any $f=\sum_{i=0}^{n}a_{i}f_{i}\in\mathrm{lin}\,\mathcal{F}$ with
$f>0$ on $[0,\infty)$ and $a_{n}>0$ there exists a unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ and $f_{*},f^{*}\geq 0$ on
$[0,\infty)$ such that the following hold:

  1. (i)

If $n=2m$ the polynomials $f_{*}$ and $f^{*}$ each possess $m$ distinct zeros
$\\{x_{i}\\}_{i=1}^{m}$ and $\\{y_{i}\\}_{i=0}^{m-1}$ satisfying

$0=y_{0}<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<\infty.$

All zeros except $y_{0}$ are double zeros.

  2. (ii)

If $n=2m+1$ the polynomials $f_{*}$ and $f^{*}$ each possess the zeros
$\\{x_{i}\\}_{i=1}^{m+1}$ and $\\{y_{i}\\}_{i=1}^{m}$ satisfying

$0=x_{1}<y_{1}<x_{2}<\dots<y_{m}<x_{m+1}<\infty.$

All zeros except $x_{1}$ are double zeros.

  3. (iii)

The coefficient of $f_{n}$ in $f_{*}$ is equal to $a_{n}$.

The proof is adapted from (KS, 66, pp. 168).

######  Proof 38.1.

By (a) there exists a function $w\in\mathcal{C}([0,\infty),\mathds{R})$ such
that $w>0$ on $[0,\infty)$ and $\lim_{x\to\infty}\frac{f_{n}(x)}{w(x)}=1$. By
(b) we define

$v_{i}(x):=\begin{cases}\frac{f_{i}(x)}{w(x)}&\text{if}\ x\in[0,\infty),\\\
\delta_{i,n}&\text{if}\ x=\infty\end{cases}$

for all $i=0,1,\dots,n$. Then by (c) and 9 we have that
$\\{v_{i}\\}_{i=0}^{n}$ is a T-system on $[0,\infty]$. With $t(x):=\tan(\pi
x/2)$ we define $g_{i}(x):=v_{i}\circ t$ for all $i=0,1,\dots,n$. Hence,
$\mathcal{G}=\\{g_{i}\\}_{i=0}^{n}$ is a T-system on $[0,1]$ by 8. We now
apply 3 to $\mathcal{G}$. Set $g:=\big{(}\frac{f}{w}\big{)}\circ t$.

(i): Let $n=2m$. Then by 3 there exist points

$0=y_{0}<x_{1}<y_{1}<\dots<x_{m}<y_{m}=1$

and unique functions $g_{*}$ and $g^{*}$ such that $g=g_{*}+g^{*}$,
$g_{*},g^{*}\geq 0$ on $[0,1]$, $x_{1},\dots,x_{m}$ are the zeros of $g_{*}$,
and $y_{0},\dots,y_{m}$ are the zeros of $g^{*}$. Then $f_{*}:=(g_{*}\circ
t^{-1})\cdot w$ and $f^{*}:=(g^{*}\circ t^{-1})\cdot w$ are the unique
components in the decomposition $f=f_{*}+f^{*}$.

(ii): Similar to (i).

(iii): From (i) (and (ii) in a similar way) we have $g_{i}(1)=0$ for
$i=0,\dots,n-1$ and $g_{n}(1)=1$. Hence, we get with $g^{*}(y_{m}=1)=0$ that
$g_{n}$ is not contained in $g^{*}$, i.e., $g_{*}$ has the only $g_{n}$
contribution because $\mathcal{G}$ is linearly independent. This is inherited
by $f_{*}$ and $f^{*}$ which proves (iii).

The transformation $g_{i}=v_{i}\circ t$ with $t$ the $\tan$-function is due to
Krein Kre (51).

If $\mathcal{F}$ in 1 is an ET-system then the $f_{*}$ and $f^{*}$ can be
written down explicitly. For that we only need $\mathcal{F}$ to be an ET-
system on $(0,\infty)$ not on all $[0,\infty)$ since at $x=0$ a possible zero
in $f_{*}$ or $f^{*}$ only has multiplicity one.

######  Corollary 2.

If in 1 we have additionally that $\mathcal{F}$ is an ET-system on
$(0,\infty)$ then the unique $f_{*}$ and $f^{*}$ are given

  1. (i)

for $n=2m$ by

$f_{*}(x)=c^{*}\cdot\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}$

and

$f^{*}(x)=-c_{*}\cdot\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m-2}&f_{2m-1}\\\
x&y_{0}&(y_{1}&y_{1})&\dots&(y_{m-1}&y_{m-1})\end{pmatrix},$

  2. (ii)

and for $n=2m+1$ by

$f_{*}(x)=-c_{*}\cdot\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m}&f_{2m+1}\\\
x&x_{1}&(x_{2}&x_{2})&\dots&(x_{m+1}&x_{m+1})\end{pmatrix}$

and

$f^{*}(x)=c^{*}\cdot\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}\\\
x&(y_{1}&y_{1})&\dots&(y_{m}&y_{m})\end{pmatrix}$

for some $c_{*},c^{*}>0$.

######  Proof 38.2.

Combine 1 with 28 and note that since $0$ is never a multiple zero we only
need $\mathcal{F}$ to be an ET-system on $(0,\infty)$.

####  39 Karlin’s Nichtnegativstellensatz for ET-Systems on $[0,\infty)$

In 1 we needed to transform the domain $[a,b]$ into $[0,\infty]$ of a
T-system. For 3 we needed an ET-system because of the additional zeros from
$f\geq 0$.

With the same technique as in the proof of 1 and 8 we get from 6 the
following.

######  Karlin’s Nichtnegativstellensatz 3 (for ET-Systems on $[0,\infty)$).

Let $n\in\mathds{N}_{0}$ and $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system of order $n$ on $[0,\infty)$ such that

  1. (a)

there exists a $C>0$ such that $f_{n}(x)>0$ for all $x\geq 0$,

  2. (b)

$\displaystyle\lim_{x\to\infty}\frac{f_{i}(x)}{f_{n}(x)}=0$ for all
$i=0,\dots,n-1$, and

  3. (c)

$\\{f_{i}\\}_{i=0}^{n-1}$ is an ET-system.

Then for any $f=\sum_{i=0}^{n}a_{i}f_{i}\in\mathrm{lin}\,\mathcal{F}$ such
that $f\geq 0$ on $[0,\infty)$, $a_{n}>0$, and $f$ has $r<n$ zeros counting
multiplicity there exists a unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that the following hold:

  1. (i)

$f_{*},\ f^{*}\geq 0$ on $[0,\infty)$,

  2. (ii)

$f_{*}$ has $n$ zeros (counting multiplicities),

  3. (iii)

$f^{*}$ has $n-1$ zeros (counting multiplicities),

  4. (iv)

the zeros of $f_{*}$ and $f^{*}$ strictly interlace if the zeros of $f$ are
removed, and

  5. (v)

the coefficient of $f_{n}$ in $f_{*}$ is equal to $a_{n}$.

######  Proof 39.1.

The conditions (a) – (c) are such that $\mathcal{F}$ on $[0,\infty]$, i.e.,
including $\infty$, is an ET-system.

With the same argument as in the proof of 1 we transform $\mathcal{F}$ on
$[0,\infty]$ into $\mathcal{G}$ on $[0,1]$ with the $\tan$-function. Here 8
ensures that also $\mathcal{G}$ is an ET-system.

Application of 6 on $[0,1]$ gives the desired decomposition $g=g_{*}+g^{*}$
with the observation that $x=1$ is a zero of at most multiplicity one by (a)
and (b). Backwards transformation into $\mathcal{F}$ on $[0,\infty]$ resp.
$[0,\infty)$ then gives the assertion.

####  40 Karlin’s Positivstellensatz for T-Systems on $\mathds{R}$

We have seen that from 3 on $[a,b]$ we get 1 on $[0,\infty)$ with the
transformation $t(x)=\tan(\pi x/2)$ from $[0,1]$ to $[0,\infty]$ and only need
to pay attention to the end point $x=1$ resp. $x=\infty$. The same
transformation however also applies going from $[-1,1]$ to $[-\infty,\infty]$
now paying attention to both end points.

######  Karlin’s Positivstellensatz 4 (for T-Systems on $\mathds{R}$; see
(Kar, 63, Thm. 10) or e.g. (KS, 66, p. 198, Thm. 8.1)).

Let $m\in\mathds{N}_{0}$ and $\mathcal{F}=\\{f_{i}\\}_{i=0}^{2m}$ be a
continuous T-system of order $2m$ on $\mathds{R}$ such that

  1. (a)

there exists a $C>0$ such that $f_{2m}(x)>0$ for all
$x\in(-\infty,-C]\cup[C,\infty)$,

  2. (b)

$\displaystyle\lim_{|x|\to\infty}\frac{f_{i}(x)}{f_{2m}(x)}=0$ for all
$i=0,\dots,2m-1$, and

  3. (c)

$\\{f_{i}\\}_{i=0}^{2m-1}$ is a continuous T-system of order $2m-1$ on
$\mathds{R}$.

Let $f=\sum_{i=0}^{2m}a_{i}f_{i}$ be such that $f>0$ on $\mathds{R}$ and
$a_{2m}>0$. Then there exists a unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ and $f_{*},f^{*}\geq 0$ on
$\mathds{R}$ such that

  1. (i)

the coefficient of $f_{2m}$ in $f_{*}$ is $a_{2m}$, and

  2. (ii)

$f_{*}$ and $f^{*}$ are non-negative polynomials having zeros
$\\{x_{i}\\}_{i=1}^{m}$ and $\\{y_{i}\\}_{i=1}^{m-1}$ with

$-\infty<x_{1}<y_{1}<x_{2}<\dots<y_{m-1}<x_{m}<\infty.$

######  Proof 40.1.

See Problem 6.

####  41 Karlin’s Nichtnegativstellensatz for ET-Systems on $\mathds{R}$

On $\mathds{R}$ we have the following Nichtnegativstellensatz for ET-systems.

######  Karlin’s Nichtnegativstellensatz 5 (for ET-Systems on $\mathds{R}$).

Let $m\in\mathds{N}_{0}$ and $\mathcal{F}=\\{f_{i}\\}_{i=0}^{2m}$ be an ET-
system of order $2m$ on $\mathds{R}$ such that

  1. (a)

there exists a $C>0$ such that $f_{2m}>0$ for all
$x\in(-\infty,-C]\cup[C,\infty)$,

  2. (b)

$\displaystyle\lim_{|x|\to\infty}\frac{f_{i}(x)}{f_{2m}(x)}=0$ for all
$i=0,\dots,2m-1$,

  3. (c)

$\\{f_{i}\\}_{i=0}^{n-1}$ is an ET-system of order $n-1$ on $\mathds{R}$.

Let $f=\sum_{i=0}^{2m}a_{i}f_{i}\in\mathrm{lin}\,\mathcal{F}$ be such that
$f\geq 0$, $a_{2m}>0$, and $f$ has $r<n$ zeros counting multiplicities. Then
there exists a unique representation

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that the following hold:

  1. (i)

$f_{*},f^{*}\geq 0$ on $\mathds{R}$,

  2. (ii)

$f_{*}$ has $2m$ zeros counting multiplicity,

  3. (iii)

$f^{*}$ has $2m-2$ zeros counting multiplicity,

  4. (iv)

the zeros of $f_{*}$ and $f^{*}$ strictly interlace if the zeros of $f$ are
removed, and

  5. (v)

the coefficient of $f_{n}$ in $f_{*}$ is equal to $a_{n}$.

######  Proof 41.1.

See Problem 7.

#### Problems

######  Problem 6.

Prove 4, i.e., adapt the proof of 1 such that both interval ends $a$ and $b$
of $[a,b]$ are mapped to $-\infty$ and $+\infty$, respectively.

######  Problem 7.

Prove 5, i.e., adapt the proof of 3 such that both interval ends $a$ and $b$
of $[a,b]$ are mapped to $-\infty$ and $+\infty$, respectively.

##  Part IV Non-Negative Algebraic Polynomials on $[a,b]$, $[0,\infty)$, and
$\mathds{R}$

I hold that it is only when we can prove everything we assert
that we understand perfectly the thing under consideration.
Gottfried Wilhelm Leibniz Lei (89)

###  Chapter 9 Non-Negative Algebraic Polynomials on $[a,b]$

We developed in the previous chapters the Positiv- and
Nichtnegativestellensätze for general T- and ET-systems due to Karlin. We will
now apply these to the algebraic polynomials, i.e., we will plug in 15 and 17.

####  42 Sparse Algebraic Positivstellensatz on $[a,b]$

At first let us have a look how all sparse strictly positive polynomials on
some interval $[a,b]\subseteq(0,\infty)$ look like.

######  Theorem 1 (Sparse Algebraic Positivstellensatz on $[a,b]$ with
$0<a<b$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in\mathds{R}$ be real
numbers with $\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$, and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$. Then for any
$f=\sum_{i=0}^{n}a_{i}x^{\alpha_{i}}\in\mathrm{lin}\,\mathcal{F}$ with $f>0$
on $[a,b]$ and $a_{n}>0$ there exists a unique decomposition

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ such that

  1. (i)

for $n=2m$ there exist points $x_{1},\dots,x_{m},y_{1},\dots,y_{m-1}\in[a,b]$
with

$a<x_{1}<y_{1}<\dots<x_{m}<b$

and constants $c_{*},c^{*}>0$ with

$f_{*}(x)=c_{*}\cdot\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\geq 0$ (60)

and

$f^{*}(x)=-c^{*}\cdot\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&a&(y_{1}&y_{1})&\dots&(y_{m-1}&y_{m-1})&b\end{pmatrix}\geq 0$ (61)

for all $x\in[a,b]$, or

  2. (ii)

for $n=2m+1$ there exist points $x_{1},\dots,x_{m},y_{1},\dots,y_{m}\in[a,b]$
with

$a<y_{1}<x_{1}<\dots<y_{m}<x_{m}<b$

and $c_{*},c^{*}>0$ with

$f_{*}(x)=-c_{*}\cdot\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\geq 0$ (62)

and

$f^{*}(x)=c^{*}\cdot\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(y_{1}&y_{1})&\dots&(y_{m}&y_{m})&b\end{pmatrix}\geq 0$ (63)

for all $x\in[a,b]$.

######  Proof 42.1.

By 17 we have that $\mathcal{F}$ on $[a,b]$ is an ET-system. Hence, 3 applies.
We check both cases $n=2m$ and $n=2m+1$ separately.

$n=2m$: By 3 we have that the zero set $\mathcal{Z}(f^{*})$ of $f^{*}$ has
index $2m$ and contains $b$ with index $1$, i.e., $a\in\mathcal{Z}(f^{*})$ and
all other zeros have index $2$. Hence,
$\mathcal{Z}(f^{*})=\\{a=y_{0}<y_{1}<\dots<y_{m-1}<y_{m}=b\\}$. By 3 we have
that $\mathcal{Z}(f_{*})$ also has index $2m$ and the zeros of $f_{*}$ and
$f^{*}$ interlace. Then the determinantal representations of $f_{*}$ and
$f^{*}$ follow from 28.

$n=2m+1$: By 3 we have that $b\in\mathcal{Z}(f^{*})$ and since the index of
$\mathcal{Z}(f^{*})$ is $2m+1$ we have that there are only double zeros
$y_{1},\dots,y_{m}\in(a,b)$ in $\mathcal{Z}(f^{*})$. Similar we find that
$a\in\mathcal{Z}(f_{*})$ since its index is odd and only double zeros
$x_{1},\dots,x_{m}\in(a,b)$ in $\mathcal{Z}(f_{*})$ remain. By 3 (iii) the
zeros $x_{i}$ and $y_{i}$ strictly interlace and the determinantal
representation of $f_{*}$ and $f^{*}$ follow again from 28.

Note, if $\alpha_{0},\dots,\alpha_{n}\in\mathds{N}_{0}$ then by 17 equation
(47) the algebraic polynomials $f_{*}$ and $f^{*}$ in (60) – (63) can be
written down with Schur polynomials.

######  Remark 2.

The condition $a_{n}>0$ in 1 is no restriction. The result also holds for
$a_{n}<0$ as long as $f>0$ on $[a,b]$. Since $[a,b]$ is compact the
polynomials $x^{\alpha_{i}}$ are bounded. In the definition of a T-system the
order of the functions $f_{i}$ can be altered since only any linear
combination has to have at most $n$ zeros. Hence, in a $f>0$ at least one
coefficient $a_{i}$ is larger then zero and we interchange $f_{i}$ with
$f_{n}$. A possible sign change in the $f_{*}$ and $f^{*}$ in (60) – (63)
might appear. $\circ$

1 does not hold for $a=0$ and $\alpha_{0}>0$ or
$\alpha_{0},\dots,\alpha_{k}<0$. In case $\alpha_{0}>0$ the determinantal
representations of $f^{*}$ for $n=2m$ and $f_{*}$ for $n=2m+1$ are the zero
polynomials. In fact, in this case $\mathcal{F}$ is not even a T-system since
in 5 the determinant contains a zero column if $x_{0}=0$. We need to have
$\alpha_{0}=0$ ($x^{\alpha_{0}}=1$) to let $a=0$. For
$\alpha_{0},\dots,\alpha_{k}<0$ we have singularities at $x=0$ and hence no
T-system.

######  Corollary 3.

If $\alpha_{0}=0$ in 1 then 1 also holds with $a=0$.

######  Proof 42.2.

The determinantal representations of $f_{*}$ for $n=2m+1$ and $f^{*}$ for
$n=2m$ in 1 continuously depend on $a$. It is sufficient to show that these
representations are non-trivial (not the zero polynomial) for $a=0$. We show
this for $f_{*}$ in case (ii) $n=2m+1$. The other cases are equivalent.

We have that $\mathcal{F}$ is a T-system on $[0,b]$ with $b>0$. For
$\varepsilon>0$ small enough we set

$\displaystyle g_{\varepsilon}(x)$
$\displaystyle=-\varepsilon^{-m}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&0&x_{1}&x_{1}+\varepsilon&\dots&x_{m}&x_{m}+\varepsilon\end{pmatrix}$
$\displaystyle=-\varepsilon^{-m}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m+1}}\\\
1&0&0&\dots&0\\\
1&x_{1}^{\alpha_{1}}&x_{1}^{\alpha_{2}}&\dots&x_{1}^{\alpha_{2m+1}}\\\
\vdots&\vdots&\vdots&&\vdots\\\
1&(x_{m}+\varepsilon)^{\alpha_{1}}&(x_{m}+\varepsilon)^{\alpha_{2}}&\dots&(x_{m}+\varepsilon)^{\alpha_{2m+1}}\end{pmatrix}$
develop with respect to the second row
$\displaystyle=\varepsilon^{-m}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}\\\
x_{1}^{\alpha_{1}}&x_{1}^{\alpha_{2}}&\dots&x_{1}^{\alpha_{2m-1}}\\\
\vdots&\vdots&&\vdots\\\
(x_{m}+\varepsilon)^{\alpha_{1}}&(x_{m}+\varepsilon)^{\alpha_{2}}&\dots&(x_{m}+\varepsilon)^{\alpha_{2m+1}}\end{pmatrix}$
$\displaystyle=\varepsilon^{-m}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&x_{1}&x_{1}+\varepsilon&\dots&x_{m}&x_{m}+\varepsilon\end{pmatrix}.$

Then $x_{1},x_{1}+\varepsilon,\dots,x_{m},x_{m}+\varepsilon\in(0,b]$, i.e.,
$\\{x^{\alpha_{i}}\\}_{i=1}^{n}$ is an ET-system on $[a^{\prime},b]$ with
$0=a<a^{\prime}<x_{1}$, see 17. By 28 the limit $\varepsilon\searrow 0$ is not
the zero polynomial which ends the proof.

######  Remark 4.

It is clear that if $\alpha_{0}>0$ then we can just factor out
$x^{\alpha_{0}}$

$f(x)=a_{0}x^{\alpha_{0}}+a_{1}x^{\alpha_{1}}+\dots+a_{n}x^{\alpha_{n}}=x^{\alpha_{0}}\cdot(\underbrace{a_{0}+a_{1}x^{\alpha_{1}-\alpha_{0}}+\dots+a_{n}x^{\alpha_{n}-\alpha_{0}}}_{=:\tilde{f}(x)})$

and apply 1 or 3 to $\tilde{f}$. $\circ$

We now prove a stronger version of (16). We only need the sparse algebraic
Positivstellensatz on $[a,b]$ (1) but not the sparse algebraic
Nichtnegativestellensatz (10) even for $p\geq 0$ on $[a,b]$. This result was
already proved in KS (53). Later the T-system approach was developed in Kar
(63) and summarized and expanded in KS (66).

We now get the strengthened version of the 7. Earlier versions are due to
Markov Mar (06) and Lukács Luk (18), see the 7 in Section 17 and the
discussion around it.

######  Lukács–Markov Theorem 5 (see (KS, 53, Thm. 10.3) or (KN, 77, p. 373,
Thm. 6.4)).

Let $p\in\mathds{R}[x]$ with $p\geq 0$ on $[a,b]$ with $-\infty<a<b<\infty$
and let $z_{1},\dots,z_{r}\in[a,b]$ be the zeros of $p$ in $[a,b]$ with
algebraic multiplicities $m_{1},\dots,m_{r}\in\mathds{N}$.

  1. (i)

If $\deg p-m_{1}-\dots-m_{r}=2m$, $m\in\mathds{N}_{0}$, is even then there
exist points $x_{1},\dots,x_{m}$ and $y_{1},\dots,y_{m-1}$ with

$a<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<b$

and constants $\alpha,\beta>0$ such that

$p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\left(\alpha\cdot\prod_{i=1}^{m}(x-x_{i})^{2}\right.\\\
\left.+\beta\cdot(x-a)\cdot(b-x)\cdot\prod_{i=1}^{m-1}(x-y_{i})^{2}\right).$

  2. (ii)

If $\deg p-m_{1}-\dots-m_{r}=2m+1$, $m\in\mathds{N}_{0}$, is odd then there
exist points $x_{1},\dots,x_{m}$ and $y_{0},\dots,y_{m-1}$ with

$a<y_{0}<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<b$

and constants $\alpha,\beta>0$ such that

$p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\left(\alpha\cdot(x-a)\cdot\prod_{i=1}^{m}(x-x_{i})^{2}\right.\\\
\left.+\beta\cdot(b-x)\cdot\prod_{i=0}^{m-1}(x-y_{i})^{2}\right).$

######  Proof 42.3.

We have $p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\tilde{p}(x)$ with
$\tilde{p}\in\mathds{R}[x]$ and $\tilde{p}>0$ on $[a,b]$. By a translation
$p(\,\cdot\,+a)$ we can assume $a=0$ and the assertion follows from 3.

Note, in 1 (and 10) we need $a\geq 0$. But in the 5 we can allow for arbitrary
$a\in\mathds{R}$ since by $p\in\mathds{R}[x]_{\leq\deg p}$ the translation
$p(\,\cdot\,+a)$ remains in $\mathds{R}[x]_{\leq\deg p}$. We see here also why
in 1 and 3 we have the restriction $a\geq 0$ since a translation can produce
monomials which are not in the family $\\{x^{\alpha_{i}}\\}_{i=0}^{n}$.

Additionally, note that in 5 we can have $z_{i}=a$ or $b$ for some $i$.

####  43 Sparse Hausdorff Moment Problem

1 is a complete description of
$\mathrm{int}\,(\mathrm{lin}\,\mathcal{F})_{+}$. Since $\mathcal{F}$ is
continuous on the compact interval $[a,b]$ and $x^{\alpha_{0}}>0$ on $[a,b]$,
we have that the truncated moment cone is closed. Hence,
$(\mathrm{lin}\,\mathcal{F})_{+}$ and the moment cone are dual to each other.
With 1 we can now write down the conditions for the sparse truncated Hausdorff
moment problem on $[a,b]$ with $a>0$. A first but insufficient attempt was
done in Hau21b  since Hausdorff did not have access to the sparse
Positivstellensatz by Karlin and therefore 1.

######  Theorem 6 (Sparse Truncated Hausdorff Moment Problem on $[a,b]$ with
$a>0$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in[0,\infty)$ with
$\alpha_{0}<\dots<\alpha_{n}$, and $a,b$ with $0<a<b$. Set
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$. Then the following are
equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a truncated $[a,b]$-moment
functional.

  2. (ii)

$L(p)\geq 0$ holds for all

$\displaystyle p(x)$
$\displaystyle:=\begin{cases}\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
-\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})&b\end{pmatrix}\end{cases}$ if $n=2m$
and $\displaystyle p(x)$
$\displaystyle:=\begin{cases}-\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
\det\begin{pmatrix}x^{\alpha_{0}}&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})&b\end{pmatrix}\end{cases}$ if $n=2m+1$

and all $x_{1},\dots,x_{m}$ with $a<x_{1}<\dots<x_{m}<b$.

######  Proof 43.1.

The implication (i) $\Rightarrow$ (ii) is clear since all given polynomials
$p$ are non-negative on $[a,b]$. It is therefore sufficient to prove (ii)
$\Rightarrow$ (i).

Since $a>0$ we have that $x^{\alpha_{0}}>0$ on $[a,b]$ and since $[a,b]$ is
compact we have that the moment cone $((\mathrm{lin}\,\mathcal{F})_{+})^{*}$
as the dual of the cone of non-negative (sparse) polynomials
$(\mathrm{lin}\,\mathcal{F})_{+}$ is a closed pointed cone.

To establish $L\in((\mathrm{lin}\,\mathcal{F})_{+})^{*}$ it is sufficient to
have $L(f)\geq 0$ for all $f\in(\mathrm{lin}\,\mathcal{F})_{+}$. Let
$f\in(\mathrm{lin}\,\mathcal{F})_{+}$. Then for all $\varepsilon>0$ we have
$f_{\varepsilon}:=f+\varepsilon\cdot x^{\alpha_{n}}>0$ on $[a,b]$, i.e., by 1
$f_{\varepsilon}$ is a conic combination of the polynomials $p$ in (ii) and
hence $L(f)+\varepsilon\cdot L(x^{\alpha_{n}})=L(f_{\varepsilon})\geq 0$ for
all $\varepsilon>0$. Since $x^{\alpha_{n}}>0$ on $[a,b]$ we also have that
$x^{\alpha_{n}}$ is a conic combination of the polynomials $p$ in (ii) and
therefore $L(x^{\alpha_{n}})\geq 0$. Then $L(f)\geq 0$ follows from
$\varepsilon\to 0$ which proves (i).

######  Corollary 7.

If $\alpha_{0}=0$ in 6 then 6 also holds with $a=0$, i.e., the following are
equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a truncated $[0,b]$-moment
functional.

  2. (ii)

$L(p)\geq 0$ holds for all

$\displaystyle p(x)$
$\displaystyle:=\begin{cases}\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})&b\end{pmatrix}\end{cases}$ if $n=2m$
and $\displaystyle p(x)$
$\displaystyle:=\begin{cases}\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})&b\end{pmatrix}\end{cases}$ if $n=2m+1$

and all $x_{1},\dots,x_{m}$ with $a<x_{1}<\dots<x_{m}<b$.

######  Proof 43.2.

Follows immediately from 3.

For the following we want to remind the reader of the _Müntz–Szász Theorem_
Mün (14); Szá (16). It states that for real exponents
$\alpha_{0}=0<\alpha_{1}<\alpha_{2}<\dots$ the vector space
$\mathrm{lin}\,\\{x^{\alpha_{i}}\\}_{i\in\mathds{N}_{0}}$ of finite linear
combinations is dense in $\mathcal{C}([0,1],\mathds{R})$ with respect to the
$\sup$-norm if and only if $\sum_{i\in\mathds{N}}\frac{1}{\alpha_{i}}=\infty$.

We state the following only for the classical case of the interval $[0,1]$.
Other cases $[a,b]\subseteq[0,\infty)$ are equivalent. Hausdorff required
$\alpha_{i}\to\infty$. The Müntz–Szász Theorem does not require
$\alpha_{i}\to\infty$. The conditions $\alpha_{0}=0$ and
$\sum_{i\in\mathds{N}}\frac{1}{\alpha_{i}}=\infty$ already appear in (Hau21b,
, eq. (17)). We can remove here the use of the Müntz–Szász Theorem and
therefore the condition $\sum_{i\in\mathds{N}}\frac{1}{\alpha_{i}}=\infty$ for
the existence of a representing measure. We need it only for uniqueness.
Additionally, we allow negative exponents. The following is an improvement of
Hau21b  and we are not aware of a reference for this result.

######  Theorem 8 (General Sparse Hausdorff Moment Problem on $[a,b]$ with
$0\leq a<b$).

Let $I\subseteq\mathds{N}_{0}$ be an index set (finite or infinite), let
$\\{\alpha_{i}\\}_{i\in I}$ be such that $\alpha_{i}\neq\alpha_{j}$ for all
$i\neq j$ and

  1. (a)

if $a=0$ then $\\{\alpha_{i}\\}_{i\in I}\subset[0,\infty)$ with $\alpha_{i}=0$
for an $i\in I$, or

  2. (b)

if $a>0$ then $\\{\alpha_{i}\\}_{i\in I}\subset\mathds{R}$.

Let $\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i\in I}$. Then the following are
equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a Hausdorff moment functional.

  2. (ii)

$L(p)\geq 0$ holds for all $p\in(\mathrm{lin}\,\mathcal{F})_{+}$.

  3. (iii)

$L(p)\geq 0$ holds for all $p\in\mathrm{lin}\,\mathcal{F}$ with $p>0$.

  4. (iv)

$L(p)\geq 0$ holds for all

$p(x)=\begin{cases}\det\begin{pmatrix}x^{\alpha_{i_{0}}}&x^{\alpha_{i_{1}}}&x^{\alpha_{i_{2}}}&\dots&x^{\alpha_{i_{2m-1}}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix},&\text{if $|I|=2m$ or
$\infty$,}\\\
\det\begin{pmatrix}x^{\alpha_{i_{1}}}&x^{\alpha_{i_{2}}}&x^{\alpha_{i_{3}}}&\dots&x^{\alpha_{i_{2m-2}}}&x^{\alpha_{i_{2m-1}}}&x^{\alpha_{i_{2m}}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})&b\end{pmatrix},&\text{if $|I|=2m$ or
$\infty$,}\\\
\det\begin{pmatrix}x^{\alpha_{i_{1}}}&x^{\alpha_{i_{2}}}&x^{\alpha_{i_{3}}}&\dots&x^{\alpha_{i_{2m}}}&x^{\alpha_{i_{2m+1}}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix},\ &\text{if $|I|=2m+1$ or
$\infty$,}\\\
\det\begin{pmatrix}x^{\alpha_{i_{0}}}&x^{\alpha_{i_{1}}}&x^{\alpha_{i_{2}}}&\dots&x^{\alpha_{i_{2m-1}}}&x^{\alpha_{i_{2m}}}&x^{\alpha_{i_{2m+1}}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})&b\end{pmatrix},&\text{if $|I|=2m+1$ or
$\infty$,}\end{cases}$

for all $m\in\mathds{N}$ if $|I|=\infty$, all $0<x_{1}<x_{2}<\dots<x_{m}<b$,
and all $\alpha_{i_{0}}<\alpha_{i_{1}}<\dots<\alpha_{i_{m}}$ with
$\alpha_{i_{0}}=0$ if $a=0$.

If additionally $\sum_{i:\alpha_{i}\neq 0}\frac{1}{|\alpha_{i}|}=\infty$ then
$L$ is determinate.

######  Proof 43.3.

The case $|I|<\infty$ is 6. We therefore prove the case $|I|=\infty$. The
choice $\alpha_{i_{0}}<\alpha_{i_{1}}<\dots<\alpha_{i_{m}}$ with
$\alpha_{i_{0}}=0$ if $a=0$ makes $\\{x^{\alpha_{i_{j}}}\\}_{j=0}^{m}$ a
T-system. The implications “(i) $\Rightarrow$ (ii) $\Leftrightarrow$ (iii)”
are clear and “(iii) $\Leftrightarrow$ (iv)” is 1. It is therefore sufficient
to show “(ii) $\Rightarrow$ (i)”. But the space $\mathrm{lin}\,\mathcal{F}$ is
an adapted space and the assertion follows therefore from the 9.

For the determinacy of $L$ split $\\{\alpha_{i}\\}_{i\in I}$ into positive and
negative exponents. If $\sum_{i:\alpha_{i}\neq
0}\frac{1}{|\alpha_{i}|}=\infty$ then the corresponding sum over at least one
group is infinite. If the sum over the positive exponents is infinite apply
the Müntz–Szász Theorem. If the sum over the negative exponents is infinite
apply the Müntz–Szász Theorem to $\\{(x^{-1})^{-\alpha_{i}}\\}_{i\in
I:\alpha_{i}<0}$ since $a>0$.

Note, since $[a,b]$ is compact the fact that $\\{x^{\alpha_{i}}\\}_{i\in I}$
is an adapted space is trivial.

######  Remark 9.

If in 8 we have $a=0$ and $\alpha_{0}>0$ then we can of course factor out
$x^{\alpha_{0}}$ and instead of determining $\mathrm{d}\mu(x)$ of the linear
functional $L$ we determine
$\mathrm{d}\tilde{\mu}(x)=x^{\alpha_{0}}~{}\mathrm{d}\mu(x)$. $\circ$

####  44 Sparse Algebraic Nichtnegativstellensatz on $[a,b]$

The non-negative polynomials are described in the following result.

######  Theorem 10 (Sparse Algebraic Nichtnegativstellensatz on $[a,b]$ with
$0<a<b$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in\mathds{R}$ be real
numbers with $\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$, and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$. Let
$f\in\mathrm{lin}\,\mathcal{F}$ with $f\geq 0$ on $[a,b]$. Then there exist
points $x_{1},\dots,x_{n},y_{1},\dots,y_{n}\in[a,b]$ (not necessarily
distinct) with $y_{n}=b$ which include the zeros of $f$ with multiplicities
such that

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$, $f_{*},f^{*}\geq 0$ on
$[a,b]$. The polynomials $f_{*}$ and $f^{*}$ are given by

$f_{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|ccc}f_{0}&\
f_{1}&\dots&f_{n}\\\ x&\
x_{1}&\dots&x_{n}\end{array}\right)\qquad\text{and}\qquad
f^{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|ccc}f_{0}&\
f_{1}&\dots&f_{n}\\\ x&\ y_{1}&\dots&y_{n}\end{array}\right)$

for all $x\in[a,b]$ and some constants $c_{*},c^{*}\in\mathds{R}$

Removing the zeros of $f$ from $x_{1},\dots,x_{n},y_{1},\dots,y_{n}$ we can
assume that the remaining $x_{i}$ and $y_{i}$ are disjoint and when grouped by
size the groups strictly interlace:

$a\leq
x_{i_{1}}=\dots=x_{i_{k}}<y_{j_{1}}=\dots=y_{j_{l}}<\dots<x_{i_{p}}=\dots=x_{i_{q}}<y_{j_{r}}=\dots=y_{j_{s}}=b.$

Each such group in $(a,b)$ has an even number of members.

######  Proof 44.1.

By 17 we have that $\mathcal{F}$ on $[a,b]$ is an ET-system. We then apply 6
similar to the proof of 1.

######  Remark 11.

The signs of $c_{*}$ and $c^{*}$ are determined by $x_{1}$ and $y_{1}$ and
their multiplicity. If $x_{1}=\dots=x_{k}<x_{k+1}\leq\dots\leq x_{n}$ then
$\mathrm{sgn}\,c_{*}=(-1)^{k}$. The same holds for $c^{*}$ from $y_{1}$.
$\circ$

######  Example 12.

Let $\alpha\in(0,\infty)$ and let $\mathcal{F}=\\{1,x^{\alpha}\\}$ on $[0,1]$.
Then we have $1=1_{*}+1^{*}$ with $1_{*}=x^{\alpha}$ and $1^{*}=1-x^{\alpha}$.
$\circ$

In 10 we can let $a=0$ if $\alpha_{0}=0$ and $f(0)>0$.

######  Theorem 13 (Sparse Algebraic Nichtnegativstellensatz on $[0,b]$ with
$0<b$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in\mathds{R}$ be real
numbers with $0=\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$, and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$ on $[0,b]$ with $b>0$. Let
$f\in\mathrm{lin}\,\mathcal{F}$ with $f\geq 0$ on $[0,b]$ and $f(0)>0$. Then
there exist points $x_{1},\dots,x_{n},y_{1},\dots,y_{n}\in[0,b]$ (not
necessarily distinct) with $y_{n}=b$ which include the zeros of $f$ with
multiplicities such that

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$, $f_{*},f^{*}\geq 0$ on $[0,b]$
and the points $x_{1},\dots,x_{n}$ are the zeros of $f_{*}$ and
$y_{1},\dots,y_{n}$ are the zeros of $f^{*}$. Removing the zeros of $f$ from
$x_{1},\dots,x_{n},y_{1},\dots,y_{n}$ we can assume that the remaining $x_{i}$
and $y_{i}$ are disjoint and when grouped by size the groups strictly
interlace:

$0\leq
x_{i_{1}}=\dots=x_{i_{k}}<y_{j_{1}}=\dots=y_{j_{l}}<\dots<x_{i_{p}}=\dots=x_{i_{q}}<y_{j_{r}}=\dots=y_{j_{s}}=b.$

Each such group in $(a,b)$ has an even number of members.

######  Proof 44.2.

See Problem 14.

#### Problems

######  Problem 14.

Prove 13, i.e., show that 10 can be extended to the case $a=0$, i.e., on
$[0,b]$ with $b>0$.

Mathematics is the tool specially suited for dealing with
abstract concepts of any kind and there is no limit to its
power in this field.
Paul Adrien Maurice Dirac (Dir, 58, p. viii)

###  Chapter 10 Non-Negative Algebraic Polynomials on $[0,\infty)$ and on
$\mathds{R}$

We went a long way to arrive here. But by using 1 and 3 on the interval
$[0,\infty)$ we can now describe all sparse algebraic strictly positive and
non-negative polynomials on $[0,\infty)$ and on $\mathds{R}$.

####  45 Sparse Algebraic Positivstellensatz on $[0,\infty)$

For the sparse algebraic Positivstellensatz on $[a,b]$ (1) we had a lot of
freedom in the exponents $\alpha_{i}$ for $a>0$. We no longer have such a
large range of freedom on $[0,\infty)$. If we now plug 16 into 1 we get the
following.

######  Theorem 1 (Sparse Algebraic Positivstellensatz on $[0,\infty)$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in[0,\infty)$ be real
numbers with $\alpha_{0}=0<\alpha_{1}<\dots<\alpha_{n}$, and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$ on $[0,\infty)$. Then for any
$f=\sum_{i=0}^{n}a_{i}f_{i}\in\mathrm{lin}\,\mathcal{F}$ with $f>0$ on
$[0,\infty)$ and $a_{n}>0$ there exists a unique decomposition

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$ and $f_{*},f^{*}\geq 0$ on
$[0,\infty)$ such that the following hold:

  1. (i)

If $n=2m$ then the polynomials $f_{*}$ and $f^{*}$ each possess $m$ distinct
zeros $\\{x_{i}\\}_{i=1}^{m}$ and $\\{y_{i}\\}_{i=0}^{m-1}$ satisfying

$0=y_{0}<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<\infty.$

The polynomials $f_{*}$ and $f^{*}$ are given by

$f_{*}(x)=c_{*}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}$

and

$\displaystyle f^{*}(x)$
$\displaystyle=c^{*}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}\\\
x&(y_{1}&y_{1})&\dots&(y_{m-1}&y_{m-1})\end{pmatrix}$

for some $c_{*},c^{*}>0$.

  2. (ii)

If $n=2m+1$ then $f_{*}$ and $f^{*}$ have zeros $\\{x_{i}\\}_{i=1}^{m+1}$ and
$\\{y_{i}\\}_{i=1}^{m}$ respectively which satisfy

$0=x_{1}<y_{1}<x_{2}<\dots<y_{m}<x_{m+1}<\infty.$

The polynomials $f_{*}$ and $f^{*}$ are given by

$\displaystyle f_{*}(x)$
$\displaystyle=c_{*}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(x_{2}&x_{2})&\dots&(x_{m+1}&x_{m+1})\end{pmatrix}$

and

$f^{*}(x)=c^{*}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(y_{1}&y_{1})&\dots&(y_{m}&y_{m})\end{pmatrix}$

for some $c_{*},c^{*}>0$.

######  Proof 45.1.

We have that $\mathcal{F}$ fulfills conditions (a) and (b) of 1 and by 15 we
known that $\mathcal{F}$ on $[0,\infty)$ is also a T-system, i.e., (c) in 1 is
fulfilled. We can therefore apply 1.

(i) $n=2m$: By 1 (i) the unique $f_{*}$ and $f^{*}$ each possess $m$ distinct
zeros $\\{x_{i}\\}_{i=1}^{m}$ and $\\{y_{i}\\}_{i=0}^{m-1}$ with $0\leq
y_{0}<x_{1}<\dots<y_{m-1}<x_{m}<\infty$. Since
$x_{1},\dots,x_{m}\in(0,\infty)$ and $\mathcal{F}$ on $[x_{1}/2,\infty)$ is an
ET-system we immediately get the determinantal representation of $f_{*}$ by 2
(combine 1 with 28). For $f^{*}$ we have $y_{0}=0$ and by 16 this is no ET-
system. Hence, we prove the representation of $f^{*}$ by hand, similar as in
the proof of 3.

Let $\varepsilon>0$ be such that
$0=y_{0}<y_{1}<y_{1}+\varepsilon<\dots<y_{m-1}<y_{m-1}+\varepsilon$ holds.
Then

$\displaystyle g_{\varepsilon}(x)$
$\displaystyle=-\varepsilon^{-m+1}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}\\\
x&0&y_{1}&y_{1}+\varepsilon&\dots&y_{m-1}&y_{m-1}+\varepsilon\end{pmatrix}$
$\displaystyle=-\varepsilon^{-m+1}\cdot\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}\\\
1&0&0&\dots&0\\\
1&y_{1}^{\alpha_{1}}&y_{1}^{\alpha_{2}}&\dots&y_{1}^{\alpha_{2m-1}}\\\
\vdots&\vdots&\vdots&&\vdots\\\
1&(y_{m-1}+\varepsilon)^{\alpha_{1}}&(y_{m-1}+\varepsilon)^{\alpha_{2}}&\dots&(y_{m-1}+\varepsilon)^{\alpha_{2m-1}}\end{pmatrix}$
expand by the second row
$\displaystyle=\varepsilon^{-m+1}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}\\\
y_{1}^{\alpha_{1}}&y_{1}^{\alpha_{2}}&\dots&y_{1}^{\alpha_{2m-1}}\\\
\vdots&\vdots&&\vdots\\\
(y_{m-1}+\varepsilon)^{\alpha_{1}}&(y_{m-1}+\varepsilon)^{\alpha_{2}}&\dots&(y_{m-1}+\varepsilon)^{\alpha_{2m-1}}\end{pmatrix}$
$\displaystyle=\varepsilon^{-m+1}\cdot\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}\\\
x&y_{1}&y_{1}+\varepsilon&\dots&y_{m-1}&y_{m-1}+\varepsilon\end{pmatrix}$

is non-negative on $[0,y_{1}]$ and every $[y_{i}+\varepsilon,y_{i+1}]$. Now
$y_{0}=0$ is removed and all $y_{i},y_{i}+\varepsilon>0$. Hence, we can work
on $[y_{1}/2,\infty)$ where $\\{x^{\alpha_{i}}\\}_{i=1}^{2m}$ is an ET-system
and we can go to the limit $\varepsilon\searrow 0$ as in 28. Then 2 proves the
representation of $f^{*}$.

(ii) $n=2m+1$: Similar to the case (i) with $n=2m$.

If all $\alpha_{i}\in\mathds{N}_{0}$ then we can express the $f_{*}$ and
$f^{*}$ in 1 also with Schur polynomials, see (47) in 17.

We now prove a stronger version of (15), i.e., $p=f^{2}+x\cdot g^{2}$ for any
$p\geq 0$ on $[0,\infty)$. It is sufficient to have only the sparse algebraic
Positivstellensatz (1). A previous version already appeared in KS (53).

######  Corollary 2 (see (KS, 66, p. 169, Cor. 8.1)).

Let $p\in\mathds{R}[x]$ with $p\geq 0$ on $[0,\infty)$. Let
$z_{1},\dots,z_{r}\in[0,\infty)$ be the zeros of $p$ in $[0,\infty)$ and let
$m_{1},\dots,m_{r}\in\mathds{N}$ be the corresponding algebraic
multiplicities.

  1. (i)

If $\deg p-m_{1}-\dots-m_{r}=2m$, $m\in\mathds{N}_{0}$, is even then there
exist points $\\{x_{i}\\}_{i=1}^{m}$ and
$\\{y_{i}\\}_{i=1}^{m-1}\subseteq(0,\infty)$ with

$0<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<\infty$

and constants $a,b>0$ such that

$p(x)=\prod_{i=1}^{r}(x-z_{i})^{m_{i}}\cdot\left(a\cdot\prod_{i=1}^{m}(x-x_{i})^{2}+b\cdot
x\cdot\prod_{i=1}^{m-1}(x-y_{i})^{2}\right).$

The constant $a$ is the leading coefficient of $p$.

  2. (ii)

If $\deg p-m_{1}-\dots-m_{r}=2m+1$, $m\in\mathds{N}_{0}$, is odd then there
exist points $\\{x_{i}\\}_{i=1}^{m}$ and
$\\{y_{i}\\}_{i=1}^{m}\subset(0,\infty)$ with

$0<x_{1}<y_{1}<\dots<x_{m}<y_{m}<\infty$

and constants $a,b>0$ such that

$p(x)=\prod_{i=1}^{r}(x-z_{i})^{m_{i}}\cdot\left(a\cdot\prod_{i=1}^{m}(x-x_{i})^{2}+b\cdot
x\cdot\prod_{i=1}^{m}(x-y_{i})^{2}\right).$

The constant $b$ is the leading coefficient of $p$.

######  Proof 45.2.

Since $z_{1},\dots,z_{r}$ are the zeros of $p$ in $[0,\infty)$ with
multiplicities $m_{1},\dots,m_{r}$ we have that
$p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\tilde{p}(x)$ with
$\tilde{p}\in\mathds{R}[x]$ and $\tilde{p}>0$ on $[0,\infty)$. Applying 1 to
$\tilde{p}$ gives the assertion.

Note, in the previous result we were able to factor out the zeros of $p$ and
were only left with $\tilde{p}>0$ on $[0,\infty)$ since we are working in
$\mathds{R}[x]_{\leq\deg p}$ where all monomials $1,x,\dots,x^{\deg p}$ are
present. In sparse systems we are not able to factor out the zeros since we no
longer know which monomials in $\tilde{p}$ will appear.

######  Remark 3.

Working in the sparse setting, i.e., in T-systems, gives us an additional
information. In (15) we only have $p(x)=x\cdot f^{2}+g^{2}$. But this also
includes that $f$ and $g$ might contain factors $((x-y_{i})^{2}+\delta_{i})$
with $\delta_{i}>0$, i.e., a pair of complex conjugated zeros can be present.
In 2 we see that this is not necessary. The polynomials $f$ and $g$ can always
be chosen such that they decompose into linear factors having only real zeros.
A similar results holds on $\mathds{R}$, see 7. $\circ$

####  46 Sparse Stieltjes Moment Problem

In Section 18 we have seen that Boas already investigated the sparse Stieltjes
moment problem Boa39a . However, the description was complicated and is even
incomplete since Boas did not had access to 1 and therefore 1. We get the
following complete and simple description. It fully solves Boa39a . We are not
aware of a reference for the following result.

######  Theorem 4 (Sparse Stieltjes Moment Problem).

Let $\\{\alpha_{i}\\}_{i\in\mathds{N}_{0}}\subseteq[0,\infty)$ be such that
$\alpha_{0}=0<\alpha_{1}<\alpha_{2}<\dots$ and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i\in\mathds{N}_{0}}$. Then the following
are equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a $[0,\infty)$-moment
functional.

  2. (ii)

$L(p)\geq 0$ for all $p\in\mathrm{lin}\,\mathcal{F}$ with $p\geq 0$.

  3. (iii)

$L(p)\geq 0$ for all $p\in\mathrm{lin}\,\mathcal{F}$ with $p>0$.

  4. (iv)

$L(p)\geq 0$ for all

$p(x)=\begin{cases}\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix},\\\
\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m-2}}&x^{\alpha_{2m-1}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})\end{pmatrix},\\\
\det\begin{pmatrix}x^{\alpha_{1}}&x^{\alpha_{2}}&x^{\alpha_{3}}&\dots&x^{\alpha_{2m}}&x^{\alpha_{2m+1}}\\\
x&(x_{2}&x_{2})&\dots&(x_{m+1}&x_{m+1})\end{pmatrix},\ \text{and}\\\
\det\begin{pmatrix}1&x^{\alpha_{1}}&x^{\alpha_{2}}&\dots&x^{\alpha_{2m-1}}&x^{\alpha_{2m}}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\end{cases}$

for all $m\in\mathds{N}_{0}$ and $0<x_{1}<\dots<x_{m}$.

######  Proof 46.1.

The implications “(i) $\Rightarrow$ (ii) $\Leftrightarrow$ (iii)” are clear
and “(iii) $\Leftrightarrow$ (iv)” is 1. It is therefore sufficient to prove
“(ii) $\Rightarrow$ (i)”.

We have
$\mathrm{lin}\,\mathcal{F}=(\mathrm{lin}\,\mathcal{F})_{+}-(\mathrm{lin}\,\mathcal{F})_{+}$,
we have $1=x^{\alpha_{0}}\in\mathrm{lin}\,\mathcal{F}$, and for any
$g=\sum_{i=0}^{m}a_{i}\cdot x^{\alpha_{i}}\in(\mathrm{lin}\,\mathcal{F})_{+}$
we have $\lim_{x\to\infty}\frac{g(x)}{x^{\alpha_{m+1}}}=0$, i.e., there exists
a $f\in(\mathrm{lin}\,\mathcal{F})_{+}$ which dominates $g$. Hence,
$\mathrm{lin}\,\mathcal{F}$ is an adapted space on $[0,\infty)$ and the
assertion follows from the 9.

In the previous result we did needed
$0=\alpha_{0}<\alpha_{1}<\alpha_{2}<\dots$. We did not needed
$\alpha_{i}\to\infty$. Hence, 4 also includes the case
$\sup_{i\in\mathds{N}_{0}}\alpha_{i}<\infty$.

4 also holds with $\alpha_{0}>0$ since we can factor out $x^{\alpha_{0}}$ and
therefore determine $x^{\alpha_{0}}~{}\mathrm{d}\mu(x)$ instead of
$\mathrm{d}\mu(x)$.

####  47 Sparse Algebraic Nichtnegativstellensatz on $[0,\infty)$

For $\\{1,x,x^{3}\\}$ we have seen in 16 that this is not an ET-systen on
$[0,\infty)$, or on any other $[0,b]$. If we remove the point $x=0$ and work
on $(0,\infty)$ then it is an ET-system and even an ECT-system (18). For a
Nichtnegativstellensatz we therefore have to exclude zeros at $x=0$ in a
sparse polynomial $p\geq 0$.

######  Theorem 5 (Sparse Algebraic Nichtnegativstellensatz on $[0,\infty)$).

Let $n\in\mathds{N}_{0}$, $\alpha_{0},\dots,\alpha_{n}\in[0,\infty)$ be real
numbers with $\alpha_{0}=0<\alpha_{1}<\dots<\alpha_{n}$, and let
$\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$. Let
$f=\sum_{i=0}^{n}a_{i}x^{\alpha_{i}}\geq 0$ on $[0,\infty)$ with $a_{n}>0$ and
$f(0)=a_{0}>0$. Then there exist points
$x_{1},\dots,x_{n},y_{1},\dots,y_{n-1}\in[0,\infty)$ (not necessarily
distinct) which include the zeros of $f$ with multiplicities and there exist
constants $c_{*},c^{*}\in\mathds{R}$ such that

$f=f_{*}+f^{*}$

with $f_{*},f^{*}\in\mathrm{lin}\,\mathcal{F}$, $f_{*},f^{*}\geq 0$ on
$[0,\infty)$, and the polynomials $f_{*}$ and $f^{*}$ are given by

$f_{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|ccc}1&\
x^{\alpha_{1}}&\dots&x^{\alpha_{n}}\\\ x&\
x_{1}&\dots&x_{n}\end{array}\right)\qquad\text{and}\qquad
f^{*}(x)=c_{*}\cdot\det\left(\begin{array}[]{c|ccc}1&\
x^{\alpha_{1}}&\dots&x^{\alpha_{n-1}}\\\ x&\
y_{1}&\dots&y_{n-1}\end{array}\right)$

for all $x\in[0,\infty)$.

######  Proof 47.1.

See Problem 9.

######  Remark 6.

Note, if $f(0)=a_{0}=0$ in 5 then

$f(x)=a_{i}x^{\alpha_{i}}+a_{i+1}x^{\alpha_{i+1}}+\dots+a_{n}x^{\alpha_{n}}=x^{\alpha_{i}}\cdot(\underbrace{a_{i}+a_{i+1}x^{\alpha_{i+1}-\alpha_{i}}+\dots+a_{n}x^{\alpha_{n}-\alpha_{i}}}_{=:\tilde{f}(x)})$

where $a_{i}$ is the first non-zero coefficient and it fulfills $a_{i}>0$
since $f\geq 0$. Then apply 5 to $\tilde{f}$ to get
$\tilde{f}=\tilde{f}_{*}+\tilde{f}^{*}$ and hence
$f=x^{\alpha_{i}}\cdot(\tilde{f}_{*}+\tilde{f}^{*})$. $\circ$

####  48 Algebraic Positiv- and Nichtnegativstellensatz on $\mathds{R}$

Since we treat $\mathcal{F}=\\{x^{i}\\}_{i=0}^{n}$ we need only 4 on
$\mathds{R}$ but not 5 on $\mathds{R}$ as we will see in the next result.

######  Theorem 7 (Algebraic Positiv- and Nichtnegativstellensatz on
$\mathds{R}$, see KS (53) or e.g. (KS, 66, p. 198, Cor. 8.1)).

Let $p\in\mathds{R}[x]$ with $p\geq 0$ on $\mathds{R}$ and let
$z_{1},\dots,z_{r}\in\mathds{R}$ be the zeros of $p$ with algebraic
multiplicities $m_{1},\dots,m_{r}\in 2\mathds{N}$. Then there exist pairwise
distinct points
$\\{x_{i}\\}_{i=1}^{m},\\{y_{i}\\}_{i=1}^{m-1}\subseteq\mathds{R}$ with
$2m=\deg p-m_{1}-\dots-m_{r}$ and

$-\infty<x_{1}<y_{1}<\dots<y_{m-1}<x_{m}<\infty$

as well as constants $a,b>0$ such that

$p(x)=\prod_{i=1}^{r}(x-z_{i})^{m_{i}}\cdot\left(a\cdot\prod_{i=1}^{m}(x-x_{i})^{2}+b\cdot\prod_{i=1}^{m-1}(x-y_{i})^{2}\right).$
(64)

The constant $a$ is the leading coefficient of $p$.

######  Proof 48.1.

We have $p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\tilde{p}(x)$ for
some $\tilde{p}\in\mathds{R}[x]$ with $\tilde{p}>0$ on $\mathds{R}$. Applying
4 to $\tilde{p}$ gives the assertion.

Like in the case on $[0,\infty)$ in 2 a factorization

$p(x)=(x-z_{1})^{m_{1}}\cdots(x-z_{r})^{m_{r}}\cdot\tilde{p}(x)$

is not possible in T-systems or sparse algebraic systems on $\mathds{R}$. But
since we are working in $\mathds{R}[x]_{\leq\deg p}$ all monomials
$1,x,\dots,x^{\deg p}$ are present.

######  Remark 8.

Similar to 3 we see that 7 gives a stronger version of (13), i.e.,
$p=f^{2}+g^{2}$. By applying only the Fundamental Theorem of Algebra $f$ and
$g$ might contain pairs of complex conjugated zeros, see e.g. (Mar, 08, Prop.
1.2.1). But by working in the T-system framework of 4 on $\mathds{R}$ we see
that $f$ and $g$ can be chosen to have only real zeros. $\circ$

#### Problems

######  Problem 9.

Use 3 to prove 5.

######  Problem 10.

Show that $a$ in (64) in 7 is the leading coefficient of $p$.

##  Part V Applications of T-Systems

Long is the way and hard, that out of Hell leads up to light.
John Milton: Paradise Lost

###  Chapter 11 Moment Problems for continuous T-Systems on $[a,b]$

In this chapter we demonstrate how e.g. 3 for general T-systems on $[a,b]$ can
be used to prove moment problems which do not live on the algebraic
polynomials $\mathds{R}[x]$.

####  49 General Moment Problems for continuous T-Systems on $[a,b]$

For T-system $\mathcal{F}$ on $[a,b]$ 3 describes all polynomials
$f\in\mathrm{lin}\,\mathcal{F}$ with $f>0$.

######  Theorem 1.

Let $n\in\mathds{N}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be a continuous
T-system on $[a,b]$ with $a<b$. The following are equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is an $[a,b]$-moment functional.

  2. (ii)

$L(f)\geq 0$ for all $f\in\mathrm{lin}\,\mathcal{F}$ such that

    1. (a)

$f\geq 0$ on $[a,b]$ and

    2. (b)

the zero set of $f$ has index $n$.

######  Proof 49.1.

The implication (i) $\Rightarrow$ (ii) is clear since $f\geq 0$. It is
therefore sufficient to prove (ii) $\Rightarrow$ (i).

Since $\mathcal{F}$ is a continuous T-system there exists a polynomial
$e\in\mathrm{lin}\,\mathcal{F}$ with $e>0$ on $[a,b]$. Since $[a,b]$ is
compact, $\mathcal{F}$ is continuous and finite dimensional, and there exists
a $e>0$ we have that the moment cone $((\mathrm{lin}\,\mathcal{F})_{+})^{*}$
is closed. Therefore, to show that $L$ is a moment functional it is sufficient
to show that $L(f)\geq 0$ for all $f\in(\mathrm{lin}\,\mathcal{F})_{+}$.

By 3 there are $e_{*},e^{*}\in\mathrm{lin}\,\mathcal{F}$ with $e_{*},e^{*}\geq
0$ and the zero sets of $e_{*}$ and of $e^{*}$ have index $n$. Hence,
$L(e)=L(e_{*})+L(e^{*})\geq 0$.

Let $f\in(\mathrm{lin}\,\mathcal{F})_{+}$ and $\varepsilon>0$. Then
$f_{\varepsilon}=f+\varepsilon\cdot e>0$ on $[a,b]$, i.e., by 3 there exist
$(f_{\varepsilon})_{*},(f_{\varepsilon})^{*}\in(\mathrm{lin}\,\mathcal{F})_{+}$
each with zero sets of index $n$. Assumption (ii) then implies
$L(f+\varepsilon\cdot e)=L((f_{\varepsilon})_{*})+L((f_{\varepsilon})^{*})\geq
0$ for all $\varepsilon>0$, i.e., $L(f)\geq 0$. That proves the assertion.

Note, that a continuous T-system on $[a,b]$ is always an adapted space.
Additionally, the use of 9 is not necessary since we only need to check in
this case $L\in((\mathrm{lin}\,\mathcal{F})_{+})^{*}$ since the moment cone is
$((\mathrm{lin}\,\mathcal{F})_{+})^{*}$ and hence it is closed.

If in the previous theorem we additionally have that $\mathcal{F}$ is an ET-
system then we can write down $f_{*}$ and $f^{*}$ explicitly in the similar
way as in 6.

######  Theorem 2.

Let $n\in\mathds{N}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-system
on $[a,b]$ with $a<b$. The following are equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a moment functional.

  2. (ii)

$L(f)\geq 0$ holds for all

$\displaystyle f(x)$
$\displaystyle:=\begin{cases}\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
-\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m-2}&f_{2m-1}&f_{2m}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})&b\end{pmatrix}\end{cases}$ if $n=2m$
and $\displaystyle f(x)$
$\displaystyle:=\begin{cases}-\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m}&f_{2m+1}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}&f_{2m+1}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})&b\end{pmatrix}\end{cases}$ if $n=2m+1$

and all $x_{1},\dots,x_{m}$ with $a<x_{1}<\dots<x_{m}<b$.

######  Proof 49.2.

Follows from 1 with 3.

####  50 A Non-Polynomial Example

In 18 we have seen that

$\mathcal{F}=\left\\{\frac{1}{x+\alpha_{0}},\frac{1}{x+\alpha_{1}},\dots,\frac{1}{x+\alpha_{n}}\right\\}$

with $n\in\mathds{N}$ and $\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$ reals is a
continuous T-system on any $[a,b]$ with $-\alpha_{0}<a<b$, see Problem 36 for
the proof. But in the proof of 18 we actually showed that this $\mathcal{F}$
is an ET-system since we multiplied with $(x+\alpha_{0})\cdots(x+\alpha_{n})$
which has no zeros on $[a,b]$ and hence the multiplicities of the zeros do not
change. Multiplicity restriction from the fundamental theorem of algebra then
shows that $\mathcal{F}$ is an ET-system.

######  Corollary 3.

Let $n\in\mathds{N}$ and $\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$ be reals.
Then

$\mathcal{F}=\left\\{\frac{1}{x+\alpha_{0}},\frac{1}{x+\alpha_{1}},\dots,\frac{1}{x+\alpha_{n}}\right\\}$

is an ET-system on any $[a,b]$ with $-\alpha_{0}<a<b$.

From 2 and 3 we therefore get the following.

######  Corollary 4.

Let $n\in\mathds{N}$, let $\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$ be reals,
and let

$\mathcal{F}=\left\\{f_{0}(x)=\frac{1}{x+\alpha_{0}},f_{1}(x)=\frac{1}{x+\alpha_{1}},\dots,f_{n}(x)=\frac{1}{x+\alpha_{n}}\right\\}$

on $[a,b]$ with $-\alpha_{0}<a<b$. Then the following are equivalent:

  1. (i)

$L:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ is a $[a,b]$-moment functional.

  2. (ii)

$L(f)\geq 0$ holds for all

$\displaystyle f(x)$
$\displaystyle:=\begin{cases}\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
-\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m-2}&f_{2m-1}&f_{2m}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m-1}&x_{m-1})&b\end{pmatrix}\end{cases}$ if $n=2m$
and $\displaystyle f(x)$
$\displaystyle:=\begin{cases}-\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&f_{3}&\dots&f_{2m}&f_{2m+1}\\\
x&a&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})\end{pmatrix}\\\
\det\begin{pmatrix}f_{0}&f_{1}&f_{2}&\dots&f_{2m-1}&f_{2m}&f_{2m+1}\\\
x&(x_{1}&x_{1})&\dots&(x_{m}&x_{m})&b\end{pmatrix}\end{cases}$ if $n=2m+1$

and all $x_{1},\dots,x_{m}$ with $a<x_{1}<\dots<x_{m}<b$.

In a similar way many other T-system moment problems can be proven from 1.

The rest is silence.
William Shakespeare: Hamlet (Act 5, Scene 2)

###  Chapter 12 Polynomials of Best Approximation and Optimization over Linear
Functionals

This last chapter is devoted to best approximation polynomials and
optimization over linear functionals.

We started in Chapter 1 with moments and moment functionals, went to the
theory of T-systems in Part II, proved Karlin’s Theorems in Part III, and
applied them to algebraic polynomials in Part IV. Now we finish our lecture by
closing the circle. We apply the previous results to best approximation in
Section 51 and to optimization over linear (moment) functionals in Section 52.

####  51 Polynomials of Best Approximation

A classical question is:

> How to approximate a given function $f\in\mathcal{C}([a,b],\mathds{R})$ in
> the $\sup$-norm by a finite linear combination $\sum_{i=0}^{n}a_{i}f_{i}$ of
> some given $f_{0},\dots,f_{n}\in\mathcal{C}([a,b],\mathds{R})$?

######  Definition 1.

Let $n\in\mathds{N}_{0}$, let
$f,f_{0},\dots,f_{n}\in\mathcal{C}([a,b],\mathds{R})$, and let
$\mathcal{F}:=\\{f_{i}\\}_{i=0}^{n}$. The polynomial
$\underline{f}\in\mathrm{lin}\,\mathcal{F}$ which solves

$\min_{a_{0},\dots,a_{n}}\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}$
(65)

is called the _polynomial of best approximation_.

Approximations (65) with the $\sup$-norm are called _Tchebycheff
approximations_.

The connection between polynomials of best approximation and T-systems is
revealed in the following result.

######  Theorem 2 (see Haa (18), Ber (26); or e.g. (Ach, 56, p. 74, §48), (KS,
66, p. 280, Thm. 1.1)).

Let $n\in\mathds{N}_{0}$, let $a,b\in\mathds{R}$ with $a<b$, and let
$\mathcal{F}:=\\{f_{i}\\}_{i=0}^{n}\subseteq\mathcal{C}([a,b],\mathds{R})$ be
a family of continuous functions. The following hold:

  1. (i)

The following are equivalent:

    1. (a)

The polynomial minimizing

$\min_{a_{0},\dots,a_{n}}\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}$
(66)

is uniquely determined for every $f\in\mathcal{C}([a,b],\mathds{R})$.

    2. (b)

The family $\mathcal{F}$ is a continuous T-system on $[a,b]$.

  2. (ii)

If $\mathcal{F}$ is a T-system then for each
$f\in\mathcal{C}([a,b],\mathds{R})$ the unique polynomial

$\underline{f}=\sum_{i=0}^{n}\underline{a}_{i}f_{i}$

minimizing (66) is characterized by the property that there exist $n+2$ points

$a\leq\quad x_{1}<x_{2}<\dots<x_{n+2}\quad\leq b$

such that

$(-1)^{i}\cdot\delta\cdot(f(x_{i})-\underline{f}(x_{i}))=\max_{a\leq x\leq
b}\left|\,f(x)-\underline{f}(x)\,\right|$

holds for all $i=1,2,\dots,n+2$ with $\delta=+1$ or $-1$.

Statement (i) of the previous theorem is essentially due to A. Haar Haa (18).
The following proof significantly differs from Haar’s proof and it is more
general. It is taken from (KS, 66, pp. 284–286), see also (Ach, 56, pp.
75–76).

######  Proof 51.1.

(a) $\Rightarrow$ (b): We prove $\neg$(b) $\Rightarrow\neg$(a).

Assume $\mathcal{F}$ is not a T-system. There exist $n+1$ distinct points
$a\leq x_{0}<x_{1}<\dots<x_{n}\leq b$ such that

$\det\left(f_{i}(x_{j})\right)_{i,j=0}^{n}=0.$ (67)

Hence, there exist real coefficients $c_{0},c_{1},\dots,c_{n}$ with
$\sum_{i=0}^{n}c_{i}^{2}>0$ with $\sum_{i=0}^{n}c_{i}f_{j}(x_{i})=0$ for all
$j=0,\dots,n$. That implies

$\sum_{i=0}^{n}c_{i}p(x_{i})=0$ (68)

for all $p\in\mathrm{lin}\,\mathcal{F}$.

The relation (67) also implies the existence of a non-trivial polynomial
$\tilde{p}=\sum_{i=0}^{n}b_{i}f_{i}\in\mathrm{lin}\,\mathcal{F}$ which
vanishes at the points $x_{0},x_{1},\dots,x_{n}$.

Let $g\in\mathcal{C}([a,b],\mathds{R})$ be such that $\|g\|_{\infty}\leq 1$
and

$g(x_{i})=\frac{c_{i}}{|c_{i}|}$

for all $i=0,1,\dots,n$ with $c_{i}\neq 0$.

Let $\lambda>0$ be such that $\|\lambda\tilde{p}\|_{\infty}<1$ then
$f:=g\cdot(1-|\lambda\tilde{p}|)$ has the same signs at the points $x_{i}$
with $c_{i}\neq 0$ as $g$.

We will now construct an infinite number of polynomials of the same minimum
deviation from $f$.

If

$\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}<1$

for some $a_{0},a_{1},\dots,a_{n}$ then

$-1<g(x_{j})\cdot(1-|\lambda\tilde{p}(x_{j})|)-\sum_{i=0}^{n}a_{i}f_{i}(x_{j})<1$

for all $j=0,1,\dots,n$ which reduces to

$-1<g(x_{j})-\sum_{i=0}^{n}a_{i}f_{i}(x_{j})<1$

for all $j=0,1,\dots,n$. Hence, if $c_{j}\neq 0$ the value of
$\sum_{i=0}^{n}a_{i}f_{i}(x_{j})$ has the sign of the $c_{j}$ so that
$\sum_{j=0}^{n}c_{j}\sum_{i=0}^{n}a_{i}f_{i}(x_{j})\neq 0$ which contradicts
(68). Therefore,

$\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}\geq 1.$

If now $|\delta|<1$ then

$\displaystyle|f(x)-\delta\lambda\tilde{p}(x)|$
$\displaystyle\leq|f(x)|+|\delta\lambda\tilde{p}(x)|$
$\displaystyle\leq|g(x)|\cdot(1-|\lambda\tilde{p}(x))+|\delta\lambda\tilde{p}(x)|$
$\displaystyle\leq 1-(1-|\delta|)\cdot|\lambda\tilde{p}(x)|$
$\displaystyle\leq 1$

so that $\delta\lambda\tilde{p}$ minimizes the distance to $f$ independent of
$\delta\in(-1,1)$. Hence, we proved $\neg$(a).

We now prove (ii) which will also establish (b) $\Rightarrow$ (a). Let
$\mathcal{F}$ be a T-system. At least one minimal polynomial exists since
$\mathrm{lin}\,\mathcal{F}$ is finite dimensional. Assume
$g=\sum_{i=0}^{n}b_{i}f_{i}$ fulfills

$\|f-g\|_{\infty}=m=\min_{a_{0},\dots,a_{n}}\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}$

and $f-g$ takes on the values $\pm m$ alternatively at only $k\leq n+1$
points. We suppose for definiteness that $f-g$ assumes the values $+m$ before
it takes the value $-m$. In this case there exist $k-1$ points

$a\leq\quad y_{1}<\dots<y_{k-1}\quad\leq b$

such that

$f(y_{i})-g(y_{i})=0$

for all $i=1,2,\dots,k-1$ and for some $d>0$ we have

$\displaystyle m$ $\displaystyle\;\geq\;f-g\;\geq\;-m+d$
$\displaystyle\text{on}\ [a,y_{1}]\cup[y_{2},y_{3}]\cup\dots$ $\displaystyle
m-d$ $\displaystyle\;\geq\;f-g\;\geq\;-m$ $\displaystyle\text{on}\
[y_{1},y_{2}]\cup[y_{3},y_{4}]\cup.$

By 30 and 27 there exists a polynomial $h$ whose _only_ zeros on the _open_
interval $(a,b)$ are the nodal zeros $y_{1},\dots,y_{k-1}$ and additionally
$h\leq 0$ on $[a,y_{1}]$. Let $\delta>0$ be such that $|\delta h|\leq d/2$
then

$|f-g+\delta h|<m$ (69)

on $(a,b)$.

Equality in (69) is possible at the end point $a$ only if $f(a)-g(a)=m$ and
$h(a)=0$ and at $b$ only if $|f(a)-g(b)|=m$ and $h(b)=0$. To repair the
situation at the points $a$ and $b$ let $\tilde{h}$ be such that
$\tilde{h}\cdot(f-g)>0$ at $a$ and $b$. Then for sufficient small $\eta$ we
have

$|f-g+\delta h-\eta\tilde{h}|<m$

on $[a,b]$. Hence, by continuity on the compact interval $[a,b]$ we have

$\min_{a_{0},\dots,a_{n}}\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}<m$

contradicting the fact that $m$ is the minimum deviation. That proves (ii)
including uniqueness in (i).

In the previous theorem we have seen the close connection between the best
approximation polynomials from the minimum problem (66) and T-systems. The
next result shows that the connection is even closer, i.e., the solution of
(66) is connected to the 4.

######  Theorem 3 (see e.g. (KS, 66, p. 283, Thm. 2.1)).

Let $n\in\mathds{N}_{0}$ and let
$f_{0},\dots,f_{n},f\in\mathcal{C}([a,b],\mathds{R})$ be such that
$\\{f_{0},\dots,f_{n}\\}$ and $\\{f_{0},\dots,f_{n},f\\}$ are continuous
T-systems on $[a,b]$ with $a<b$. Let

$f^{*}=c\cdot f+\sum_{i=0}^{n}c_{i}\cdot f_{i}$

be the $f^{*}$ from the 4 with $g_{1}=-1$ and $g_{2}=1$, i.e., $f^{*}$ is
uniquely characterized by the following conditions:

  1. (a)

$-1\leq f^{*}\leq 1$ on $[a,b]$, and

  2. (b)

there exist $n+2$ points $x_{1}<x_{2}<\dots<x_{n+2}$ in $[a,b]$ such that

$f^{*}(x_{i})=(-1)^{n+1-i}$

for all $i=1,\dots,n+2$.

Then $c\neq 0$ and the polynomial

$\underline{f}:=-\frac{1}{c}\cdot\sum_{i=0}^{n}c_{i}f_{i}$

is the unique minimizer of

$d=\min_{a_{0},\dots,a_{n}}\left\|\;f-\sum_{i=0}^{n}a_{i}f_{i}\;\right\|_{\infty}$

and the minimum deviation is $d=|c|^{-1}$.

The proof is taken from (KS, 66, pp. 283–284).

######  Proof 51.2.

The coefficient $c$ can not be zero. Otherwise the polynomial
$\sum_{i=0}^{n}c_{i}f_{i}$ vanishes at $n+1$ points in the T-system
$\\{f_{0},\dots,f_{n}\\}$ by (b) and would therefore be equal to zero by 5.

From (a) we get

$\left\|\;f-\left(-\frac{1}{d}\sum_{i=0}^{n}c_{i}f_{i}\right)\;\right\|_{\infty}\leq\frac{1}{|d|}.$

Since $\underline{f}$ fulfills (b) we get from 2 (ii) uniqueness of
$\underline{f}$ and $d=|c|^{-1}$.

Finding approximations is also done with respect to the
$\mathcal{L}^{p}$-norms

$\min_{a_{0},\dots,a_{n}}\int\left|\;f(x)-\sum_{i=0}^{n}a_{i}f_{i}(x)\;\right|^{p}~{}\mathrm{d}\mu(x)$
(70)

with a fixed measure $\mu$ and $p\geq 1$. For $p=2$ this leads to the well-
studied _orthogonal polynomials_ , a special branch of the theory of moments.

For $p=1$ in (70) this also is connected to T-systems. D. Jackson Jac (24)
showed that if $\mathcal{F}=\\{f_{0},\dots,f_{n}\\}$ is a T-system then the
best approximation of (70) is unique, see also (Ach, 56, p. 77).

####  52 Optimization over Linear Functionals

In optimization one often encounters the problem of having only a linear
functional $L:\mathcal{V}\to\mathds{R}$, e.g. a moment functional, and one
wants to minimize $L(f)$ over $\mathcal{V}_{+}$. By removing the dependency on
the scaling of $f$ we get the following result.

######  Theorem 4 (see e.g. (KS, 66, p. 312, Thm. 9.1)).

Let $n\in\mathds{N}_{0}$, let $\mathcal{F}=\\{f_{i}\\}_{i=0}^{n}$ be an ET-
system on $[a,b]$ with $a<b$, and let
$L,S:\mathrm{lin}\,\mathcal{F}\to\mathds{R}$ be two linear functionals such
that $S$ is strictly positive on $(\mathrm{lin}\,\mathcal{F})_{+}$, i.e.,
$S(f)>0$ for all $f\in\mathrm{lin}\,\mathcal{F}\setminus\\{0\\}$ with $f\geq
0$. Then

$\min_{f\in(\mathrm{lin}\,\mathcal{F})_{+}\setminus\\{0\\}}\frac{L(f)}{S(f)}\qquad\text{and}\qquad\max_{f\in(\mathrm{lin}\,\mathcal{F})_{+}\setminus\\{0\\}}\frac{L(f)}{S(f)}$
(71)

are attained at non-negative polynomials possessing $n$ zeros counting
multiplicities.

The proof is taken from (KS, 66, p. 312).

######  Proof 52.1.

Since $\mathrm{lin}\,\mathcal{F}$ is finite dimensional the values in (71) are
attained.

It is sufficient to prove the statement for the maximum. But maximizing
$\frac{L(f)}{S(f)}$ over $(\mathrm{lin}\,\mathcal{F})_{+}\setminus\\{0\\}$ is
equivalent to maximize $L(f)$ over
$f\in(\mathrm{lin}\,\mathcal{F})_{+}\setminus\\{0\\}$ with $S(f)=1$.

Let $f\geq 0$ be such that $S(f)=1$ and suppose $f$ has at most $n-1$ zeros
counting multiplicities. Then by 6 there is a unique decomposition
$f=f_{*}+f^{*}$ where $f_{*}$ and $f^{*}$ differ, are non-negative, and both
have $n$ zeros counting multiplicities. Set $\alpha:=S(f_{*})$ and
$\beta:=S(f^{*})$. Then $\alpha,\beta>0$ since $S$ is strictly positive and
$\alpha+\beta=S(f_{*})+S(f^{*})=S(f)=1$. Then

$f=\alpha\cdot\frac{f_{*}}{\alpha}+\beta\cdot\frac{f^{*}}{\beta}$

and by linearity

$L(f)\leq\max\left(\frac{L(f_{*})}{\alpha},\frac{L(f^{*})}{\beta}\right)$

which proves the statement.

More results on best approximation and optimization over linear functionals
can already be found in Ber (26), Ach (56), and KS (66). Let alone the
enormous literature after that.

## Appendices

Solutions

### Problems of Chapter 1

prob:determinacy The 0.3 states that for a compact set
$K\subset\mathds{R}^{n}$ the polynomials $\mathds{R}[x_{1},\dots,x_{n}]$ are
dense in $\mathcal{C}(K,\mathds{R})$ with respect to the $\sup$-norm. Let
$A\in\mathfrak{B}(K)$ be a Borel measurable set, let $\varepsilon>0$, and let
$\mu_{1}$ and $\mu_{2}$ be two representing measures of $L$. Set
$A_{\delta}:=(A+B_{\delta}(0))\cap K$ for all $\delta>0$. Then for any
$\varepsilon>0$ there exists a $\delta=\delta(\varepsilon)>0$ such that