$\mu_{1}(A_{\delta}\setminus A),\mu_{2}(A_{\delta}\setminus A)<\varepsilon$.

By 0.2 there exists a $\varphi_{\varepsilon}\in\mathcal{C}(K,[0,1])$ such that

$\varphi_{\varepsilon}(x)=\begin{cases}1&\text{for}\ x\in A\\\ 0&\text{for}\
x\in K\setminus A_{\varepsilon}\end{cases}$

and since $\mathds{R}[x_{1},\dots,x_{n}]$ is dense in
$\mathcal{C}(K,\mathds{R})$ there exists a family of polynomials
$(p_{i}^{\varepsilon})_{i\in\mathds{N}}\subseteq\mathds{R}[x_{1},\dots,x_{n}]$
such that

$\|p_{i}^{\varepsilon}-\varphi_{\varepsilon}\|_{\infty}\xrightarrow{i\to\infty}0\quad\text{and
hence}\quad\int_{K}p_{i}^{\varepsilon}(x)~{}\mathrm{d}\mu_{j}(x)\xrightarrow{i\to\infty}\int_{K}\varphi_{\varepsilon}(x)~{}\mathrm{d}\mu_{j}(x)$

for $j=1,2$. Then we have

$\displaystyle\mu_{1}(A)$ $\displaystyle=\lim_{\varepsilon\searrow
0}\int_{K}\varphi_{\varepsilon}(x)~{}\mathrm{d}\mu_{1}(x)$
$\displaystyle=\lim_{\varepsilon\searrow
0}\lim_{i\to\infty}\int_{K}p_{i}^{\varepsilon}(x)~{}\mathrm{d}\mu_{1}(x)$
$\displaystyle=\lim_{\varepsilon\searrow
0}\lim_{i\to\infty}L(p_{i}^{\varepsilon})$
$\displaystyle=\lim_{\varepsilon\searrow
0}\lim_{i\to\infty}\int_{K}p_{i}^{\varepsilon}(x)~{}\mathrm{d}\mu_{2}(x)$
$\displaystyle=\lim_{\varepsilon\searrow
0}\int_{K}\varphi_{\varepsilon}(x)~{}\mathrm{d}\mu_{2}(x)\qquad=\mu_{2}(A).$

Since $A\in\mathfrak{B}(K)$ was arbitrary we have $\mu_{1}=\mu_{2}$, i.e., $L$
has a unique representing measure and is therefore determinate.

prob:convexML Proof of 1.3
Let $\mu_{1},\mu_{2}\in\mathcal{M}(L)$ and $\lambda\in[0,1]$. Then

$\displaystyle\int p(x)~{}\mathrm{d}[\lambda\mu_{1}+(1-\lambda)\mu_{2}](x)$
$\displaystyle=\lambda\int p(x)~{}\mathrm{d}\mu_{1}(x)+(1-\lambda)\int
p(x)~{}\mathrm{d}\mu_{2}(x)$ $\displaystyle=\lambda L(p)+(1-\lambda)L(p)$
$\displaystyle=L(p)$

and hence $\lambda\mu_{1}+(1-\lambda)\mu_{2}\in\mathcal{M}(L)$ which proves
convexity.

prob:indeter Proof of 1.10
Let $\mu_{0},\mu_{1}\in\mathcal{M}(L)$ with $\mu_{0}\neq\mu_{1}$, i.e., there
exists a $A\in\mathfrak{A}$ such that $\mu_{0}(A)\neq\mu_{1}(A)$ and without
loss of generality we have $\mu_{0}(A)<\mu_{1}(A)$. Hence, for all
$\lambda\in[0,1]$ we set $\mu_{\lambda}:=\lambda\mu_{1}+(1-\lambda)\mu_{0}$
and we have

$\mu_{\lambda_{0}}(A)<\mu_{\lambda_{1}}(A)$

for all $0\leq\lambda_{0}<\lambda_{1}\leq 1$ which proves that
$\mu_{\lambda_{0}}\neq\mu_{\lambda_{1}}$ for all $\lambda_{0}\neq\lambda_{1}$.
Hence, we have at least $|[0,1]|=|\mathds{R}|$ many representing measures for
$L$.

### Problems of Chapter 2

prob:linearCone Proof of 1
The proof is taken from (Cho, 69, Vol. 2, p. 268).

(i) $\Rightarrow$ (ii): If $F+C$ is a vector space then $-(F+C)=(F+C)$ and
$-(F+C)=F-C$ since $-F=F$.

(ii) $\Rightarrow$ (iii): If $x\in F+C$, i.e., $x=y^{\prime}+z$ for some
$y^{\prime}\in F$ and $z\in C$, then $x\geq y^{\prime}$. Similarly, if $x=y-w$
then $y\geq x$.

(iii) $\Rightarrow$ (i): First note that $F+C$ is a convex cone. So if
suffices to show that $F+C=-(F+C)$, i.e., $F+C=F-C$. But if $x\in F+C$ and
$x\leq y$ then $x=y-z$ for some $z\in C$, or $x\in F-C$. Similarly, if $x\in
F-C$ and $x=y^{\prime}+w$ for some $w\in C$ then $x\in F+C$.

prob:adaptedLem Proof of 6

(i) $\Rightarrow$ (ii): Set $K_{\varepsilon}=\mathrm{supp}\,h_{\varepsilon}$.

(ii) $\Rightarrow$ (iii): Chose by 0.2 a
$\eta_{\varepsilon}\in\mathcal{C}_{c}(\mathcal{X},\mathds{R})$ with
$\eta_{\varepsilon}|_{K_{\varepsilon}}=1$.

(iii) $\Rightarrow$ (i): Take $h_{\varepsilon}=\eta_{\varepsilon}\cdot
g\in\mathcal{C}_{c}(\mathcal{X},\mathds{R})$.

prob:compactAdapted Since $\mathcal{X}$ is compact for every $f\in E$ we have
$m_{f}:=\min_{x\in\mathcal{X}}f(x)>-\infty$ and
$M_{f}:=\max_{x\in\mathcal{X}}f(x)<\infty$, especially for $f=e>0$ we have
$m_{e}>0$. Then for every $f$ there exists a $d_{f}>0$ such that
$f=(f+d_{f}e)-d_{f}e$ such that $f+d_{f}e,d_{f}e\in E_{+}$ and hence
$E=E_{+}-E_{+}$ proving (i) in 7.

Since $e>0$ we also have (ii) in 7.

For (iii) in 7 it is sufficient to note that $\mathcal{X}$ is compact, i.e.,
for every $g$ there is a $c_{g}>0$ such that $g\leq c_{g}e$.

prob:adaptedPolynomials Let $E=\mathds{R}[x_{1},\dots,x_{n}]$ on
$\mathcal{X}$. Then (i) $E=E_{+}-E_{+}$ follows immediately from the fact that
for every $f\in E$ there is a $g\in E_{+}$ such that $f=f+g-g$ with $f+g\in
E_{+}$.

For (ii) we take $f=1>0$ on $\mathcal{X}$.

For (iii) take the $g$ from (i).

prob:adaptedPolynomials2 Since $E$ is finite dimensional we can equip it with
a norm, e.g. the $l^{2}$-norm in the coefficients of $f$. Assume $\mathcal{X}$
is not compact then there exists an unbounded sequence
$(x_{i})_{i\in\mathds{N}_{0}}$ and a $f\in E$ with $\|f\|\leq 1$ such that
$(f(x_{i}))_{i\in\mathds{N}_{0}}$ grows faster than any other
$(g(x_{i}))_{i\in\mathds{N}_{0}}$. Hence, $f$ can not be dominated by any $g$.

prob:adaptedCompact Proof of 8
Since $K=\mathrm{supp}\,g$ is compact and $E$ is an adapted space, i.e., there
exists a $f\in E_{+}$ with $f>0$ we have that $\min_{x\in K}f(x)>0$ and hence
there exists a $c>0$ such that $cf>g$ on $K$ and hence on all $\mathcal{X}$.

### Problems of Chapter 3

prob:stieltjes Proof of 1
We have (iii) $\Leftrightarrow$ (iv) $\Leftrightarrow$ (v) by the definition
of the Hankel matrix and also (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii).
Additionally, we have (iii) $\Rightarrow$ (ii) by (15) since
$L(p)=L(f^{2})+L(xg^{2})\geq 0$. At last (ii) $\Rightarrow$ (i) holds by the 9
since $\mathds{R}[x]$ on $[0,\infty)$ is an adapted space.

prob:hamburger Proof of 2
We have (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) and additionally (iii)
$\Leftrightarrow$ (iv) $\Leftrightarrow$ (v) by the definition of the Hankel
matrix. The implication (iii) $\Rightarrow$ (ii) follows from Equation 13 by
$L(p)=L(f^{2}+g^{2})\geq 0$. At last (ii) $\Rightarrow$ (i) holds by the 9
since $\mathds{R}[x]$ on $\mathds{R}$ is an adapted space.

prob:hausdorff Proof of 3
We have (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) and additionally (iii)
$\Leftrightarrow$ (iv) $\Leftrightarrow$ (v) by the definition of the Hankel
matrix. The implication (iii) $\Rightarrow$ (ii) follows from (20) since it is
sufficient to look only at $f(x)^{2}+xg(x)^{2}+(1-x)h(x)^{2}$. At last (ii)
$\Rightarrow$ (i) holds by the 9 since $\mathds{R}[x]$ on $[0,1]$ is an
adapted space.

prob:haviland Proof of 4
Since (i) $\Rightarrow$ (ii) is clear it is sufficient to show (ii)
$\Rightarrow$ (i). But since $E=\mathds{R}[x_{1},\dots,x_{n}]$ on $K$, is an
adapted space (see Problem 21) and since $E_{+}=\mathrm{Pos}(K)$ by definition
the 9 applies and gives the assertion.

prob:bernstein Proof of 6
We have that (ii) $\Rightarrow$ (i) is clear since $x^{k}\cdot(1-x)^{l}>0$ on
$(0,1)$ and at least one $c_{k^{\prime},l^{\prime}}>0$. It remains to prove
(i) $\Rightarrow$ (ii).

Let $f\in\mathds{R}[x]\setminus\\{0\\}$ with $f>0$ on $(0,1)$ then we can
write $f$ as

$f(x)=x^{p}\cdot(1-x)^{q}\cdot\tilde{f}(x)$

with $\tilde{f}\in\mathds{R}[x]$, $\tilde{f}>0$ on $[0,1]$, and
$p,q\in\mathds{N}_{0}$, i.e., by the fundamental theorem of algebra we can
factor out the zeros at $x=0$ and at $x=1$. Applying 5 (ii) to $\tilde{f}$
then gives the assertion.

prob:boundaryCone Proof of 9
Since the moment cone $\mathcal{S}_{\mathcal{F}}$ and the hyperplane $H$ are
convex we have that $\mathcal{S}_{\mathcal{F}}\cap H$ is a convex cone, i.e.,
it is a moment cone and there exists a family
$\mathcal{G}\subsetneq\mathrm{lin}\,\mathcal{F}$ of $m<n$ elements which spans
$\mathcal{S}_{\mathcal{F}}\cap H$. It is sufficient to show that $\mathcal{G}$
lives on $(\mathcal{Y},\mathfrak{A}|_{\mathcal{Y}})$ for some
$\mathcal{Y}\subseteq\mathcal{X}$.

For the hyperplane $H$ there exists a function $h\in\mathrm{lin}\,\mathcal{F}$
such that $L_{s}(h)\geq 0$ for all $s\in\mathcal{S}_{\mathcal{F}}$. Note, that
$\mathcal{N}=\cap_{k\in\mathds{N}}\\{x\in\mathcal{X}\,|\,f_{1}(x)^{2}+\dots+f_{n}(x)^{2}\geq
k\\}$ has measure zero for any representing measure $\mu_{s}$ on $\mathcal{X}$
of a moment sequence $s\in\mathcal{S}_{\mathcal{F}}$ since the moments are
finite, i.e., the $f_{i}$ are $\mu_{s}$-integrable. Without loss of generality
we can therefore work on $\mathcal{X}\setminus\mathcal{N}$. Hence, all
$\delta_{x}$ with $x\in\mathcal{X}\setminus\mathcal{N}$ are moment measures
and $L_{s}(h)\geq 0$ implies $h\geq 0$ on $\mathcal{X}\setminus\mathcal{N}$.

Then $s\in\mathcal{S}_{\mathcal{F}}\cap H\Leftrightarrow L_{s}(h)=0$ implies
that all representing measures $\mu$ of all $s\in\mathcal{S}_{\mathcal{F}}\cap
H$ have the support in
$\mathcal{Y}:=\\{x\in\mathcal{X}\setminus\mathcal{N}\,|\,h(x)=0\\}$.

prob:richterFromRosen Let $\mathcal{F}=\\{f_{1},\dots,f_{n}\\}$ be measurable
functions on $(\mathcal{X},\mathfrak{A})$ which are not necessarily bounded.
Set

$I:=\bigcap_{k\in\mathds{N}}\\{x\in\mathcal{X}\,|\,|f_{i}(x)|>k\ \text{for
all}\ i=1,\dots,n\\}.$

Then $I$ is measurable. Let $s$ be a moment sequence with representing measure
$\mu$. Since all $f_{i}$ are $\mu$-measurable we have $\mu(I)=0$. Therefore,
by working on $\mathcal{X}\setminus I$ we can assume without loss of
generality that $|f_{i}(x)|<\infty$ for all $x\in\mathcal{X}$.

Define $\mathcal{G}=\\{g_{1},\dots,g_{n}\\}$ with $g_{i}:=\frac{g_{i}}{f}$ and
$f:=1+\sum_{i=1}^{n}f_{i}^{2}$.

At first we note that from

$\int_{\mathcal{X}}f_{i}(x)~{}\mathrm{d}\mu(x)=\int_{\mathcal{X}}g_{i}(x)\cdot
f(x)~{}\mathrm{d}\mu=\int_{\mathcal{X}}g_{i}(x)~{}\mathrm{d}\nu(x),$ (72)

we have that every sequence $s=(s_{1},\dots,s_{n})$ is a moment sequence with
respect to $\mathcal{G}$ if and only if it is moment sequence with respect to
$\mathcal{F}$.

Since all $g_{i}$ are bounded we have by Rosenbloom’s Theorem that there is a
$k$-atomic representing measure $\nu=\sum_{i=1}^{k}c_{i}\cdot\delta_{x_{i}}$
which represents the moment sequence $s$. Then by (72) we find that
$\mu=\sum_{i=1}^{k}c_{i}\cdot f(x_{i})^{-1}\cdot\delta_{x_{i}}$ is a
representing measure of $s$ with respect to $\mathcal{F}$ which proves the
statement.

### Problems of Chapter 4

prob:restriction Proof of 3
Let $f\in\mathrm{lin}\,\mathcal{F}$. Then $f$ has at most $n$ zeros in
$\mathcal{X}$ and hence $f|_{\mathcal{Y}}$ has at most $n$ zeros in
$\mathcal{Y}\subset\mathcal{X}$. Since for any $g\in\mathrm{lin}\,\mathcal{G}$
there is a $f\in\mathrm{lin}\,\mathcal{F}$ such that $g=f|_{\mathcal{Y}}$ we
have the assertion.

prob:transf Proof of 8
Let $w_{0},\dots,w_{n}\in\mathcal{W}$ be pairwise distinct. Since $g$ is
injective we have that also $g(w_{0}),\dots,g(w_{n})\in\mathcal{X}$ are
pairwise distinct. Hence,

$\det\begin{pmatrix}g_{0}&g_{1}&\dots&g_{n}\\\
w_{0}&w_{1}&\dots&w_{n}\end{pmatrix}=\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
g(w_{0})&g(w_{1})&\dots&g(w_{n})\end{pmatrix}\neq 0$

and the statement follows from 5.

prob:scaling Proof of 9
Let $x_{0},\dots,x_{n}\in\mathcal{X}$ be pairwise distinct. Then

$\det\begin{pmatrix}g_{0}&g_{1}&\dots&g_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}=\det\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{0}&x_{1}&\dots&x_{n}\end{pmatrix}\cdot g(x_{1})\cdots g(x_{n})\neq 0$

and the statement follows from 5.

prob:4.1 Proof of 10

  1. (i)

Assume $f_{0},\dots,f_{n}$ are linearly dependent, i.e., there are
$a_{0},\dots,a_{n}\in\mathds{R}$ not all zero such that
$a_{0}f_{0}+\dots+a_{n}f_{n}$ is the zero polynomial. Hence, $f$ has at least
$n+1$ zeros. But since $\mathcal{F}$ is a T-system this is a contradiction.

  2. (ii)

Let $x_{0},\dots,x_{n}\in\mathcal{X}$ be $n+1$ pairwise distinct points. Then
by 4 we have

$\begin{pmatrix}f(x_{0})\\\ \vdots\\\
f(x_{n})\end{pmatrix}=\underbrace{\begin{pmatrix}f_{0}&\dots&f_{n}\\\
x_{0}&\dots&x_{n}\end{pmatrix}}_{=:M}\cdot\begin{pmatrix}a_{0}\\\ \vdots\\\
a_{n}\end{pmatrix}$

and since $\mathcal{F}$ is a T-system we have that $M$ has full rank by 5.
Hence, the coefficients $a_{0},\dots,a_{n}$ are unique.

prob:fraction Proof of 18
Set $f_{i}(x):=(x+\alpha_{i})^{-1}$ and
$g(x)=(x+\alpha_{0})\cdots(x+\alpha_{n})$. Then $g>0$ on $[a,b]$ since
$-\alpha_{0}<a<b$. Hence, $\mathcal{F}$ is a T-system on $[a,b]$ if and only
if $\mathcal{G}=\\{g_{i}:=g\cdot f_{i}\\}_{i=0}^{n}$ is a T-system on $[a,b]$
by 9.

We have
$g_{i}(x)=(x+\alpha_{0})\cdots(x+\alpha_{i-1})\cdot(x+\alpha_{i+1})\cdots(x+\alpha_{n})$
and $\deg g_{i}=n$. It is now sufficient to show that $\mathcal{G}$ is a
T-system on $\mathds{R}$ by 3 since then it will also be a T-system on
$[a,b]$.

Since $g_{i}(\alpha_{j})=0$ for all $i\neq j$ we have that the $g_{i}$ are
linearly independent. Hence, $\mathrm{lin}\,\mathcal{G}=\mathds{R}[x]_{\leq
n}$. But since $\\{x^{i}\\}_{i=0}^{n}$ is a T-system so is $\mathcal{G}$ since
every non-trivial $f\in\mathrm{lin}\,\mathcal{G}=\mathds{R}[x]_{\leq n}$ has
at most $n$ zeros.

In summary, we have that $\\{x^{i}\\}_{i=0}^{n}$ is a T-system on $\mathds{R}$
$\Rightarrow$ $\mathcal{G}$ on $\mathds{R}$ is a T-system $\Rightarrow$
$\mathcal{G}$ on $[a,b]$ is a T-system $\Rightarrow$ $\mathcal{F}$ on $[a,b]$
is a T-system.

prob:4.2 To the points $x_{0},\dots,x_{k+l}\in[a,b]$ add pairwise distinct
points $x_{k+l+1},\dots,x_{n}\in[a,b]\setminus\\{x_{0},\dots,x_{k+l}$. Then
the matrix

$\begin{pmatrix}f_{0}(x_{0})&\dots&f_{n}(x_{0})\\\ \vdots&&\vdots\\\
f_{0}(x_{n})&\dots&f_{n}(x_{n})\end{pmatrix}$ (73)

has full rank since $\mathcal{F}$ is a T-system, i.e., every vector,
especially

$(m,\dots,m,-m,\dots,-m,0,\dots,0,*,\dots,*)^{T}\in\mathds{R}^{n+1}$

is in its image. But the matrix

$\begin{pmatrix}f_{0}(x_{1})&\dots&f_{n}(x_{1})\\\ \vdots&&\vdots\\\
f_{0}(x_{k+l})&\dots&f_{n}(x_{k+l})\end{pmatrix}$

in (28) only contains the first $k+l$ rows of (73), i.e., (28) has at least
one solution.

prob:kreinError By 27 only the case $n=2m+2p$ and one end point is contained.
But then we can apply 26 to $\tilde{\mathcal{F}}=\\{f_{i}\\}_{i=0}^{n-1}$
which ensures by the same arguments in 27 that $x_{1},\dots,x_{p}$ are the
only zeros of some $f\geq 0$.

### Problems of Chapter 5

prob:etMultiplication Proof of 7
Set $g_{i}:=g\cdot f_{i}$. Then we have to check that

$\mathcal{W}(g_{0},\dots,g_{k})(x)=\det\begin{pmatrix}g_{0}(x)&g_{1}(x)&\dots&g_{n}(x)\\\
g_{0}^{\prime}(x)&g_{1}^{\prime}(x)&\dots&g_{n}^{\prime}(x)\\\
\vdots&\vdots&&\vdots\\\
g_{0}^{(n)}(x)&g_{1}^{(n)}(x)&\dots&g_{n}^{(n)}(x)\end{pmatrix}\neq 0$

holds for all $x\in[a,b]$. Since $g_{i}=g\cdot f_{i}$ we apply the product
rule and get

$\displaystyle\mathcal{W}(g_{0},\dots,g_{k})(x)$
$\displaystyle=g^{2}\cdot\det\begin{pmatrix}f_{0}(x)&f_{1}(x)&\dots&f_{n}(x)\\\
f_{0}^{\prime}(x)&f_{1}^{\prime}(x)&\dots&f_{n}^{\prime}(x)\\\
g_{0}^{\prime\prime}(x)&g_{1}^{\prime\prime}(x)&\dots&g_{n}^{\prime\prime}(x)\\\
\vdots&\vdots&&\vdots\\\
g_{0}^{(n)}(x)&g_{1}^{(n)}(x)&\dots&g_{n}^{(n)}(x)\end{pmatrix}$

since in the first line we factored out $g$ and then subtracted
$g^{\prime}$-times the first line from the second, and factored out $g$ from
the remaining second line. For the second derivatives in the third line we
have

$(g\cdot f_{i})^{\prime\prime}=g^{\prime\prime}\cdot f_{i}+2g^{\prime}\cdot
f_{i}^{\prime}+g\cdot f_{i}^{\prime\prime}$

and hence subtracting $g^{\prime\prime}$-times the first row,
$2g^{\prime}$-times the second row, and finally factoring out $g$ from the
remaining third row we get

$\displaystyle\mathcal{W}(g_{0},\dots,g_{k})(x)$
$\displaystyle=g^{3}\cdot\det\begin{pmatrix}f_{0}(x)&f_{1}(x)&\dots&f_{n}(x)\\\
f_{0}^{\prime}(x)&f_{1}^{\prime}(x)&\dots&f_{n}^{\prime}(x)\\\
f_{0}^{\prime\prime}(x)&f_{1}^{\prime\prime}(x)&\dots&f_{n}^{\prime\prime}(x)\\\
g_{0}^{\prime\prime\prime}(x)&g_{1}^{\prime\prime\prime}(x)&\dots&g_{n}^{\prime\prime\prime}(x)\\\
\vdots&\vdots&&\vdots\\\
g_{0}^{(n)}(x)&g_{1}^{(n)}(x)&\dots&g_{n}^{(n)}(x)\end{pmatrix}.$

Proceeding in this manner we arrive at

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

for all $x\in[a,b]$ which proves the statement.

prob:wronsTrans Proof of 8
We proceed similar to Problem/Solution 21 but now with the rule of
differentiation for $f_{i}\circ g$. We have

$(f_{i}\circ g)^{\prime}=g^{\prime}\cdot(f_{i}^{\prime}\circ g)$

and hence

$\mathcal{W}(g_{0},\dots,g_{n})=g^{\prime}\cdot\det\begin{pmatrix}f_{0}\circ
g&\dots&f_{n}\circ g\\\ f_{0}^{\prime}\circ g&\dots&f_{n}^{\prime}\circ g\\\
(f_{0}\circ g)^{\prime\prime}&\dots&(f_{n}\circ g)^{\prime\prime}\\\
\vdots&&\vdots\\\ (f_{0}\cdot g)^{(n)}&\dots&(f_{n}\circ
g)^{(n)}\end{pmatrix}$

by factoring out $g^{\prime}$ from the second row. Then we have

$\displaystyle(f_{i}\circ g)^{\prime\prime}$
$\displaystyle=(g^{\prime}\cdot(f_{i}^{\prime}\circ
g))^{\prime}=g^{\prime\prime}\cdot(f_{i}^{\prime}\circ
g)+(g^{\prime})^{2}\cdot(f_{i}^{\prime\prime}\circ g),$

i.e., we subtract $g^{\prime\prime}$-times the second row and factor out
$(g^{\prime})^{2}$ to get

$\mathcal{W}(g_{0},\dots,g_{n})=(g^{\prime})^{3}\cdot\det\begin{pmatrix}f_{0}\circ
g&\dots&f_{n}\circ g\\\ f_{0}^{\prime}\circ g&\dots&f_{n}^{\prime}\circ g\\\
f_{0}^{\prime\prime}\circ g&\dots&f_{n}^{\prime\prime}\circ g\\\ (f_{0}\circ
g)^{\prime\prime\prime}&\dots&(f_{n}\circ g)^{\prime\prime\prime}\\\
\vdots&&\vdots\\\ (f_{0}\cdot g)^{(n)}&\dots&(f_{n}\circ
g)^{(n)}\end{pmatrix}.$

Proceeding in this manner with

$(f_{i}\circ g)^{(k)}=(g^{\prime})^{(k)}\cdot(f_{i}^{(k)}\circ
g)+{\dots}+g^{(k)}\cdot(f_{i}^{\prime}\circ g)$

we get

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

with proves the assertion.

prob:wronskiReduction Proof of 9
Set $\mathcal{H}=\\{h_{i}\\}_{i=0}^{n}$ with $h_{i}:=\frac{f_{i}}{f_{0}}$.
Then by 7 we have

$\displaystyle\mathcal{W}(f_{0},\dots,f_{n})$
$\displaystyle=f_{0}^{n+1}\cdot\mathcal{W}(h_{0},\dots,h_{n})$ and since
$h_{0}=1$ we have $h_{0}^{\prime}=h_{0}^{\prime\prime}=\dots=0$ and
$\displaystyle=f_{0}^{n+1}\cdot\det\begin{pmatrix}1&h_{1}&\dots&h_{n}\\\
0&h_{1}^{\prime}&\dots&h_{n}^{\prime}\\\ \vdots&\vdots&&\vdots\\\
0&h_{1}^{(n)}&\dots&h_{n}^{(n)}\end{pmatrix}$ which gives by expanding along
the first column
$\displaystyle=f_{0}^{n+1}\cdot\det\begin{pmatrix}h_{1}^{\prime}&\dots&h_{n}^{\prime}\\\
\vdots&&\vdots\\\ h_{1}^{(n)}&\dots&h_{n}^{(n)}\end{pmatrix}$
$\displaystyle=f_{0}^{n+1}\cdot\mathcal{W}(h_{1}^{\prime},\dots,h_{n}^{\prime})$
and with $g_{i}=h_{i+1}^{\prime}$ for $i=0,\dots,n-1$ we get
$\displaystyle=f_{0}^{n+1}\cdot\mathcal{W}(g_{0},\dots,g_{n-1})$

which proves the statement.

prob:ecRestriction (a) Since $\mathcal{F}$ is an ET-system on $[a,b]$ we have

$\mathcal{W}(f_{0},\dots,f_{n})(x)\neq 0$

for all $x\in[a,b]$, i.e., also for all
$x\in[a^{\prime},b^{\prime}]\subseteq[a,b]$ and hence it is an ET-system on
$[a^{\prime},b^{\prime}]$.

(b) Since $\mathcal{F}$ is an ECT-system on $[a,b]$ we have

$\mathcal{W}(f_{0},\dots,f_{k})(x)\neq 0$

for all $x\in[a,b]$ and $k=0,\dots,n$, i.e., also for all
$x\in[a^{\prime},b^{\prime}]\subseteq[a,b]$ and $k=0,\dots,n$ and hence it is
an ECT-system on $[a^{\prime},b^{\prime}]$.

prob:xf Proof of 19
We already know that $\\{1,x,x^{2},\dots,x^{k}\\}$ is an ET-system for any
$k=0,1,\dots,n$ since

$cW(1,x,\dots,x^{k})(x)=1\cdot 1!\cdot{\cdots}\cdot k!>0.$

From the Wronskian determinant

$\mathcal{W}(1,x,\dots,x^{n},f)(x)=1\cdot 1!\cdot 2!\cdot{\dots}\cdot n!\cdot
f^{(n)}(x)>0$

we then get that $\mathcal{F}$ is an ECT-system on $[a,b]$ by 12.

prob:etSystExm Proof of 18
By 8 we only need to prove the statement for one case, say case (b)
$\mathcal{G}=\\{e^{\alpha_{i}x}\\}_{i=0}^{n}$. Let $k\in\\{0,1,\dots,n\\}$.
Then

$\displaystyle\mathcal{W}(g_{0},\dots,g_{k})$
$\displaystyle=\det\begin{pmatrix}g_{0}&g_{1}&\dots&g_{k}\\\
g_{0}^{\prime}&g_{1}^{\prime}&\dots&g_{k}^{\prime}\\\ \vdots&\vdots&&\vdots\\\
g_{0}^{(k)}&g_{1}^{(k)}&\dots&g_{k}^{(k)}\end{pmatrix}$ and with
$g_{i}^{(j)}=\alpha_{i}^{j}\cdot g_{i}$ we get
$\displaystyle=\deg\begin{pmatrix}g_{0}&g_{1}&\dots&g_{k}\\\
\alpha_{0}g_{0}&\alpha_{1}g_{1}&\dots&\alpha_{k}g_{k}\\\
\vdots&\vdots&&\vdots\\\
\alpha_{0}^{k}g_{0}&\alpha_{1}^{k}g_{1}&\dots&\alpha_{k}^{k}g_{k}\end{pmatrix}=g_{0}\cdot
g_{1}\cdots g_{n}\cdot\det\begin{pmatrix}1&1&\dots&1\\\
\alpha_{0}&\alpha_{1}&\dots&\alpha_{k}\\\ \vdots&\vdots&&\vdots\\\
\alpha_{0}^{k}&\alpha_{1}^{k}&\dots&\alpha_{k}^{k}\end{pmatrix}$
$\displaystyle=g_{0}\cdot g_{1}\cdots g_{k}\cdot\prod_{0\leq i<j\leq
k}(\alpha_{j}-\alpha_{i})\neq 0$

which proves the statement.

prob:exampleECTpolynomial To construct the non-negative polynomial on
$[0,\infty)$ with the double zero $x_{1}=1$ and the zero $x_{2}=2$ with
algebraic multiplicity $m_{2}=4$ we need $7$ monomials. We chose
$f_{0}(x)=1,f_{1}(x)=x^{2},f_{2}(x)=x^{3},f_{3}(x)=x^{5},f_{4}(x)=x^{8},f_{5}(x)=x^{11},f_{6}(x)=x^{13}$
and leave out $x^{42}$. With (38) we get

$\displaystyle f(x)$
$\displaystyle=\det\left(\begin{array}[]{c|cccccc}f_{0}&\,f_{1}&f_{2}&f_{3}&f_{4}&f_{5}&f_{6}\\\
x&\,1&1&2&2&2&2\end{array}\right)$
$\displaystyle=\det\begin{pmatrix}f_{0}(x)&f_{1}(x)&f_{2}(x)&f_{3}(x)&f_{4}(x)&f_{5}(x)&f_{6}(x)\\\
f_{0}(x_{1})&f_{1}(x_{1})&f_{2}(x_{1})&f_{3}(x_{1})&f_{4}(x_{1})&f_{5}(x_{1})&f_{6}(x_{1})\\\
f_{0}^{\prime}(x_{1})&f_{1}^{\prime}(x_{1})&f_{2}^{\prime}(x_{1})&f_{3}^{\prime}(x_{1})&f_{4}^{\prime}(x_{1})&f_{5}^{\prime}(x_{1})&f_{6}^{\prime}(x_{1})\\\
f_{0}(x_{2})&f_{1}(x_{2})&f_{2}(x_{2})&f_{3}(x_{2})&f_{4}(x_{2})&f_{5}(x_{2})&f_{6}(x_{2})\\\
f_{0}^{\prime}(x_{2})&f_{1}^{\prime}(x_{2})&f_{2}^{\prime}(x_{2})&f_{3}^{\prime}(x_{2})&f_{4}^{\prime}(x_{2})&f_{5}^{\prime}(x_{2})&f_{6}^{\prime}(x_{2})\\\
f_{0}^{\prime\prime}(x_{2})&f_{1}^{\prime\prime}(x_{2})&f_{2}^{\prime\prime}(x_{2})&f_{3}^{\prime\prime}(x_{2})&f_{4}^{\prime\prime}(x_{2})&f_{5}^{\prime\prime}(x_{2})&f_{6}^{\prime\prime}(x_{2})\\\
f_{0}^{\prime\prime\prime}(x_{2})&f_{1}^{\prime\prime\prime}(x_{2})&f_{2}^{\prime\prime\prime}(x_{2})&f_{3}^{\prime\prime\prime}(x_{2})&f_{4}^{\prime\prime\prime}(x_{2})&f_{5}^{\prime\prime\prime}(x_{2})&f_{6}^{\prime\prime\prime}(x_{2})\\\
f_{0}^{(4)}(x_{2})&f_{1}^{(4)}(x_{2})&f_{2}^{(4)}(x_{2})&f_{3}^{(4)}(x_{2})&f_{4}^{(4)}(x_{2})&f_{5}^{(4)}(x_{2})&f_{6}^{(4)}(x_{2})\end{pmatrix}$
$\displaystyle=\det\begin{pmatrix}1&x^{2}&x^{3}&x^{5}&x^{8}&x^{11}&x^{13}\\\
1&1&1&1&1&1&1\\\ 0&2&3&5&8&11&13\\\ 1&4&8&32&256&2\,048&8\,192\\\
0&4&12&80&1\,024&11\,264&53\,248\\\ 0&2&12&160&3\,584&56\,320&319\,488\\\
0&0&6&240&10\,752&253\,440&1\,757\,184\end{pmatrix}$ $\displaystyle f(x)$
$\displaystyle=48\cdot\left(14\,980\,788x^{13}-184\,325\,420x^{11}+2\,421\,354\,616x^{8}-26\,336\,028\,160x^{5}\right.$
$\displaystyle\qquad\quad\,\left.+112\,945\,898\,496x^{3}-112\,347\,781\,120x^{2}+23\,485\,900\,800\right).$

The function $f$ is shown in Figure 3.

Figure 3: The function $f$ from the solution of Problem 27.

This function $f$ we gave here is not unique. Of course every multiple of $f$
also fulfills the requirements but we also made the restrictions to use all
monomials except $x^{42}$. We get another polynomial when we e.g. leave out
$x^{13}$ (or any other monomial except $1$) instead of $x^{42}$. Then any
conic linear combination of these functions also fulfills the requirements.

We can not leave out $1$ since any linear combination has the additional zero
$x=0$.

### Problems of Chapter 6

prob:smoothing Proof of 8
Since $\mathcal{F}$ is a continuous T-system we can assume that

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

for all $a\leq x_{0}<x_{1}<\dots<x_{n}\leq b$. Since the Gaussian kernel is
ETPk for every $k\in\mathds{N}_{0}$, see 6, we have

$K_{\sigma}^{*}\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{n}\\\
y_{1}&y_{2}&\dots&y_{n}\end{pmatrix}>0$

for all $x_{1}<x_{2}<\dots<x_{n}$ and $y_{1}\leq y_{2}\leq\dots\leq y_{n}$ in
$\mathds{R}$ as well as $\sigma>0$. Hence, in
$\mathcal{W}(f_{\sigma,0},f_{\sigma,1},\dots,f_{\sigma,n})(x)=$ (53) in 7 we
are integrating over a non-negative functions with respect to the Lebesgue
measure $\mu=\lambda$, i.e.,
$\mathcal{W}(f_{\sigma,0},f_{\sigma,1},\dots,f_{\sigma,n})(x)>0$ for all
$x\in[a,b]$ which proves the statement.

### Problems of Chapter 7

prob:etNeighborhood The family $\mathcal{F}$ on $[a,b]$ needs for a fixed
$f\geq 0$ only be an ET-system around the zeros of $f$ but otherwise the proof
of 1 is employed, i.e., there we only need $\mathcal{F}$ to be a T-system.

### Problems of Chapter 8

prob:posR Proof of 4 on $\mathds{R}$
By (a) there exists a function $w\in\mathcal{C}(\mathds{R},\mathds{R})$ such
that $w>0$ on $\mathds{R}$ and

$\lim_{x\to\infty}\frac{f_{n}(x)}{w(x)}=1.$

By (b) we define

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

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 $[-1,1]$ by 8. We now apply 3 to $\mathcal{G}$. Set
$g:=(\frac{f}{w})\circ t$.

(ii): By 3 on $[a,b]$ there exist points

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

and unique functions $g_{*}$ and $g^{*}$ such that $g=g_{*}+g^{*}$,
$g_{*},g^{*}\geq 0$ on $[-1,1]$, $x_{1},\dots,x_{m}$ 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 $f=f_{*}+f^{*}$.

(i): Since $g^{*}(y_{0})=g^{*}(y_{m})=0$ we have that $g^{*}$ contains no
$g_{2m}$ and hence the coefficient of $g_{2m}$ in $g_{*}$ is $a_{2m}$.

prob:nonnegR Proof of 5 on $\mathds{R}$
Similar to the proof of 3 on $[0,\infty)$ and hence Problem/Solution 6.

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

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

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

### Problems of Chapter 9

prob:a0 Proof of 13
10 can in general not be extended to $[0,b]$ since
$\\{x^{\alpha_{0}},\dots,x^{\alpha_{n}}\\}$ is not an ET-system. This fails at
$x=0$. But on $(0,b]$ it is an ET-system. We can therefore factor out the
zeros of $f\geq 0$ at $x=0$

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

to get some $\tilde{f}$ with $\tilde{f}\geq 0$ on $[0,b]$ and
$\tilde{f}(0)>0$. To $\tilde{f}$ we can then apply 10 with $a=0$.

In summary, 10 on $[0,b]$ holds if $f(0)>0$, see also 5 and 6 for the
corresponding version on $[0,\infty)$.

### Problems of Chapter 10

prob:nonneg0infty Proof of 5
To prove 5 we have to note that $\mathcal{F}=\\{x^{\alpha_{i}}\\}_{i=0}^{n}$
with $\alpha_{0}=1$ is an ET-system on $(0,\infty)$. The only difficulty is
$x=0$ where $\mathcal{F}$ fails to be a ET-system.

But looking closely at the proof of 5 (see Problem/Solution 8) the ET-system
property is only required in a neighborhood of the zeros of $f$ and otherwise
it is the proof of 1 for T-systems. Since $f(0)>0$ we have no zero at $x=0$
where $\mathcal{F}$ fails to be a T-system. In fact, we have $f(x)>0$ for all
$x\in[0,\varepsilon)$ for some $\varepsilon>0$. Hence, we can apply 3 since
its proof requires for our $f$ with $f(0)>0$ only that $\mathcal{F}$ to be an
ET-system on $(0,\infty)$ which is fulfilled.

prob:leadingcoeffR By expanding

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

we see that $a\cdot x^{m_{1}+\dots+m_{r}+2m}$ is the monomial with the highest
degree $m_{1}+\dots+m_{r}+2m=\deg p$ and the coefficient is $a$.

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List of Symbols

### Matrices

$\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{1}&x_{2}&\dots&x_{n}\end{pmatrix}$: 4, eq. (25)4

$\begin{pmatrix}f_{0}&\dots&f_{i-1}&f_{i}&\dots&f_{i+p}&f_{i+p+1}&\dots&f_{n}\\\
x_{0}&\dots&x_{i-1}&(x_{i}&\dots&x_{i})&x_{i+p+1}&\dots&x_{n}\end{pmatrix}$:
Eq. (36) 36

$\begin{pmatrix}f_{0}&f_{1}&\dots&f_{n}\\\
x_{1}&x_{2}&\dots&x_{n}\end{pmatrix}^{*}$: Eq. (37) 37

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

### Determinants

$K\\!\begin{pmatrix}x_{0}&x_{1}&\dots&x_{n}\\\
y_{0}&y_{1}&\dots&y_{n}\end{pmatrix}$: Eq. (48) 48

$K^{*}\\!\begin{pmatrix}x_{1}&x_{2}&\dots&x_{i}\\\
y_{1}&y_{2}&\dots&y_{i}\end{pmatrix}$: 3, eq. (49) 49

$\mathcal{W}(f_{0},\dots,f_{k})$: 6, eq. (40) 40

### Further Mathematical Symbols

$\leq$ 1

$B_{f,d}$: Eq. (18) 18

$h_{n}\nearrow g$ 7

$\mathfrak{B}(\mathds{R}^{n})$ 6

$\mathcal{C}(\mathcal{X},\mathcal{Y})$ 2

$\mathcal{C}_{c}(\mathcal{X},\mathds{R})$ 2

$\mathrm{conv}\,A$ 4

$E_{+}$ 1

see also $(\mathrm{lin}\,\mathcal{F})_{+}$ 25

$E^{*}$ 1

$f_{+}$, $f_{-}$, $|f|$ 0.11

$\mathcal{H}(s)$: Eq. (1) 1

$\mathrm{int}\,A$ 2

$L_{\mu}$ 1.4

$\mathcal{L}^{p}(\mathcal{X},\mu)$ 6

$\varepsilon(x)$: 24, eq. (29) 24

$\varepsilon(\mathcal{X})$: 24, eq. (30) 30

$K_{\sigma}$: Eq. (50) 50

$\mathrm{lin}\,\mathcal{F}$: Eq. (24) 24

$(\mathrm{lin}\,\mathcal{F})_{+}$: 25 25

$(\mathrm{lin}\,\mathcal{F})^{e}$: 25 25

$(\mathrm{lin}\,\mathcal{F})_{+}^{e}$: 25 25

$L_{\mu}$: 1.4 1.4

$L_{s}$: 1.6 1.6

$\mathcal{M}(L)$: 1.21.2

$\mathcal{M}(\mathcal{X})_{+}$ 6

$\mathds{N}$ 1

$\mathds{N}_{0}$ 1

$\mathrm{Pos}(K)$: Eq. (12) 12

$\mathrm{Pos}(\mathds{R})$: Eq. (13) 13

$\mathrm{Pos}([0,\infty))$: Eqs. (14) and (15) 14

$\mathrm{Pos}([-1,1])$: Eq. (16) 16

$\mathrm{Pos}([a,b])$: Eq. (20) 20

$\mathcal{P}(\mathcal{X})$ 1

$\mathds{Q}$ 1

$\mathds{R}$ 1

$\mathcal{S}_{\mathcal{F}}$ 9

$\sigma(A)$ 6

$\mathds{T}$ 1

$\Xi^{m}$: Eq. (54) 54

$\mathds{Z}$ 1

### Index

A

  * adapted
    * cones25
    * space23
  * algebra4, 27
    * $*$-27
      * Fréchet topological27
  * approximation
    * best
      * polynomial117
    * Tchebycheff117 B
  * Banach, S.2
  * basic composition formulas73
  * Basic Representation Theorem24
  * Bernstein
    * polynomial32
    * Theorem32
  * Bernstein, S. N.32, 44
  * best approximation
    * polynomial117
  * Boas’ Theorem38
  * Boas, R. P.35, 38, 108
  * Borel
    * measure4
    * $\sigma$-algebra4
  * Brickman, L.33 C
  * Carathéodory
    * Theorem3
  * Carroll, L.57
  * compact
    * locally
      * Hausdorff space2
    * set2
  * cone3
    * adapted25
    * negative1
    * positive1
      * linear functional21
  * Conic Extension Theorem26
  * continuous2
  * convex3
  * cubature formula
    * Gaussian37
  * Curtis–Mairhuber–Sieklucki Theorem47 D
  * Daniell’s Representation Theorem6
  * Daniell’s Signed Representation Theorem9
  * Daniell, P. J.6
  * decomposition
    * Riesz property12
  * determinant
    * representation as a49, 68
    * Vandermonde44, 48
    * Wronskian60
  * determinate
    * moment functional19
  * Dirac, P. A. M.105
  * dominate23
    * cone25
  * dual2 E
  * ECT-system60
  * Einstein, A.1
  * ET-system57
  * ETPk72
  * Euripides71
  * extended
    * totally
      * positive72 F
  * Fil’štinskiǐ, V. A.34
  * Fixed Point Theorem of Brouwer79
  * Fréchet space27
  * function
    * measurable4
    * $\mu$-integrable4
  * functional
    * $K$-moment18
    * linear27
      * cone positive21
      * non-negative27
    * moment17
    * Riesz18
    * sublinear2
    * superlinear2
  * Fundamental Theorem of Algebra110 G
  * Gauß, C. F.37
  * Gaussian kernel72 H
  * Haar, A.118
  * Hahn, H.2
  * Hahn–Banach Theorem2
  * Hamburger moment problem _see_ Theorem, Hamburger31
  * Hamburger’s Theorem30
  * Hamburger, H. L.30, 39
  * Hankel matrix3
    * of a sequence3, 18
  * Hardy, G. H.77
  * Hausdorff
    * moment problem
      * sparse102
    * space2
    * truncated moment problem
      * sparse101
  * Hausdorff moment problem _see_ Theorem, Hausdorff32
  * Hausdorff’s Theorem30
  * Hausdorff, F.30, 35, 101
  * Haviland’s Theorem31
  * Haviland, E. K.31
  * Haviland–Hildebrandt–Schoenberg–Wintner Theorem34
  * Haviland–Wintner Theorem34
  * Helly, E.2
  * Herglotz
    * moment problem31
  * Hilbert, D.32
  * Hildebrandt, T. H.34
  * hull
    * convex3 I
  * identity
    * Sylvester62
  * indeterminacy
    * Stieltjes example19
  * indeterminate
    * moment
      * functional19
  * index52
    * of a set52
  * interior
    * of a set2
  * Interlacing Theorem _see_ Snake Theorem83 J
  * Jackson, D.121 K
  * $K$-moment functional18
  * Kakutani, S.11
  * Karlin
    * Nichtnegativstellensatz
      * on $[0,\infty)$91
      * on $[a,b]$86
      * on $\mathds{R}$92
    * Positivstellensatz
      * on $[0,\infty)$89
      * on $[a,b]$82
      * on $\mathds{R}$92
    * Theorem
      * for $f>0$ on $[a,b]$77
      * for $f\geq 0$ on $[a,b]$84
  * Karlin, S.vii
  * kernel71
    * Gaussian72
  * Krein, M. G.33, 83 L
  * lattice
    * of functions5
    * space5
  * Leibniz, G. W.97
  * Lemma
    * Markov82
  * Lewin, K.43
  * locally compact2
  * Lukács Theorem _see_ Lukács–Markov Theorem33
  * Lukács, F.33
  * Lukács–Markov Theorem33, 99 M
  * majorized21
  * Markov Lemma82
  * Markov’s Theorem _see_ Lukács–Markov Theorem33
  * Markov, A. A.11, 33
  * Markov–Lukács Theorem _see_ Lukács–Markov Theorem34
  * measurable
    * function4
    * space4
  * measure4
    * Borel4
    * Carathéodory outer4
    * outer4
    * Radon4
    * regular5
    * representing17
    * space4
  * Milton, J.113
  * minorized21
  * moment
    * $f$-moment of $\mu$17
    * functional17
      * determinate19
      * generated by $\mu$18
      * indeterminate19
      * truncated19
    * problem
      * Hamburger _see_ Theorem, Hamburger30
      * Hausdorff _see_ Theorem, Hausdorff30
      * Herglotz31
      * Stieltjes _see_ Theorem, Stieltjes29
      * Švenco _see_ Theorem, Švenco34
      * trigonometric31
    * sequence18
  * $\mu$-integrable4
  * $\mu$-measurable4 N
  * negative cone1
  * neighborhood2
  * Nichtnegativstellensatz
    * Karlin
      * on $[0,\infty)$91
      * on $[a,b]$86
      * on $\mathds{R}$92
  * Nudel’man, A. A.33, 83 O
  * optimization
    * over linear functionals121
  * order
    * partial1
    * total1
    * vector space1
  * orthogonal
    * polynomial121 P
  * Pólya, G.32, 39, 89
  * partial order1
  * polynomial44
    * Bernstein32
    * best approximation117
    * orthogonal121
    * Schur68
  * Pope, A.17
  * positive
    * totally71
      * extended72
      * strictly71
  * positive cone1
  * Positivstellensatz
    * Karlin
      * on $[0,\infty)$89
      * on $[a,b]$82
      * on $\mathds{R}$92 R
  * Radau, M. R.33
  * Radon
    * measure4
  * reduced
    * system61, 63
  * regular
    * measure5
  * representation
    * as a determinant49, 68
    * Theorem
      * Basic24
      * Daniell6
      * Daniell, signed9
      * Riesz12
      * Riesz, signed11
      * Riesz–Markov–Kakutani11
  * representing
    * measure17
  * Richter’s Theorem36
  * Richter, H.35
  * Richter–Rogosinski–Rosenbloom Theorem38
  * Riesz
    * decomposition property12
  * Riesz functional18
  * Riesz’ Representation Theorem12
    * Signed11
  * Riesz, F.11
  * Riesz, M.31
  * Riesz–Markov–Kakutani Representation Theorem11
  * Rogosinski, W. W.38
  * Rosenbloom, P. C.37 S
  * Santayana, G.29
  * Schmüdgen, K.39
  * Schoenberg, I. J.34
  * Schur polynomial68
  * sequence
    * moment18
  * Shakespeare, W.117
  * Sherman, T.39
  * $\sigma$-algebra4
    * Borel4
  * Snake Theorem83
  * space
    * adapted23
    * Hausdorff2
      * locally compact2
    * lattice5
    * measurable4
    * measure4
    * topological2
  * sparse
    * algebraic Nichtnegativstellensatz
      * on $[0,\infty)$109
      * on $[0,b]$104
      * on $[a,b]$103
      * on $\mathds{R}$110
    * algebraic Positivstellensatz
      * on $[0,\infty)$105
      * on $[a,b]$97
      * on $\mathds{R}$110
    * Hausdorff moment problem101, 102
      * truncated101
    * Stieltjes moment problem108
  * Stieltjes
    * example
      * indeterminacy19
  * Stieltjes moment problem _see_ Theorem, Stieltjes32
  * Stieltjes’ Theorem29
  * Stieltjes, T. J.19, 29
  * Stochel, J.39
  * Stone–Weierstrass Theorem3
  * STPk71
  * Strassen’s Theorem22
  * strictly
    * totally
      * positive71
  * sublinear
    * functional2
  * suits6
  * superlinear
    * functional2
  * Švenco’s Theorem34
  * Švenco, K. I.34
  * Sylvester’s identity62
  * system
    * ECT-60
    * ET-57
    * reduced61, 63
    * T-44
    * Tchebycheff _see_ T-44
  * Szegö, G.32 T
  * T-system44
    * continuous45
    * extended57
      * complete60
    * periodic45
  * Tchakaloff, M. V.37
  * Tchebycheff approximation117
  * Tchebycheff system _see_ T-system44
  * Tchebycheff, P. L.43
  * Theorem
    * Algebra
      * Fundamental110
    * Basic Representation24
    * Bernstein32
    * Boas38
    * Brouwer Fixed Point79
    * Carathéodory3
    * Conic Extension26
    * Curtis–Mairhuber–Sieklucki47
    * Daniell’s Representation6
      * Signed9
    * Hahn–Banach2
    * Hamburger30
    * Hausdorff30
      * sparse101
    * Haviland31
    * Haviland–Hildebrandt–Schoenberg–Wintner34
    * Haviland–Wintner34
    * Interlacing _see_ Snake83
    * Karlin
      * for $f>0$ on $[a,b]$77
      * for $f\geq 0$ on $[a,b]$84
      * Nichtnegativstellensatz on $[0,\infty)$91
      * Nichtnegativstellensatz on $[a,b]$86
      * Nichtnegativstellensatz on $\mathds{R}$92
      * Positivstellensatz on $[0,\infty)$89
      * Positivstellensatz on $[a,b]$82
      * Positivstellensatz on $\mathds{R}$92
    * Lukács _see_ Lukács–Markov33
    * Lukács–Markov33, 99
    * Markov _see_ Lukács–Markov33
    * Markov–Lukács _see_ Lukács–Markov34
    * Müntz–Szász102
    * Richter36
    * Richter–Rogosinski–Rosenbloom38
    * Riesz’ Representation12
      * Signed11
    * Riesz–Markov–Kakutani Representation11
    * Rogosinski38
    * Rosenbloom37
    * Snake83
    * Stieltjes29
    * Stone–Weierstrass3
    * Strassen22
    * Švenco34
    * Tchakaloff37
    * Wald37
  * topology2
  * total order1
  * totally
    * positive71
  * TPk71
  * trigonometric
    * moment problem31
  * truncated
    * moment functional19 U
  * unit element27 V
  * Vandermonde
    * determinant44, 48
  * vector space
    * ordered1 W
  * Wald, A.37
  * Wiener, N.21
  * Wintner, A.31
  * Wronskian60
  * Wronskian determiant _see_ Wronskian60 Z
  * zero
    * multiplicity57
    * nodal50
    * non-nodal50