# Non Oscillatory Functions and A Fourier Inversion Theorem for Some Functions
of Very Moderate Decrease

Tristram de Piro  Flat 3, Redesdale House, 85 The Park, Cheltenham, GL50 2RP
<EMAIL_ADDRESS>
###### Abstract.

We consider non oscillatory functions and prove an everywhere Fourier
Inversion Theorem for functions of very moderate decrease. The proofs rely on
some ideas in nonstandard analysis.

Non oscillatory functions of very moderate decrease are quite typical in the
class $L^{2}(\mathcal{R})$ and are important in Physics. The first result
presented here improves slightly on a result due to Carlsen, see [2], in that
the convergence is everywhere rather than almost everywhere. This property is
useful in verifying certain differentiability criteria in Physics, arising
mainly from Maxwell’s equations, and in showing decay properties of fields
produced in connection with Jefimenko’s equations, see [8] and [9]. At the end
of the paper, we show how the result can be improved to functions of just very
moderate decrease, without the non oscillatory assumption. The proofs rely on
some ideas from nonstandard analysis, contained in the papers [6] and [7].

######  Definition 0.1.

We say that $f\in C(\mathcal{R})$ is of very moderate decrease if there exists
a constant $C\in\mathcal{R}_{>0}$ with $|f(x)|\leq{C\over|x|}$, for $|x|>1$.
We say that $f\in C(\mathcal{R})$ is of moderate decrease if there exists a
constant $C\in\mathcal{R}_{>0}$ with $|f(x)|\leq{C\over|x|^{2}}$, for $|x|>1$.
We say that $f\in C(\mathcal{R})$ is non-oscillatory if there exist finitely
many points $\\{y_{i}:1\leq i\leq n\\}$, for which $f|_{(y_{i},y_{i+1})}$ is
monotone, $1\leq i\leq n-1$, and $f|_{(-\infty,y_{1})}$, $f|_{(y_{n},\infty)}$
are monotone. We say that $f\in C(\mathcal{R})$ is oscillatory if there exists
an infinite sequence of points $\\{y_{i}:i\in\mathcal{Z}\\}$, for which
$f|_{(y_{i},y_{i+1})}$, $i\in\mathcal{Z}$, is monotone and there exists
$\delta\in\mathcal{R}_{>0}$, with $|y_{i+1}-y_{i}|\geq\delta$, for
$i\in\mathcal{Z}$.

######  Lemma 0.2.

Let $f\in C(\mathcal{R})$ and ${df\over dx}\in C(\mathcal{R})$ be of very
moderate decrease, with $f$ and ${df\over dx}$ non-oscillatory, then defining
the Fourier transform by;

$\mathcal{F}(f)(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$ $(k\neq 0)$

$\mathcal{F}({df\over dx})(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{df\over dx}(y)e^{-iky}dy$ $(k\neq 0)$

we have that $\mathcal{F}(f)(k)$ and $\mathcal{F}({df\over dx})(k)$ are
bounded, for $|k|>k_{0}>0$, and there exists a constant
$G\in\mathcal{R}_{>0}$, such that;

$|\mathcal{F}(f)(k)|\leq{G\over|k|^{2}}$

for sufficiently large $k$.

###### Proof.

As $f$ is of very moderate decrease, we have that $f$ is continuous and
$lim_{|x|\rightarrow\infty}f(x)=0$. Similarly, ${df\over dx}$ is continuous
and $lim_{|x|\rightarrow\infty}{df\over dx}=0$. As
$lim_{|x|\rightarrow\infty}f(x)=0$, and $f$ is non-oscillatory, we have that,
for $k\neq 0$, the indefinite integral;

$lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

$=lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)cos(ky)dy-
ilim_{r\rightarrow\infty}\int_{-r}^{r}f(y)sin(ky)dy$

exists. As $f$ is of very moderate decrease and non-oscillatory, we have that
$|f(x)|\leq{D\over|x|}$, for $|x|>E$, and monotone in the intervals
$(-\infty,E)$ and $(E,\infty)$. Using the method of [7], letting
$K=max(|f||_{[-E,E]})$, we have that;

$|lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)cos(ky)dy|\leq
2KE+2(|\int_{E}^{{\pi\over 2|k|}+{n_{k}\pi\over|k|}}{Dcos(ky)\over
y}dy|+|\int_{{\pi\over 2|k|}+{n_{k}\pi\over|k|}}^{{\pi\over
2|k|}+{(n_{k}+1)\pi\over|k|}}{Dcos(ky)\over y}dy|)$

$\leq 2KE+{2D\over E}({\pi\over
2|k|}+{n_{k}\pi\over|k|}-E)+2\int_{E}^{E+{\pi\over|k|}}{Dsin(|k|(y-E))\over
y}dy$

$\leq 2KE+{2D\pi\over E|k|}+2\int_{E}^{E+{\pi\over|k|}}{Dsin(|k|(y-E))\over
y}dy$

$|lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)sin(ky)dy|\leq
2KE+2(|\int_{E}^{{m_{k}\pi\over|k|}}{Dsin(ky)\over
y}dy|+|\int_{{m_{k}\pi\over|k|}}^{{(m_{k}+1)\pi\over|k|}}{Dsin(ky)\over
y}dy|)$

$\leq 2KE+{2D\over
E}({m_{k}\pi\over|k|}-E)+2\int_{E}^{E+{\pi\over|k|}}{Dsin(|k|(y-E))\over y}dy$

$\leq 2KE+{2D\pi\over E|k|}+2\int_{E}^{E+{\pi\over|k|}}{Dsin(|k|(y-E))\over
y}dy$

where $n_{k}=\mu n({\pi\over 2|k|}+{n\pi\over|k|}\geq E:n\in\mathcal{Z}_{\geq
0})$ and $m_{k}=\mu n({n\pi\over|k|}\geq E:n\in\mathcal{Z}_{\geq 0})$

so that;

$|lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy|\leq 4KE+{4D\pi\over
E|k|}+4\int_{E}^{E+{\pi\over|k|}}{Dsin(|k|(y-E))\over y}dy$

$=4KE+{4D\pi\over
E|k|}+4D([{-cos(|k|(y-E))\over|k|y}]^{E+{\pi\over|k|}}_{E}-\int_{E}^{E+{\pi\over|k|}}{cos(|k|(y-E))\over
y^{2}}dy)$

$=4KE+{4D\pi\over E|k|}+4D({1\over E|k|+\pi}+{1\over
E|k|}+\int_{E}^{E+{\pi\over|k|}}{cos(|k|(y-E))\over y^{2}}dy)$

$\leq 4KE+{4D\pi\over E|k|}+4D({1\over E|k|+\pi}+{1\over
E|k|}+\int_{E}^{\infty}{1\over y^{2}}dy)$

$\leq 4KE+{4D\pi\over E|k|}+4D({2\over E|k|}+{1\over E})=N_{k}$

so that $\mathcal{F}(f)(k)$ and, similarly, $\mathcal{F}({df\over dx})(k)$ are
bounded, for $|k|>k_{0}>0$.

We have, using integration by parts, that;

$\mathcal{F}({df\over dx})(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{df\over dx}(y)e^{-iky}dy$

$={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}([f(y)e^{-iky}]^{r}_{-r}+ik\int_{-r}^{r}f(y)e^{-iky}dy)$

$={1\over(2\pi)^{1\over
2}}[f(y)e^{-iky}]^{\infty}_{-\infty}+ik{1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

$=ik\mathcal{F}(f)(k)$

so that, for $|k|>1$;

$|\mathcal{F}(f)(k)|\leq{|\mathcal{F}({df\over dx})(k)|\over|k|}$,
$({\dagger})$

As ${df\over dx}$ is continuous and non-oscillatory, by the proof of Lemma 0.9
in [7], using underflow, for $r\in\mathcal{R}_{>0}$, we can find
$\\{F_{r},G_{r}\\}\subset\mathcal{R}_{>0}$, such that, for all $|k|>F_{r}$, we
have that;

$|{1\over(2\pi)^{1\over 2}}\int_{-r}^{r}{df\over
dx}(y)e^{-iky}dy|<{G_{r}\over|k|}$, $(**)$

It is easy to see from the proof, that $\\{F_{r},G_{r}\\}$ can be chosen
uniformly in $r$. Then, from $(**)$, we obtain that, for $|k|>F$;

$|\mathcal{F}({df\over dx})(k)|<{G\over|k|}$, for $|k|>F$

and, from $({\dagger})$, for $|k|>max(F,1)$, that;

$|\mathcal{F}(f)(k)|\leq{|\mathcal{F}({df\over
dx})(k)|\over|k|}<{G\over|k|^{2}}$

∎

######  Definition 0.3.

Let $f\in C^{3}(\mathcal{R})$, with $f,f^{\prime},f^{\prime\prime}$ and
$f^{\prime\prime\prime}$ bounded, then we define an approximating sequence
$\\{f_{m}:m\in\mathcal{N}\\}$ by the requirements;

$(i)$. $f_{m}\in C^{2}(\mathcal{R})$, for $m\in\mathcal{N}$.

$(ii)$. $f_{m}|_{[-m,m]}=f|_{[-m,m]}$.

$(iii)$. $f_{m}$ is of uniform moderate decay, in the sense that there exists
a constant $C\in\mathcal{R}_{>0}$, independent of $m$, with;

$|f_{m}(x)|\leq{C\over|x|^{2}}$, for $x\in(-\infty,-m-{1\over
m})\cup(m+{1\over m},\infty)$

$(iv)$. There exists constants $\\{D,E\\}\subset\mathcal{R}_{>0}$, with
$\int_{-m-{1\over m}}^{m}|f_{m}(x)|dx\leq{D\over m}$ and $\int_{m}^{m+{1\over
m}}|f_{m}(x)|dx\leq{D\over m}$.

######  Lemma 0.4.

Let $f\in C(\mathcal{R})$ and ${df\over dx}\in C(\mathcal{R})$ be of very
moderate decrease, with $f$ and ${df\over dx}$ non-oscillatory. Let
$\\{f_{m};m\in\mathcal{N}\\}$ be an approximating sequence. Let $\mathcal{F}$
be the ordinary Fourier transform, defined for each $f_{m}$, then, for any
$k_{0}>0$, the sequence $\\{\mathcal{F}(f_{m}):m\in\mathcal{N}\\}$ converges
pointwise and uniformly to $\mathcal{F}(f)$ on
${\mathcal{R}\setminus(|k|<k_{0})}$, where $\mathcal{F}(f)$ is defined in
Lemma 0.2. In particularly, $\mathcal{F}(f)\in
C({\mathcal{R}\setminus\\{0\\}})$.

###### Proof.

For $g\in C(\mathcal{R})$ and $n\in\mathcal{N}$, define;

$\mathcal{F}_{n}(g)(k)={1\over(2\pi)^{1\over 2}}\int_{-n}^{n}f(y)e^{-iky}dy$

For $k\in{\mathcal{R}\setminus(|k|<k_{0})}$, $\\{m,n\\}\subset\mathcal{N}$,
and $m\geq n$, $\epsilon>0$,$\delta>0$, we have;

$|\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k)|\leq|\mathcal{F}(f)(k)-\mathcal{F}_{n}(f)(k)|+|\mathcal{F}_{m}(f)(k)-\mathcal{F}_{m}(f_{m})(k)|$

$+|\mathcal{F}_{m}(f_{m})(k)-\mathcal{F}(f_{m})(k)|$

$=|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+|\mathcal{F}_{m}(f_{m})(k)-\mathcal{F}(f_{m})(k)|$

$\leq|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+\int_{-\infty}^{-m}|f_{m}(x)|dx+|+\int_{m}^{\infty}|f_{m}(x)|dx$

$=|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+\int_{-\infty}^{-m-{1\over
m}}|f_{m}(x)dx+\int_{-m-{1\over m}}^{-m}|f_{m}(x)|dx$

$+\int_{m}^{m+{1\over m}}|f_{m}(x)dx+\int_{m+{1\over m}}^{\infty}|f_{m}(x)|dx$

$\leq|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+{D+E\over
m}+\int_{-\infty}^{-m-{1\over m}}{C\over x^{2}}dx+\int_{m+{1\over
m}}^{\infty}{C\over x^{2}}dx$

$\leq|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+{D+E\over m}+{2C\over m+{1\over
m}}$

$\leq|\mathcal{F}(f)(k)-\mathcal{F}_{m}(f)(k)|+{2C+D+E\over m}$

$\leq{C_{k_{0}}\over m}+{2C+D+E\over m}$

$\leq\epsilon+\delta$, for $m\geq
max({C_{k_{0}}\over\epsilon},{2C+D+E\over\delta})$. As $\epsilon>0$ and
$\delta>0$ were arbitrary, we obtain the first result. The fact that each
$\mathcal{F}(f_{m})$ in the approximating sequence is continuous follows from
$(iii)$ in definition 0.3 and the DCT. The last result then follows
immediately from the fact that $k_{0}>0$ is arbitrary and the uniform limit of
continuous functions is continuous.

∎

######  Lemma 0.5.

If $m\in\mathcal{R}_{>0}$ is sufficiently large,
$\\{a_{0},a_{1},a_{2}\\}\subset\mathcal{R}$, there exists $h\in\mathcal{R}[x]$
of degree $5$, with the property that;

$h(m)=a_{0}$, $h^{\prime}(m)=a_{1}$, $h^{\prime\prime}(m)=a_{2}$, $(i)$

$h(m+{1\over m})=h^{\prime}(m+{1\over m})=h^{\prime\prime}(m+{1\over m})=0$
$(ii)$

$|h_{[m,m+{1\over m}]}|\leq C$

for some $C\in\mathcal{R}_{>0}$, independent of $m$ sufficiently large, and,
if $h^{\prime\prime\prime}(m)>0$, $h^{\prime\prime\prime}(x)|_{[m,m+{1\over
m}]}>0$, if $h^{\prime\prime\prime}(m)<0$,
$h^{\prime\prime\prime}|_{[m,m+{1\over m}]}<0$. In particularly;

$\int_{m}^{m+{1\over m}}|h^{\prime\prime\prime}(x)|dx=|a_{2}|$

###### Proof.

If $p(x)$ is any polynomial, we have that $h(x)=(x-(m+{1\over m}))^{3}p(x)$
satisfies condition $(ii)$. Then;

$h^{\prime}(x)=3(x-(m+{1\over m}))^{2}p(x)+(x-(m+{1\over
m}))^{3}p^{\prime}(x)$

$h^{\prime\prime}(x)=6(x-(m+{1\over m}))p(x)+6(x-(m+{1\over
m}))^{2}p^{\prime}(x)+(x-(m+{1\over m}))^{3}p^{\prime\prime}(x)$

$h^{\prime\prime\prime}(x)=6p(x)+18(x-(m+{1\over
m}))p^{\prime}(x)+9(x-(m+{1\over m}))^{2}p^{\prime\prime}(x)$

so we can satisfy $(i)$, by requiring that;

$(a)$. $-{p(m)\over m^{3}}=a_{0}$

$(b)$. ${3p(m)\over m^{2}}-{p^{\prime}(m)\over m^{3}}=a_{1}$

$(c)$. ${-6p(m)\over m}+{6p^{\prime}(m)\over m^{2}}-{p^{\prime\prime}(m)\over
m^{3}}=a_{2}$

which has the solution;

$p(m)=-a_{0}m^{3}$, $p^{\prime}(m)=-3a_{0}m^{4}-a_{1}m^{3}$,
$p^{\prime\prime}(m)=-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3}$

and can be satisfied by the polynomial;

$p(x)={1\over 2}(-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3})(x-m)^{2}$

$+(-3a_{0}m^{4}-a_{1}m^{3})(x-m)+(-a_{0}m^{3})$

$={1\over
2}(-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3})x^{2}+(-m(-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3})$

$+(-3a_{0}m^{4}-a_{1}m^{3}))x+({m^{2}\over
2}(-12a{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3})$

$-m(-3a_{0}m^{4}-a_{1}m^{3})-a_{0}m^{3})$

$=(-6a_{0}m^{5}-3a_{1}m^{4}-{a_{2}\over
2}m^{3})x^{2}+(12a_{0}m^{6}+6a_{1}m^{5}+a_{2}m^{4}-3a_{0}m^{4}$

$-a_{1}m^{3})x+(-6a_{0}m^{7}-3a_{1}m^{6}-{a_{2}\over
2}m^{5}+3a_{0}m^{5}+a_{1}m^{4}-a_{0}m^{3})$

$=(-6a_{0}m^{5}-3a_{1}m^{4}-{a_{2}\over
2}m^{3})x^{2}+(12a_{0}m^{6}+6a_{1}m^{5}+(a_{2}-3a_{0})m^{4}$

$-a_{1}m^{3})x+(-6a_{0}m^{7}-3a_{1}m^{6}+(3a_{0}-{a_{2}\over
2})m^{5}+3a_{0}m^{5}+a_{1}m^{4}-a_{0}m^{3})$

$=ax^{2}+bx+c$ $(*)$

so that;

$h^{\prime\prime\prime}(x)=6(ax^{2}+bx+c)+18(x-(m+{1\over
m}))(2ax+b)+9(x-(m+{1\over m}))^{2}2a$

$=(60a)x^{2}+(24b-72a(m+{1\over m}))x+(6c-18(m+{1\over m})b+18a(m+{1\over
m})^{2})$

and, using the computation $(*)$

$h^{\prime\prime\prime}(x)=(60(-6a_{0}m^{5})+O(m^{4}))x^{2}+(24.12a_{0}m^{6}-72m(-6a_{0}m^{5})$

$+O(m^{5}))x+(6.-6a_{0}m^{7}-18m(12a_{0}m^{6})+18m^{2}(-6a_{0}m^{5})+O(m^{6}))$

$=(-360a_{0}m^{5}+O(m^{4}))x^{2}+(740a_{0}m^{6}+O(m^{5}))x+$

$(-360a_{0}m^{7}+O(m^{6}))$

which, by the quadratic formula, has roots when;

$x={-740a_{0}m^{6}+\\-\sqrt{740^{2}a_{0}^{2}m^{12}-4(-360a_{0}m^{5})(-360a_{0}m^{7})}\over
2.-360a_{0}m^{5}}+O(1)$

$={740m\over 720}+\\-{170m\over 720}+O(1)$

$={19m\over 24}+O(1)$ or ${91m\over 72}+O(1)$

We have that $m>{19m\over 24}$ and $m+{1\over m}<{91m\over 72}$ iff
$m>\sqrt{72\over 19}$, and, clearly, we can ignore the $O(1)$ term for $m$
sufficiently large. In particularly, for sufficiently large $m$,
$h^{\prime\prime\prime}(x)$ has no roots in the interval $[m,m+{1\over m}]$,
so $h^{\prime\prime\prime}|_{[m,m+{1\over m}]}>0$ or
$h^{\prime\prime\prime}|_{[m,m+{1\over m}]}<0$.

We calculate that;

$|h_{[m,m+{1\over m}]}|=|(x-(m+{1\over m}))^{3}p(x)|_{[m,m+{1\over m}]}|$

$\leq{1\over m^{3}}|p(x)|_{[m,m+{1\over m}]}$

$={1\over m^{3}}|[{1\over 2}(-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3})(x-m)^{2}$

$+(-3a_{0}m^{4}-a_{1}m^{3})(x-m)+(-a_{0}m^{3})]|_{[m,m+{1\over m}]}$

$\leq{1\over m^{3}}[{1\over 2}|-12a_{0}m^{5}-6a_{1}m^{4}-a_{2}m^{3}|{1\over
m^{2}}+|-3a_{0}m^{4}-a_{1}m^{3}|{1\over m}+|-a_{0}m^{3}|]$

$\leq{12|a_{0}|m^{5}+6|a_{1}|m^{4}+|a_{2}|m^{3}|\over
m^{5}}+{3|a_{0}|m^{4}+|a_{1}|m^{3}\over m^{4}}+{|a_{0}|m^{3}\over m^{3}}$

$\leq 12|a_{0}|+6|a_{1}|+|a_{2}|+3|a_{0}|+|a_{1}|+|a_{0}|$ $(m>1)$

$\leq 16|a_{0}|+7|a_{1}|+|a_{2}|$

For the final claim, we have, as $h^{\prime\prime\prime}|_{[m,m+{1\over
m}]}>0$ or $h^{\prime\prime\prime}|_{[m,m+{1\over m}]}<0$, that, using the
fundamental theorem of calculus;

$\int_{m}^{m+{1\over m}}|h^{\prime\prime\prime}(x)|dx=|\int_{m}^{m+{1\over
m}}h^{\prime\prime\prime}(x)dx|$

$=|h^{\prime\prime}(m+{1\over
m})-h^{\prime\prime}(m)|=|-h^{\prime\prime}(m)|=|a_{2}|$
∎

######  Lemma 0.6.

If $m\in\mathcal{R}_{>0}$, $\\{a_{0},a_{1},a_{2},a_{3}\\}\subset\mathcal{R}$,
there exists $h\in C^{3}(\mathcal{R})$, with the property that;

$h(m)=a_{0}$, $h^{\prime}(m)=a_{1}$, $h^{\prime\prime}(m)=a_{2}$,
$h^{\prime\prime\prime}(m)=a_{3}$, $(i)$

$h(m+{1\over m})=h^{\prime}(m+{1\over m})=h^{\prime\prime}(m+{1\over
m})=h^{\prime\prime\prime}(m+{1\over m})=0$ $(ii)$

$|h|_{[m,m+1]}\leq C$

where $C\in\mathcal{R}_{>0}$ is independent of $m>1$, and, if $a_{3}>0$,
$h^{\prime\prime\prime}(x)|_{[m,m+{1\over m}]}\geq 0$, $a_{3}<0$,
$h^{\prime\prime\prime}(x)|_{[m,m+{1\over m}]}\leq 0$. In particularly;

$\int_{m}^{m+{1\over m}}|h^{\prime\prime\prime}(x)|dx=|a_{2}|$

###### Proof.

Let $g(x)$ be a polynomial, then it is clear that the polynomial
$h_{1}(x)=(x-(m+{1\over m})^{n}g(x)$, for $n\geq 4$, has the property $(ii)$,
that $h_{1}(m+{1\over m})=h_{1}^{\prime}(m+{1\over
m})=h_{1}^{\prime\prime}(m+{1\over m})=h_{1}^{\prime\prime\prime}(m+{1\over
m})=0$. The condition $(i)$, then amounts to the equations;

$(i)^{\prime}$ ${g(m)\over(-1)^{n}m^{n}}=a_{0}$

$(ii)^{\prime}$
${ng(m)\over(-1)^{n-1}m^{n-1}}+{g^{\prime}(m)\over(-1)^{n}m^{n}}=a_{1}$

$(iii)^{\prime}$
${n(n-1)g(m)\over(-1)^{n-2}m^{n-2}}+{2ng^{\prime}(m)\over(-1)^{n-1}m^{n-1}}+{g^{\prime\prime}(m)\over(-1)^{n}m^{n}}=a_{2}$

$(iv)^{\prime}$
${n(n-1)(n-2)g(m)\over(-1)^{n-3}m^{n-3}}+{3n(n-1)g^{\prime}(m)\over(-1)^{n-2}m^{n-2}}+{3ng^{\prime\prime}(m)\over(-1)^{n-1}m^{n-1}}+{g^{\prime\prime\prime}(m)\over(-1)^{n}m^{n}}=a_{3}$

which we can solve, by requiring that;

$(i)^{\prime\prime}$ $g(m)=(-1)^{n}a_{0}m^{n}$

$(ii)^{\prime\prime}$ $g^{\prime}(m)=(-1)^{n}a_{1}m^{n}+(-1)^{n}a_{0}nm^{n+1}$

$(iii)^{\prime\prime}$
$g^{\prime\prime}(m)=(-1)^{n}a_{2}m^{n}+2(-1)^{n}na_{1}m^{n+1}+(-1)^{n}n(n+1)a_{0}m^{n+2}$

$(iv)^{\prime\prime}$
$g^{\prime\prime\prime}(m)=(-1)^{n}a_{3}m^{n}+3n(-1)^{n}a_{2}m^{n+1}+(-1)^{n}a_{1}n(n+3)m^{n+2}$

$+n(n+1)(n+2)(-1)^{n}a_{0}m^{n+3}$ $(*)$

Let;

$g_{1}(x)=((-1)^{n}a_{3}m^{n}+3n(-1)^{n}a_{2}m^{n+1}+(-1)^{n}a_{1}n(n+3)m^{n+2}$

$+n(n+1)(n+2)(-1)^{n}a_{0}m^{n+3})(x-m)^{3}+((-1)^{n}a_{2}m^{n}+2(-1)^{n}na_{1}m^{n+1}$

$+(-1)^{n}n(n+1)a_{0}m^{n+2})(x-m)^{2}+((-1)^{n}a_{1}m^{n}+(-1)^{n}a_{0}nm^{n+1})$

$(x-m)+((-1)^{n}a_{0}m^{n})$

Then $g_{1}(x)$ satisfies $(*)$, and so does any function of the form
$g_{2}(x)+g_{1}(x)$ where;

$g_{2}(m)=g_{2}^{\prime}(m)=g_{2}^{\prime\prime}(m)=g_{2}^{\prime\prime\prime}(m)=0$

provided $g_{2}\in C^{3}(\mathcal{R})$. In this case, if;

$h(x)=(x-(m+{1\over m})^{n}(g_{2}(x)+g_{1}(x))$

then $h$ satisfies $(i),(ii)$. We have that;

$|x-(m+{1\over m})^{n}g_{1}(x)|_{[m,m+{1\over m}]}\leq{1\over
m^{n}}(|g_{2}|_{[m,m+{1\over m}]}+|g_{1}|_{[m,m+{1\over m}]})$

$\leq{1\over m^{n}}(|g_{2}|_{[m,m+{1\over m}]}+{1\over
m^{n}}|((-1)^{n}a_{3}m^{n}+3n(-1)^{n}a_{2}m^{n+1}+(-1)^{n}a_{1}n(n+3)m^{n+2}$

$+n(n+1)(n+2)(-1)^{n}a_{0}m^{n+3}){1\over
m^{3}}+((-1)^{n}a_{2}m^{n}+2(-1)^{n}na_{1}m^{n+1}$

$+(-1)^{n}n(n+1)a_{0}m^{n+2}){1\over
m^{2}}+((-1)^{n}a_{1}m^{n}+(-1)^{n}a_{0}nm^{n+1})$

${1\over m}+((-1)^{n}a_{0}m^{n})|$

$=|((-1)^{n}a_{3}m^{n}+3n(-1)^{n}a_{2}m^{n+1}+(-1)^{n}a_{1}n(n+3)m^{n+2}$

$+n(n+1)(n+2)(-1)^{n}a_{0}m^{n+3}){1\over
m^{n+3}}+((-1)^{n}a_{2}m^{n}+2(-1)^{n}na_{1}m^{n+1}$

$+(-1)^{n}n(n+1)a_{0}m^{n+2}){1\over
m^{n+2}}+((-1)^{n}a_{1}m^{n}+(-1)^{n}a_{0}nm^{n+1})$

${1\over m^{n+1}}+((-1)^{n}a_{0})|$

$\leq|a_{3}|+3n|a_{2}|+n(n+3)|a_{1}|+n(n+1)(n+2)|a_{0}|+|a_{2}|+2n|a_{1}|+n(n+1)|a_{0}|+|a_{1}|+n|a_{0}|+|a_{0}|$,
$(m\geq 1)$

$={1\over m^{n}}(|g_{2}|_{[m,m+{1\over
m}]}+(n+1)(n^{2}+3n+1)|a_{0}|+(n^{2}+5n+1)|a_{1}+(3n+1)|a_{2}|+|a_{3}|=F$
$(F)$

where $F\in\mathcal{R}_{>0}$ is independent of $m$. Using the product rule,
the condition that $h^{\prime\prime\prime}(x)=0$ in the interval $(m,m+{1\over
m})$, is given by;

$n(n-1)(n-2)(x-(m+{1\over m}))^{n-3}(g_{2}+g_{1})(x)+3n(n-1)(x-(m+{1\over
m}))^{n-2}(g_{2}+g_{1})^{\prime}(x)$

$+3n(x-(m+{1\over m}))^{n-1}(g_{2}+g_{1})^{\prime\prime}(x)+(x-(m+{1\over
m}))^{n}(g_{2}+g_{1})^{\prime\prime\prime}(x)=0$

which, dividing by $(x-(m+{1\over m}))^{n-3}$, reduces to;

$n(n-1)(n-2)(g_{2}+g_{1})(x)+3n(n-1)(x-(m+{1\over
m}))(g_{2}+g_{1})^{\prime}(x)+$

$3n(x-(m+{1\over m}))^{2}(g_{2}+g_{1})^{\prime\prime}(x)+(x-(m+{1\over
m}))^{3}(g_{2}+g_{1})^{\prime\prime\prime}(x)=0$

and;

$n(n-1)(n-2)g_{2}(x)+3n(n-1)(x-(m+{1\over
m}))g_{2}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{2}^{\prime\prime}(x)$

$+(x-(m+{1\over
m}))^{3}g_{2}^{\prime\prime\prime}(x)=-(n(n-1)(n-2)g_{1}(x)+3n(n-1)(x-(m+{1\over
m}))g_{1}^{\prime}(x)$

$+3n(x-(m+{1\over m}))^{2}g_{1}^{\prime\prime}(x)+(x-(m+{1\over
m}))^{3}g_{1}^{\prime\prime\prime}(x))$ $(A)$

Without loss of generality, assuming that;

$-(n(n-1)(n-2)g_{1}(x)+3n(n-1)(x-(m+{1\over
m}))g_{1}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{1}^{\prime\prime}(x)$

$+(x-(m+{1\over
m}))^{3}g_{1}^{\prime\prime\prime}(x))|_{m}=-(n(n-1)(n-2)a_{0}-{3n(n-1)a_{1}\over
m}+{3na_{2}\over m^{2}}$

$-{a_{3}\over m^{3}}\geq 0$

we can choose an analytic function $\phi(x)$ on $[m,m+{1\over m}]$ with;

$(a)$. $\phi(x)\leq-(n(n-1)(n-2)g_{1}(x)+3n(n-1)(x-(m+{1\over
m}))g_{1}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{1}^{\prime\prime}(x)$

$+(x-(m+{1\over m}))^{3}g_{1}^{\prime\prime\prime}(x))$

$(b)$. $\phi(m)=0$

The third order differential equation for $g_{2}$;

$n(n-1)(n-2)g_{2}(x)+3n(n-1)(x-(m+{1\over
m}))g_{2}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{2}^{\prime\prime}(x)$

$+(x-(m+{1\over m}))^{3}g_{2}^{\prime\prime\prime}(x)=\phi(x)$, on $[m,1+m]$
$(B)$

with the requirement that
$g_{2}(m)=g_{2}^{\prime}(m)=g_{2}^{\prime\prime}(m)=0$, has a solution in
$C^{3}([m,m+{1\over m}))$ by Peano’s existence theorem. By the fact $(b)$, we
must have that $g_{2}^{\prime\prime\prime}(m)=0$. Writing the power series for
$\phi$ on $[m,m+{1\over m}]$, as;

$\phi(x)=\sum_{j=0}^{\infty}b_{j}(x-(m+{1\over m}))^{j}$

we can use the method of equating coefficients, to obtain a particular
solution, with;

$g_{2,part}(x)=\sum_{j=0}^{\infty}a_{j,part}(x-(m+{1\over m}))^{j}$, with;

$a_{j,part}={b_{j}\over n(n-1)(n-2)+3n(n-1)j+3nj(j-1)+j(j-1)(j-2)}$, $(j\geq
3)$

$a_{2,part}={b_{2}\over n(n-1)(n-2)+6n(n-1)+3n}$ $a_{1,part}={b_{1}\over
n(n-1)(n-2)+3n(n-1)}$ $a_{0,part}={b_{0}\over n(n-1)(n-2)}$

so that $g_{2,part}$ is analytic as $|a_{j,0}|\leq{|b_{j}|\over n(n-1)(n-2)}$
for $j\geq 0$.

To solve the homogenous Euler equation;

$n(n-1)(n-2)g_{2}(x)+3n(n-1)(x-(m+{1\over
m}))g_{2}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{2}^{\prime\prime}(x)$

$+(x-(m+{1\over m}))^{3}g_{2}^{\prime\prime\prime}(x)=0$ on $[m,m+{1\over m}]$

we can make the substitution $y=m+{1\over m}-x$, to reduce to the equation;

$n(n-1)(n-2)g_{2,m}(y)+3n(n-1)yg_{2,m}^{\prime}(y)+3ny^{2}g_{2,m}^{\prime\prime}(y)+y^{3}g_{2,m}^{\prime\prime\prime}(y)=0$
on $[0,{1\over m}]$

with $g_{2,m}(y)=g_{2}(m+{1\over m}-y)$. Making the further substitution
$y=e^{u}$, and letting $r_{2,m}(u)=g_{2,m}(e^{u})$, we have that;

$r_{2,m}^{\prime}(u)=g_{2,m}^{\prime}(e^{u})e^{u}$

$r_{2,m}^{\prime\prime}(u)=g_{2,m}^{\prime\prime}(e^{u})e^{2u}+g_{2,m}^{\prime}(e^{u})e^{u}$

$r_{2,m}^{\prime\prime\prime}(u)=g_{2,m}^{\prime\prime\prime}(e^{3u})+3g_{2,m}^{\prime\prime}(e^{u})e^{2u}+g_{2,m}^{\prime}(e^{u})e^{u}$

so that;

$n(n-1)(n-2)g_{2,m}(e^{u})+3n(n-1)e^{u}g_{2,m}^{\prime}(e^{u})+3ne^{2u}g_{2,m}^{\prime\prime}(e^{u})+e^{3u}g_{2,m}^{\prime\prime\prime}(e^{u})$

$=n(n-1)(n-2)r_{2,m}(u)+3n(n-1)e^{u}(r_{2,m}^{\prime}(u)e^{-u})+3ne^{2u}((r_{2,m}^{\prime\prime}(u)-g_{2,m}^{\prime}(e^{u})e^{u})e^{-2u})$

$+e^{3u}((r_{2,m}^{\prime\prime\prime}(u)-3g_{2,m}^{\prime\prime}(e^{u})e^{2u}-g_{2,m}^{\prime}(e^{u})e^{u})e^{-3u})$

$=n(n-1)(n-2)r_{2,m}(u)+3n(n-1)r_{2,m}^{\prime}(u)+3nr_{2,m}^{\prime\prime}(u)-3ng_{2,m}^{\prime}(e^{u})e^{u}+r_{2,m}^{\prime\prime\prime}(u)-3g_{2,m}^{\prime\prime}(e^{u})e^{2u}$

$-g_{2,m}^{\prime}(e^{u})e^{u}$

$=n(n-1)(n-2)r_{2,m}(u)+3n(n-1)r_{2,m}^{\prime}(u)+3nr_{2,m}^{\prime\prime}(u)+r_{2,m}^{\prime\prime\prime}(u)-(3n+1)g_{2,m}^{\prime}(e^{u})e^{u}-3g_{2,m}^{\prime\prime}(e^{u})e^{2u}$

$=n(n-1)(n-2)r_{2,m}(u)+3n(n-1)r_{2,m}^{\prime}(u)+3nr_{2,m}^{\prime\prime}(u)+r_{2,m}^{\prime\prime\prime}(u)-(3n+1)r_{2,m}^{\prime}(u)$

$-3e^{2u}((r_{2,m}^{\prime\prime}(u)-g_{2,m}^{\prime}(e^{u})e^{u})e^{-2u})$

$=n(n-1)(n-2)r_{2,m}(u)+(3n^{2}-6n-1)r_{2,m}^{\prime}(u)+3nr_{2,m}^{\prime\prime}(u)+r_{2,m}^{\prime\prime\prime}(u)-3r_{2,m}^{\prime\prime}(u)+3g_{2,m}^{\prime}(e^{u})e^{u}$

$=n(n-1)(n-2)r_{2,m}(u)+(3n^{2}-6n-1)r_{2,m}^{\prime}(u)+3(n-1)r_{2,m}^{\prime\prime}(u)+r_{2,m}^{\prime\prime\prime}(u)+3r_{2,m}^{\prime}(u)$

$=n(n-1)(n-2)r_{2,m}(u)+(3n^{2}-6n+2)r_{2,m}^{\prime}(u)+(3n-3)r_{2,m}^{\prime\prime}(u)+r_{2,m}^{\prime\prime\prime}(u)=0$
$(C)$

We have that;

$(\lambda^{3}+3(n-1)\lambda^{2}+(3n^{2}-6n+2)\lambda+n(n-1)(n-2))^{\prime}=3\lambda^{2}+6(n-1)\lambda+(3n^{2}-6n+2)$

which has roots when $\lambda=-(n-1)+\\-{1\over\sqrt{3}}$, so that, for large
n, the characteristic polynomial of $(C)$ has exactly one real root
$\lambda_{1}$ and 2 complex conjugate non-real roots,
$\\{\lambda_{2}+i\lambda_{3},\lambda_{2}-i\lambda_{3}\\}$. It follows, the
general solution of $(C)$ is given by;

$r_{2,m}(u)=A_{1}e^{\lambda_{1}u}+A_{2}e^{\lambda_{2}u+i\lambda_{3}}+A_{3}e^{\lambda_{2}u-i\lambda_{3}}$

where $\\{A_{1},A_{2},A_{3}\\}\subset\mathcal{C}$, and, we can obtain a real
solution, fitting the corresponding initial conditions, of the form;

$r_{2,m}(u)=B_{1}e^{\lambda_{1}u}+B_{2}e^{\lambda_{2}u}cos(\lambda_{3}u)+B_{3}e^{\lambda_{2}u}sin(\lambda_{3}u)$

where $\\{B_{1},B_{2},B_{3}\\}\subset\mathcal{R}$. It follows that;

$g_{2,m}(y)=r_{2,m}(ln(y))$

$g_{2}(x)=g_{2,m}(m+{1\over m}-x)+g_{2,part}(x)=r_{2,m}(ln(m+{1\over
m}-x))+g_{2,part}(x)$

$=B_{1}e^{\lambda_{1}ln(m+{1\over m}-x)}+B_{2}e^{\lambda_{2}ln(m+{1\over
m}-x)}cos(\lambda_{3}ln(m+{1\over m}-x))$

$+B_{3}e^{\lambda_{2}ln(m+{1\over m}-x)}sin(\lambda_{3}ln(m+{1\over
m}-x))+g_{2,part}(x)$ (on $[m,m+{1\over m}]$)

We have that;

$\lambda_{1}|\lambda_{2}+i\lambda_{3}|^{2}=-n(n-1)(n-2)$

$\lambda_{1}+\lambda_{2}+i\lambda_{3}+\lambda_{2}-i\lambda_{3}=\lambda_{1}+2\lambda_{2}=-3(n-1)$

Computing the highest degree in $n$ term of the characteristic polynomial, we
obtain that, for $\lambda=\alpha n$;

$\alpha^{3}n^{3}+3n(\alpha n)^{2}+3n^{2}(\alpha
n)+n^{3}=n^{3}(\alpha+3)^{3}=0$

so that $\alpha=-3$, $\lambda_{1}=-3n+O(1)$ and
$2\lambda_{2}=-3(n-1)-(-3n+O(1))=3-O(1)=O(1)$

Then, if $B_{1}=0$, we can see that $g_{2}(x)$ has at most a ${1\over
x^{O(1)}}$ singularity at $(m+{1\over m})$, which we can achieve with a
2-parameter family choice for the initial conditions of
$\\{\phi(m),\phi^{\prime}(m),\phi^{\prime\prime}(m)\\}$. If;

$-(n(n-1)(n-2)a_{0}-{3n(n-1)a_{1}\over m}+{3na_{2}\over m^{2}}-{a_{3}\over
m^{3}}\neq 0$

we can clearly achieve this, while satisfying $(a),(b)$. If;

$-(n(n-1)(n-2)a_{0}-{3n(n-1)a_{1}\over m}+{3na_{2}\over m^{2}}-{a_{3}\over
m^{3}}=0$

by requiring the the additional property $(c)$;

$\phi^{\prime}(m)<-(n(n-1)(n-2)g_{1}(x)+3n(n-1)(x-(m+{1\over
m}))g_{1}^{\prime}(x)+3n(x-(m+{1\over m}))^{2}g_{1}^{\prime\prime}(x)$

$+(x-(m+{1\over m}))^{3}g_{1}^{\prime\prime\prime}(x))^{\prime}|_{m}$

we can clearly satisfy $(a),(b)$ as well.

Then, as, for sufficiently large $n$;

$lim_{x\rightarrow 0}({B_{2}x^{n}\over
x^{O(1)}}sin(\lambda_{3}ln(x))+{B_{3}x^{n}\over
x^{O(1)}}cos(\lambda_{3}ln(x)))$

$=lim_{x\rightarrow 0}({B_{2}x^{n}\over
x^{O(1)}}sin(\lambda_{3}ln(x))+{B_{3}x^{n}\over
x^{O(1)}}cos(\lambda_{3}ln(x)))^{\prime}$

$=lim_{x\rightarrow 0}({B_{2}x^{n}\over
x^{O(1)}}sin(\lambda_{3}ln(x))+{B_{3}x^{n}\over
x^{O(1)}}cos(\lambda_{3}ln(x)))^{\prime\prime}$

$=lim_{x\rightarrow 0}({B_{2}x^{n}\over
x^{O(1)}}sin(\lambda_{3}ln(x))+{B_{3}x^{n}\over
x^{O(1)}}cos(\lambda_{3}ln(x)))^{\prime\prime\prime}=0$

we obtain that $(x-(m+{1\over m}))^{n}g_{2}(x)$ extends to a solution in
$C^{3}([m,m+{1\over m}])$, and $(x-(m+{1\over m}))^{n}(g_{2}+g_{1})(x)\in
C^{3}([m,m+{1\over m}])$. By the fact $(a)$, $(A)$ has no solutions in
$(m,m+{1\over m})$, so that $h^{\prime\prime\prime}(x)\geq 0$.

We have that;

$|(x-(m+{1\over m})^{n}g_{2}(x)|_{[m,m+{1\over m}]}=|(x-(m+{1\over
m})^{n}(B_{2}e^{\lambda_{2}ln(m+{1\over m}-x)}cos(\lambda_{3}ln(m+{1\over
m}-x))$

$+B_{3}e^{\lambda_{2}ln(m+{1\over m}-x)}sin(\lambda_{3}ln(m+{1\over
m}-x))+g_{2,part}(x))|$

$\leq|B_{2}|m^{\lambda_{2}-n}+|B_{3}|m^{\lambda_{2}-n}+m^{-n}|g_{2,part}(x)|$

Noting the right hand side of $(a)$ is bounded by $O(m^{n})$ on $[m,m+{1\over
m}]$, we can also choose $\phi(x)$ and $g_{2,part}(x)$ to be of $O(m^{n})$ on
$[m,m+{1\over m}]$, irrespective of the choice of initial conditions
$\\{\phi(m),\phi^{\prime}(m),\phi^{\prime\prime}(m)\\}$. We have that
$\phi^{\prime}(m)=O(m^{n+1})$, in the special case, so that choosing
$\\{B_{2},B_{3}\\}$ sufficiently small, noting;

$(x-(m+{1\over m})^{n}(B_{2}e^{\lambda_{2}ln(m+{1\over
m}-x)}cos(\lambda_{3}ln(m+{1\over m}-x))$

$+B_{3}e^{\lambda_{2}ln(m+{1\over m}-x)}sin(\lambda_{3}ln(m+{1\over
m}-x)))^{\prime}|_{m}=O(max(B_{2}m^{n-\lambda_{2}-1},B_{3}m^{n-\lambda_{2}-1}))$

we can assume that;

$|(x-(m+{1\over m})^{n}g_{2}(x)|_{[m,m+{1\over m}]}\leq D$

where $D\in\mathcal{R}_{>0}$ is independent of $m$, so that, using $(F)$;

$|h(x)|_{[m,m+{1\over m}]}\leq|(x-(m+{1\over m})^{n}g_{1}(x)|_{[m,m+{1\over
m}]}+|(x-(m+{1\over m})^{n}g_{2}(x)|_{[m,m+{1\over m}]}\leq F+D$

For the final claim, we have, as $h^{\prime\prime\prime}|_{[m,m+{1\over
m}]}\geq 0$ or $h^{\prime\prime\prime}|_{[m,m+{1\over m}]}\leq 0$, that, using
the fundamental theorem of calculus, that;

$\int_{m}^{m+{1\over m}}|h^{\prime\prime\prime}(x)|dx=|\int_{m}^{m+{1\over
m}}h^{\prime\prime\prime}(x)dx|$

$=|h^{\prime\prime}(m+{1\over
m})-h^{\prime\prime}(m)|=|-h^{\prime\prime}(m)|=|a_{2}|$

∎

######  Lemma 0.7.

Let $f$ be as in Definition 0.3, then there exists an approximating sequence
$\\{f_{m}:m\in\mathcal{N}\\}$. Moreover, for sufficiently large $m$,
$|\mathcal{F}(f_{m})(k)|\leq{Cm\over|k|^{3}}$, for $|k|>1$, where
$C\in\mathcal{R}_{>0}$, independent of $m$.

###### Proof.

Define $f_{m}$ by setting;

$f_{m}(x)=f(x)$ for $x\in[-m,m]$

$f_{m}(x)=h_{1,m}(x)$, for $x\in[-m-{1\over m},-m]$

$f_{m}(x)=h_{2,m}(x)$, for $x\in[m,m+{1\over m}]$

$f_{m}(x)=0$, for $x\in(-\infty,-m-{1\over m}]$

$f_{m}(x)=0$, for $x\in[m+{1\over m},\infty)$

where $\\{h_{1,m},h_{2,m}\\}\subset C^{2}([-m-{1\over m},-m]\cup[m,m+{1\over
m}])$ are generated by the data $a_{1,m,0}=f(-m)$, $a_{1,m,1}=f^{\prime}(-m)$,
$a_{1,m,2}=f^{\prime\prime}(-m)$, $a_{2,m,0}=f(m)$, $a_{2,m,1}=f^{\prime}(m)$,
$a_{2,m,2}=f^{\prime\prime}(m)$, guaranteed by Lemma 0.5 or Lemma 0.6. By the
constructions of Lemmas 0.5 and 0.6, we have that $(i)$ in Definition 0.3
holds. By the definition, we have $(ii)$. As $f_{m}$ is identically zero on
$(-\infty,-m-{1\over m}]\cup[m+{1\over m},\infty)$, we have that $(iii)$
holds. By the proof of Lemma 0.5, or using Lemma 0.6, we have that;

$max(|h_{1,m}|_{[m,m+{1\over m}]}|,|h_{2,m}|_{[-m-{1\over m},-m]}|)\leq
16||f||_{\infty}+7||f^{\prime}||_{\infty}+||f^{\prime\prime}||_{\infty}$

It follows that;

$\int_{-m-{1\over
m}}^{-m}|f_{m}(x)|dx\leq(16||f||_{\infty}+7||f^{\prime}||_{\infty}+||f^{\prime\prime}||_{\infty})(-m-(-m-{1\over
m}))$

$\leq{D\over m}$

$\int_{m}^{m+{1\over
m}}|f_{m}(x)|dx\leq(16||f||_{\infty}+7||f^{\prime}||_{\infty}+||f^{\prime\prime}||_{\infty})((m+{1\over
m})-m)$

$\leq{E\over m}$

where
$D=E=(16||f||_{\infty}+7||f^{\prime}||_{\infty}+||f^{\prime\prime}||_{\infty})$

proving $(iv)$. For the second claim, we have that;

$\mathcal{F}(f_{m}^{\prime\prime\prime})(k)={1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}f_{m}^{\prime\prime\prime}(x)e^{-ikx}dx$

$={1\over(2\pi)^{1\over
2}}([f_{m}^{\prime\prime}(x)e^{-ikx}]^{\infty}_{-\infty}-ik\int_{-\infty}^{\infty}f_{m}^{\prime\prime}(x)e^{-ikx}dx$

$={-ik\over(2\pi)^{1\over
2}}([f_{m}^{\prime}(x)e^{-ikx}]^{\infty}_{-\infty}-ik\int_{-\infty}^{\infty}f_{m}^{\prime}(x)e^{-ikx}dx)$

$={-k^{2}\over(2\pi)^{1\over
2}}([f_{m}(x)e^{-ikx}]^{\infty}_{-\infty}-ik\int_{-\infty}^{\infty}f_{m}(x)e^{-ikx}dx)$

$={ik^{3}\over(2\pi)^{1\over 2}}\int_{-\infty}^{\infty}f_{m}(x)e^{-ikx}dx)$

so that, for $|k|>1$;

$|\mathcal{F}(f_{m})(k)|=|{1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}f_{m}(x)e^{-ikx}dx|$

$={|{1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}f_{m}^{\prime\prime\prime}(x)e^{-ikx}dx|\over|k|^{3}}$

$\leq{1\over|k|^{3}(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}|f_{m}^{\prime\prime\prime}(x)e^{-ikx}|dx$

$={1\over|k|^{3}(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}|f_{m}^{\prime\prime\prime}(x)|dx$

$={1\over|k|^{3}(2\pi)^{1\over 2}}(\int_{-m-{1\over
m}}^{-m}|h_{1,m}^{\prime\prime\prime}(x)|dx+\int_{-m}^{m}|f^{\prime\prime\prime}(x)|dx+\int_{m}^{m+{1\over
m}}|h_{2,m}^{\prime\prime\prime}(x)|dx)$

$\leq{1\over|k|^{3}(2\pi)^{1\over
2}}(|f^{\prime\prime}(-m)|+2m|f^{\prime\prime\prime}|_{\infty}+|f^{\prime\prime}(m))$

$\leq{1\over|k|^{3}(2\pi)^{1\over
2}}(2||f^{\prime\prime}||_{\infty}+2m||f^{\prime\prime\prime}||_{\infty})$

$\leq{1\over|k|^{3}(2\pi)^{1\over
2}}(2m+2m||f^{\prime\prime\prime}||_{\infty})$,
$(m>||f^{\prime\prime}||_{\infty}$

$={Cm\over|k|^{3}}$

where $C={1\over(2\pi)^{1\over 2}}(2+2||f^{\prime\prime\prime}||_{\infty})$

∎

######  Lemma 0.8.

Let $f\in C^{3}(\mathcal{R})$, with $f$ and ${df\over dx}$ non-oscillatory and
of very moderate decrease, with
$\\{f,f^{\prime},f^{\prime\prime},f^{\prime\prime\prime}\\}$ bounded, then
$\mathcal{F}(f)\in L^{1}(\mathcal{R})$, and we have that;

$f(x)=\mathcal{F}^{-1}(\mathcal{F}(f))(x)$

where, for $g\in L^{1}(\mathcal{R})$;

$\mathcal{F}^{-1}(g)(x)={1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}g(k)e^{ikx}dk$

###### Proof.

By Lemma 0.2, we have that there exists $C\in\mathcal{R}_{>0}$, with
$|\mathcal{F}(f)(k)|\leq{C\over|k|^{2}}$, for sufficiently large $k$, $(*)$.
As $f$ is of very moderate decrease, we have that
$|f|^{2}\leq{D\over|x|^{2}}$, for $|x|>1$, so that, as $f\in
C^{0}(\mathcal{R})$, we have that $f\in L^{2}(\mathcal{R})$. It follows that
$\mathcal{F}(f)\in L^{2}(\mathcal{R})$, and $\mathcal{F}(f)|_{[-n,n]}\in
L^{1}(\mathcal{R})$, for any $n\in\mathcal{N}$, $(**)$. Combining $(*),(**)$,
we obtain that $\mathcal{F}(f)\in L^{1}(\mathcal{R})$. Let
$\\{f_{m}:m\in\mathcal{N}\\}$ be the approximating sequence, given by Lemma
0.7, then, as $f_{m}\in L^{1}(\mathcal{R})$, $\mathcal{F}(f_{m})$ is
continuous and, by Lemma 0.4, converges uniformly to $\mathcal{F}(f)$ on
${\mathcal{R}\setminus\\{0\\}}$. It follows that $\mathcal{F}(f)\in
C^{0}({\mathcal{R}\setminus\\{0\\}})$. As $f_{m}\in C^{2}(\mathcal{R})$ and
$f_{m}^{\prime\prime}\in L^{1}(\mathcal{R})$, we have that there exists
constants $D_{m}\in\mathcal{R}_{>0}$, such that
$|\mathcal{F}(f_{m})(k)\leq{D_{m}\over|k|^{2}}$, for sufficiently large $k$.
Moreover, as $x^{n}f_{m}(x)\in L^{1}(\mathcal{R})$, for $n\in\mathcal{N}$,
$\mathcal{F}(f_{m})\in C^{\infty}(\mathcal{R})$. It follows, the Fourier
inversion theorem $f_{m}=\mathcal{F}^{-1}(\mathcal{F}(f_{m}))$, $(***)$, holds
for each $f_{m}$, see the proof in [6]. By Lemma 0.4, we have that, for
$k_{0}>0$, $|\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k)|\leq{E_{k_{0}}\over m}$,
for $|k|>k_{0}$. As $f$ is of very moderate decrease, we have that,
$f-f_{m}\in L^{2}(\mathcal{R})$ and
$||f-f_{m}||_{L^{2}(\mathcal{R})}\rightarrow 0$,
$||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{2}(\mathcal{R})}\rightarrow 0$ as
$m\rightarrow\infty$, in particularly for $m$ sufficiently large, we have that
$||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{2}(\mathcal{R})}\leq 1$. We then
have that, for $\epsilon>0$;

$||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{1}((-\epsilon,\epsilon)}$

$=<|\mathcal{F}(f)-\mathcal{F}(f_{m})|,1_{(-\epsilon,\epsilon)}>_{L^{2}((-\epsilon,\epsilon))}$

$\leq||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{2}((-\epsilon,\epsilon))}||1_{(-\epsilon,\epsilon)}||_{L^{2}((-\epsilon,\epsilon))}$

$=\sqrt{2}\epsilon^{1\over
2}||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{2}((-\epsilon,\epsilon))}$

$\leq\sqrt{2}\epsilon^{1\over 2}$ $(A)$

for $m$ sufficiently large. As $f$ is oscillatory and of very moderate
decrease, $f$ is monotone in the intervals $(-\infty,-E)$ and $(E,\infty)$,
and, we have that for $|k|>\epsilon$, $m>max(E,1)$, $f$ is monotone in the
intervals $(-\infty,-m)$ and $(m,\infty)$, with
$|f|_{(-\infty,-m)}|\leq{C\over m}$, $|f|_{(m,\infty)}|\leq{C\over m}$, for
some $C\in\mathcal{R}_{>0}$. Then, using the proof of the alternating series
test, see [1] ;

$|\mathcal{F}(f)(k)-\mathcal{F}(m)(k)|={1\over(2\pi)^{1\over
2}}|lim_{r\rightarrow\infty}\int_{m<|x|<r}f(x)e^{-ikx}dx|$

$\leq{1\over(2\pi)^{1\over
2}}|lim_{r\rightarrow\infty}\int_{m<|x|<r}f(x)cos(kx)dx|+{1\over(2\pi)^{1\over
2}}|lim_{r\rightarrow\infty}\int_{m<|x|<r}f(x)sin(kx)dx|$

$={1\over(2\pi)^{1\over 2}}|\int_{m}^{{\pi(n_{k}+{1\over
2})\over|k|}}f(x)cos(kx)dx+\int_{-{\pi(n_{k}+{1\over
2})\over|k|}}^{-m}f(x)cos(kx)dx+\int_{m}^{{m_{k}\pi\over|k|}}f(x)sin(kx)dx$

$+\int_{-{m_{k}\pi\over|k|}}^{-m}f(x)sin(kx)dx+lim_{r\rightarrow\infty}\int_{{\pi(n_{k}+{1\over
2})\over|k|}<|x|<r}f(x)cos(kx)dx$

$+lim_{r\rightarrow\infty}\int_{{m_{k}\pi\over|k|}<|x|<r}f(x)sin(kx)dx|$

$\leq{4\over(2\pi)^{1\over 2}}{C\over m}{\pi\over|k|}+{1\over(2\pi)^{1\over
2}}|lim_{r\rightarrow\infty}\int_{{\pi(n_{k}+{1\over
2})\over|k|}<|x|<r}f(x)cos(|k|x)dx|$

$+{1\over(2\pi)^{1\over
2}}|lim_{r\rightarrow\infty}\int_{{m_{k}\pi\over|k|}<|x|<r}f(x)sin(|k|x)dx|$

$\leq{4\over(2\pi)^{1\over 2}}{C\over m}{\pi\over k_{0}}+{1\over(2\pi)^{1\over
2}}|\int_{{\pi(n_{k}+{1\over 2})\over|k|}}^{{\pi(n_{k}+{3\over
2})\over|k|}}f(x)cos(|k|x)dx|+{1\over(2\pi)^{1\over
2}}|\int_{{m_{k}\pi\over|k|}}^{{(m_{k}+1)\pi\over|k|}}f(x)sin(|k|x)dx|$

$\leq{4\over(2\pi)^{1\over 2}}{C\over m}{\pi\over k_{0}}+{4\over(2\pi)^{1\over
2}}{C\over m}{\pi\over k_{0}}$

$={8\pi^{1\over 2}C\over 2^{1\over 2}k_{0}m}={E_{k_{0}}\over m}$ $(B)$

where $n_{k}=\mu n(n\in\mathcal{Z}_{>0},{\pi(n+{1\over 2})\over|k|}\geq m)$,
$m_{k}=\mu n(n\in\mathcal{Z}_{>0},{\pi n\over|k|}\geq m)$. Then, for
$n\in\mathcal{N}$, $m\in\mathcal{N}$, with $m=n^{3\over 2}$, using Lemma 0.7
and $(A),(B)$, we have, for $x\in\mathcal{R}$, that;

$|\mathcal{F}^{-1}(\mathcal{F}(f))(x)-\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)|=|\mathcal{F}^{-1}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))|$

$={1\over(2\pi)^{1\over
2}}|\int_{-n}^{n}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))e^{ikx}dk+\int_{|k|>n}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))e^{ikx}dk|$

$\leq{1\over(2\pi)^{1\over
2}}(\int_{-n}^{n}|\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k)|dk+\int_{|k|>n}|\mathcal{F}(f)(k)|dk+\int_{|k|>n}|\mathcal{F}(f_{m})(k)|dk)$

$\leq{1\over(2\pi)^{1\over
2}}(\int_{-\epsilon}^{\epsilon}|\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k)|dk+{2nE_{\epsilon}\over
m}+\int_{|k|>n}{C\over|k|^{2}}dk+\int_{|k|>n}{Cm\over|k|^{3}}dk)$

$\leq{1\over(2\pi)^{1\over 2}}(\sqrt{2}\epsilon^{1\over
2}+{2nE_{\epsilon}\over n^{3\over 2}}+{2C\over n}+{Cn^{3\over 2}\over n^{2}})$

$<2\epsilon^{1\over 2}$

for sufficiently large $n$, so that, as $\epsilon>0$ was arbitrary, for
$x\in\mathcal{R}$;

$lim_{m\rightarrow\infty}\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)=\mathcal{F}^{-1}\mathcal{F}(f)(x)$,
$(****)$

and, by Definition 0.3, $(***)$, $(****)$;

$f(x)=lim_{m\rightarrow\infty}f_{m}(x)=lim_{m\rightarrow\infty}\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)=\mathcal{F}^{-1}\mathcal{F}(f)(x)$

∎

######  Remarks 0.9.

The previous lemma proves an inversion theorem for non-oscillatory functions
with very moderate decrease. Such functions belong to $L^{2}(\mathcal{R})$ and
an analogous result for Fourier series can be found in [2], where convergence
is proved almost everywhere rather than everywhere. The corresponding result
for transforms is that;

If $f\in L^{p}(\mathcal{R})$, $p\in(1,2]$, then;

$f(x)=lim_{R\rightarrow\infty}{1\over 2\pi}\int_{|k|\leq
R}\mathcal{F}(f)(k)e^{ixk}dk$

for almost every $x\in\mathcal{R}$.

We can define the function $\mathcal{F}_{1}(f)(k)$, for $k\in\mathcal{R}$,
using the usual Fourier transform transform method, when $f\in
L^{1}(\mathcal{R})$, see [10], and, we can define the function
$\mathcal{F}_{2}(f)(k)$, for $k\in\mathcal{R}_{\neq 0}$, using the limit
definition when $f$ is of very moderate decrease and non-oscillatory, a
particular case of $f\in L^{2}(\mathcal{R})$. However, the operators
$\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ need not commute, so that even if we
show that $\mathcal{F}_{2}\circ\mathcal{F}_{1}=Id$, it doesn’t necessarily
follow that $\mathcal{F}_{1}\circ\mathcal{F}_{2}=Id$. The first claim is, in a
sense, shown in [4];

If $f\in L^{1}(\mathcal{R})\cap C^{0}(\mathcal{R})$ and
$|\mathcal{F}(f)(k)|\leq{A\over|k|}$, for all $k\neq 0$, and
$A\in\mathcal{R}_{\geq 0}$, then;

$f(x)=lim_{R\rightarrow\infty}{1\over 2\pi}\int_{|k|\leq
R}\mathcal{F}(f)(k)e^{ixk}dk$

for every $x\in\mathcal{R}$.

######  Definition 0.10.

We say that $f:\mathcal{R}\rightarrow\mathcal{R}$ is analytic at infinity, if
$f({1\over x})$ has a convergent power series expansion for $|x|<\epsilon$,
$\epsilon>0$. We say that $f$ is eventually monotone, if there exists
$y_{0}\in\mathcal{R}_{>0}$ such that $f|_{(-\infty,-y_{0})}$ and
$f|_{(y_{0},\infty)}$ are monotone.

######  Remarks 0.11.

The class of functions which are analytic at infinity and of very moderate
decrease is important in Physics. The components of the causal field generated
by Jefimenko equations can be shown to have this property if the corresponding
charge and current $(\rho,\overline{J})$ are smooth and have compact support.

A criteria for a function being non-oscillatory or eventually monotone is
provided by the following lemma;

######  Lemma 0.12.

If $f:\mathcal{R}\rightarrow\mathcal{R}$, $f\neq 0$ is analytic and analytic
at infinity, then it has finitely many zeroes. If
$f:\mathcal{R}\rightarrow\mathcal{R}$, ${df\over dx}$ is analytic and analytic
at infinity, and $f\neq c$, where $c\in\mathcal{R}$, then $f$ is non-
oscillatory. If $f:\mathcal{R}\rightarrow\mathcal{R}$, $f$ is analytic for
$|x|>a$, where $a\in\mathcal{R}_{\geq 0}$, analytic at infinity, and
$f|_{|x|>a}\neq 0$ then $f$ has finitely many zeroes in the region $|x|>a+1$.
If $f:\mathcal{R}\rightarrow\mathcal{R}$, ${df\over dx}$ is analytic for
$|x|>a$, analytic at infinity, and $f|_{|x|>a}\neq c$, where
$c\in\mathcal{R}$, then $f$ is eventually monotone.

###### Proof.

For the first claim, suppose that $f$ has infinitely many zeroes. Then we can
find a sequence $\\{y_{i};i\in\mathcal{N}\\}$ with $f(y_{i})=0$. If the
sequence is bounded, then by the Bolzano-Weierstrass Theorem, we can find a
subsequence $\\{y_{i_{k}};k\in\mathcal{N}\\}$, with $f(y_{i_{k}})=0$,
converging to $y\in\mathcal{R}$. By continuity, we have that $f(y)=0$ and $y$
is a limit point of zeroes. As $f$ is analytic, by the identity theorem, it
must be identically zero, contradicting the hypothesis. If the sequence is
unbounded, then we can find a subsequence $\\{y_{i_{k}};k\in\mathcal{N}\\}$,
with $f(y_{i_{k}})=0$, such that $lim_{k\rightarrow\infty}y_{i_{k}}=\infty$ or
$lim_{k\rightarrow\infty}y_{i_{k}}=-\infty$. As $f$ is analytic at $\infty$,
we can find $\epsilon>0$, such that $f(y)=0$ for $|y|>{1\over\epsilon}$. By
the identity theorem again, $f$ is identically zero, contradicting the
hypothesis. It follows that $f$ has finitely many zeroes. For the second
claim, as ${df\over dx}\neq 0$, by the first part, there exist finitely many
points $\\{y_{1},\ldots,y_{n}\\}$, with ${df\over dx}|_{y_{i}}=0$, for $1\leq
i\leq n$, and with $y_{i}<y_{i+1}$, for $1\leq i\leq n-1$. In particularly, we
have that $f|_{(-\infty,y_{1})}$, $f|_{(y_{n},\infty)}$ and
$f|_{(y_{i},y_{i+1})}$ is monotone for $1\leq i\leq n-1$, so that $f$ is non-
oscillatory. For the third claim, suppose that $f$ has infinitely many zeroes
in the region $|x|>a+1$, then we can find a sequence
$\\{y_{i};i\in\mathcal{N}\\}$ with $f(y_{i})=0$ and $|y_{i}>a+1$. As above, if
the sequence is bounded, we can find a subsequence
$\\{y_{i_{k}};k\in\mathcal{N}\\}$, with $f(y_{i_{k}})=0$, converging to
$y\in\mathcal{R}$, with $|y|\geq a+1>a$. As $f$ is analytic for $|x|>a$, by
the identity theorem, it must be identically zero in the region $|x|>a$,
contradicting the hypothesis. If the sequence is unbounded, by the same
argument as above, $f$ must be identically zero in the region $|x|>a$,
contradicting the hypothesis. It follows that $f$ has finitely many zeroes in
the region $|x|>a+1$. For the fourth claim, as ${df\over dx}|_{|x|>a}\neq 0$,
by the first part, there exist finitely many points
$\\{y_{1},\ldots,y_{n}\\}$, with ${df\over dx}|_{y_{i}}=0$, and $|y_{i}|>a+1$,
for $1\leq i\leq n$. Choose $y_{0}>max_{1\leq i\leq n}(|y_{i}|)$, then
${df\over dx}|_{|x|>y_{0}}\neq 0$, so that $f|_{|x|>y_{0}}$ is monotone.

∎

######  Lemma 0.13.

If $f\in C^{0}(\mathcal{R})$ and $f$ is of very moderate decrease, then
defining the Fourier transform by;

$\mathcal{F}(f)(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

we have that $\mathcal{F}(f)(k)$ defines a function $\mathcal{F}(f)(k)\in
L^{2}(\mathcal{R})$, with
$||\mathcal{F}(f)||_{L^{2}(\mathcal{R})}=||f||_{L^{2}(\mathcal{R})}$. There
exists a sequence $\\{r_{n}:n\in\mathcal{N}\\}$ with
$r_{n}\in\mathcal{R}_{>0}$ such that;

${1\over(2\pi)^{1\over
2}}lim_{n\rightarrow\infty}\int_{-r_{n}}^{r_{n}}f(y)e^{-iky}dy$

converges almost everywhere.

###### Proof.

By the hypotheses, we have that $f\in L^{2}(\mathcal{R})$ and if
$f_{r}=f\chi_{(-r,r)}$, where $\chi_{(=r,r)}$ is the characteristic function
on $(-r,r)$, then $f_{r}\in L^{1}(\mathcal{R})\cap L^{2}(\mathcal{R})$ and
$||f-f_{r}||_{L^{2}(\mathcal{R})}\rightarrow 0$, as $r\rightarrow\infty$, in
particular the sequence $\\{f_{r}:r\in\mathcal{R}\\}$ is Cauchy. By a result
in [3], we have that the usual Fourier transform
$\mathcal{F}:L^{1}(\mathcal{R})\cap L^{2}(\mathcal{R})\rightarrow
L^{2}(\mathcal{R})$ is an isometry, so that the sequence
$\\{\mathcal{F}(f_{r}):r\in\mathcal{R}\\}$ is also Cauchy. By the completeness
of $L^{2}(\mathcal{R})$, there exists $g\in L^{2}(\mathcal{R})$, such that
$||g-\mathcal{F}(f_{r})||_{L^{2}(\mathcal{R})}\rightarrow 0$, as
$r\rightarrow\infty$. We have that;

$||\mathcal{F}(f)(k)||_{L^{2}(\mathcal{R})}=||lim_{r\rightarrow\infty}\mathcal{F}(f_{r})||_{L^{2}(\mathcal{R})}$

$=lim_{r\rightarrow\infty}||\mathcal{F}(f_{r})||_{L^{2}(\mathcal{R})}$

$=lim_{r\rightarrow\infty}||f_{r}||_{L^{2}(\mathcal{R})}$

$=||lim_{r\rightarrow\infty}f_{r}||_{L^{2}(\mathcal{R})}$

$=||f||_{L^{2}(\mathcal{R})}$

Finally, by a result in [5], If we choose $r_{n}\in\mathcal{R}_{>0}$ such that
$||f-f_{r_{n}}||_{L^{2}(\mathcal{R})}\leq 2^{-n}$, then
$||\mathcal{F}(f)-\mathcal{F}(f_{r_{n}})||_{L^{2}(\mathcal{R})}\leq 2^{-n}$
and $|\mathcal{F}(f_{r_{n}})(k)-\mathcal{F}(f)(k)|\rightarrow 0$ almost
everywhere.

∎

######  Remarks 0.14.

We will use $\mathcal{F}(f)(k)$ to denote this a.e limit.

######  Lemma 0.15.

Let $f\in C^{2}(\mathcal{R})$ and $\\{f,{df\over dx}\\}\subset C(\mathcal{R})$
be of very moderate decrease, with ${d^{2}f\over dx^{2}}$ of moderate
decrease, then there exists a constant $G\in\mathcal{R}_{>0}$, such that;

$|\mathcal{F}(f)(k)|\leq{G\over|k|^{2}}$

for $|k|>1$. In particularly, $\mathcal{F}(f)(k)\in L^{1}(\mathcal{R})$ is
defined everywhere for $k\neq 0$.

Let $f\in C^{1}(\mathcal{R})$ be of very moderate decrease, with ${df\over
dx}$ of very moderate decrease and oscillatory, then there exists a constant
$G\in\mathcal{R}_{>0}$, such that;

$|\mathcal{F}(f)(k)|\leq{G\over|k|^{2}}$

for sufficiently large $|k|$. In particularly, $\mathcal{F}(f)(k)\in
L^{1}(\mathcal{R})$ is defined for $k\neq 0$.

###### Proof.

We have that ${d^{2}f\over dx^{2}}\in L^{1}(\mathcal{R})$ and;

$|\mathcal{F}({d^{2}f\over dx^{2}})(k)|=|{1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}{d^{2}f\over dx^{2}}(y)e^{-iky}dy|$

$\leq{1\over(2\pi)^{1\over 2}}\int_{-\infty}^{\infty}|{d^{2}f\over
dx^{2}}(y)e^{-iky}|dy$

$\leq{1\over(2\pi)^{1\over 2}}\int_{-\infty}^{\infty}|{d^{2}f\over
dx^{2}}(y)|dy$

$={1\over(2\pi)^{1\over 2}}||{d^{2}f\over dx^{2}}||_{L^{1}(\mathcal{R})}$

$\leq G$

where $G\in\mathcal{R}_{>0}$. Then, using integration by parts and the DCT, we
have, for $k\neq 0$, that;

$\mathcal{F}({d^{2}f\over dx^{2}})(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{d^{2}f\over dx^{2}}(y)e^{-iky}dy$

$={1\over(2\pi)^{1\over 2}}lim_{r\rightarrow\infty}([{df\over
dx}(y)e^{-iky}]^{r}_{-r}+ik\int_{-r}^{r}{df\over dx}(y)e^{-iky}dy)$

$={1\over(2\pi)^{1\over 2}}[{df\over
dx}e^{-iky}]^{\infty}_{-\infty}+ik{1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{df\over dx}e^{-iky}dy$

$={ik\over(2\pi)^{1\over 2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{df\over
dx}e^{-iky}dy$

$={ik\over(2\pi)^{1\over
2}}(lim_{r\rightarrow\infty}([f(y)e^{-iky}]^{r}_{-r}+ik\int_{-r}^{r}f(y)e^{-iky}dy))$

$=-k^{2}({1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

$=-k^{2}\mathcal{F}(f)(k)$

The calculation shows that $\mathcal{F}(f)(k)$ and $\mathcal{F}({df\over
dx})(k)$ are defined for $k\neq 0$, and, for $|k|>1$;

$|\mathcal{F}(f)(k)|\leq{|\mathcal{F}({d^{2}f\over dx^{2}})(k)|\over|k|^{2}}$

$\leq{G\over|k|^{2}}$

We have that $\mathcal{F}(f)(k)\in L^{2}(\mathcal{R})$, so that
$\mathcal{F}(f)|_{(-1,1)}(k)\in L^{2}(\mathcal{R})\subset L^{1}(\mathcal{R})$,
and $\mathcal{F}(f)(k)\in L^{1}(\mathcal{R})$
.

For the second claim, we have that, ${df\over dx}\in L^{1}(\mathcal{R})$, and,
similarly to the above, $|\mathcal{F}({df\over dx})(k)|\leq H$, for some
$H\in\mathcal{R}_{>0}$ and $k\in\mathcal{R}$. Using the DCT, for $k\neq 0$;

$\mathcal{F}({df\over dx})(k)={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}{df\over dx}(y)e^{-iky}dy$

$={1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}([f(y)e^{-iky}]^{r}_{-r}+ik\int_{-r}^{r}f(y)e^{-iky}dy)$

$={1\over(2\pi)^{1\over
2}}[f(y)e^{-iky}]^{\infty}_{-\infty}+ik{1\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

$={ik\over(2\pi)^{1\over
2}}lim_{r\rightarrow\infty}\int_{-r}^{r}f(y)e^{-iky}dy$

$=ik\mathcal{F}(f)(k)$

showing that $\mathcal{F}(f)(k)$ is defined for $k\neq 0$, and, for $|k|>1$;

$|\mathcal{F}(f)(k)|\leq{|\mathcal{F}({df\over dx})(k)|\over|k|}$ $(*)$

$\leq{H\over|k|}$

We have that ${df\over dx}|_{(-1,1)}$ is non-oscillatory, and, moreover, there
are at most ${2\over\delta}$ monotone intervals. Using the proof in [7], we
get, for sufficiently large $|k|$;

$|\int_{-1}^{1}{df\over dx}e^{-iky}dy|<{E_{1}\over|k|}$

where $E_{1}\leq{4C\pi\over\delta}$ and $C=max_{x\in\mathcal{R}}|{df\over
dx}|$. As ${df\over dx}$ is of moderate decrease, we have that $|{df\over
dx}|\leq{D\over|x|^{2}}$, for $|x|>1$, where $D\in\mathcal{R}_{>0}$. Then,
using the proof in [7] again, and the definition of oscillatory, we have that,
for sufficiently large $|k|$;

$|\int_{|y|>1}{df\over
dx}e^{-iky}dy|<({2\pi\over|k|}\sum_{|y_{i}|>1}{D\over|y_{i}|^{2}})$

$\leq({4\pi\over|k|}\sum_{n\in\mathcal{Z}_{\geq
0}}{D\over(y_{i_{0}}+n\delta)^{2}})$

$\leq{4\pi D\over\delta|k|}\int_{y_{i_{0}}}^{\infty}{dx\over x^{2}}$

$={4\pi D\over\delta|k|y_{i_{0}}}$

$\leq{4\pi D\over\delta|k|}$

where $|y_{i_{0}}|\geq 1$ and $|y_{i_{0}}|\leq|y_{i}|$, for all $|y_{i}|\geq
N$. It follows that;

$|\mathcal{F}({df\over dx})(k)|=|\int_{-1}^{1}{df\over
dx}e^{-iky}dy+\int_{|y|>1}{df\over dx}e^{-iky}dy|$

$\leq|\int_{-1}^{1}{df\over dx}e^{-iky}dy|+|\int_{|y|>1}{df\over
dx}e^{-iky}dy|$

$\leq{E_{1}\over|k|}+{4\pi D\over\delta|k|}$

$\leq{4C\pi\over\delta|k|}+{4\pi D\over\delta|k|}$

$={R\over|k|}$ $(**)$

where $R={4\pi(C+D)\over\delta}$. Combining $(*)$ and $(**)$, we obtain that,
for sufficiently large $|k|$;

$|\mathcal{F}(f)(k)|\leq{|\mathcal{F}({df\over dx})(k)|\over|k|}$

$\leq{G\over|k|^{2}}$

where $G={1\over(2\pi)^{1\over 2}}R$

∎

######  Lemma 0.16.

Let $f\in C^{3}(\mathcal{R})$, with $\\{f,{df\over dx}\\}$ of very moderate
decrease, ${d^{2}f\over dx^{2}}$ of very moderate decrease, and
$\\{f,f^{\prime},f^{\prime\prime},f^{\prime\prime\prime}\\}$ bounded, then
$\mathcal{F}(f)\in L^{1}(\mathcal{R})$, and we have that;

$f(x)=\mathcal{F}^{-1}(\mathcal{F}(f))(x)$

where, for $g\in L^{1}(\mathcal{R})$;

$\mathcal{F}^{-1}(g)(x)={1\over(2\pi)^{1\over
2}}\int_{-\infty}^{\infty}g(k)e^{ikx}dk$

The same result holds if $f\in C^{3}(\mathcal{R})$, with $f$ of very moderate
decrease, ${df\over dx}$ of very moderate decrease and oscillatory, and
$\\{f,f^{\prime},f^{\prime\prime},f^{\prime\prime\prime}\\}$ bounded.

###### Proof.

By Lemma 0.15, we have that there exists $E\in\mathcal{R}_{>0}$, with
$|\mathcal{F}(f)(k)|\leq{E\over|k|^{2}}$, for sufficiently large $k$, $(*)$.
As $f$ is of very moderate decrease, we have that
$|f|^{2}\leq{D\over|x|^{2}}$, for $|x|>1$, so that, as $f\in
C^{0}(\mathcal{R})$, we have that $f\in L^{2}(\mathcal{R})$. It follows that
$\mathcal{F}(f)\in L^{2}(\mathcal{R})$, and $\mathcal{F}(f)|_{[-n,n]}\in
L^{1}(\mathcal{R})$, for any $n\in\mathcal{N}$, $(**)$. Combining $(*),(**)$,
we obtain that $\mathcal{F}(f)\in L^{1}(\mathcal{R})$. Let
$\\{f_{m}:m\in\mathcal{N}\\}$ be the approximating sequence, given by Lemma
0.7, then, as $f_{m}\in L^{1}(\mathcal{R})$, $\mathcal{F}(f_{m})$ is
continuous. As $f_{m}\in C^{2}(\mathcal{R})$ and $f_{m}^{\prime\prime}\in
L^{1}(\mathcal{R})$, we have that there exists constants
$D_{m}\in\mathcal{R}_{>0}$, such that
$|\mathcal{F}(f_{m})(k)\leq{D_{m}\over|k|^{2}}$, for sufficiently large $k$.
Moreover, as $x^{n}f_{m}(x)\in L^{1}(\mathcal{R})$, for $n\in\mathcal{N}$,
$\mathcal{F}(f_{m})\in C^{\infty}(\mathcal{R})$. It follows, the Fourier
inversion theorem $f_{m}=\mathcal{F}^{-1}(\mathcal{F}(f_{m}))$, $(***)$, holds
for each $f_{m}$, see the proof in [6]. As $f$ is of very moderate decrease,
we have that, $f-f_{m}\in L^{2}(\mathcal{R})$ and
$||f-f_{m}||_{L^{2}(\mathcal{R})}\rightarrow 0$,
$||\mathcal{F}(f)-\mathcal{F}(f_{m})||_{L^{2}(\mathcal{R})}\rightarrow 0$ as
$m\rightarrow\infty$. We then have that, for $n\in\mathcal{N}$,
$m\in\mathcal{N}$;

$|\int_{-n}^{n}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))dk|$

$\leq(2n)^{1\over 2}||\mathcal{F}(f-f_{m})||_{L}^{2}$

$\leq(2n)^{1\over 2}||f-f_{m}||_{L^{2}}$

$=(2n)^{1\over 2}(\int_{|x|>m}f^{2}(x)dx)^{1\over 2}$

$\leq(2n)^{1\over 2}(\int_{|x|>m}({D^{2}\over x^{2}})(x)dx)^{1\over 2}$

$=(2n)^{1\over 2}({2D^{2}\over m})^{1\over 2}$

$=2D{n^{1\over 2}\over m^{1\over 2}}$ $(A)$

where $D\in\mathcal{R}_{>0}$. Then, for $n\in\mathcal{N}$, $m\in\mathcal{N}$,
with $m=[n^{3\over 2}]$, using Lemma 0.7 and $(A)$, we have, for
$x\in\mathcal{R}$, that;

$|\mathcal{F}^{-1}(\mathcal{F}(f))(x)-\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)|=|\mathcal{F}^{-1}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))|$

$={1\over(2\pi)^{1\over
2}}|\int_{-n}^{n}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))e^{ikx}dk+\int_{|k|>n}(\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k))e^{ikx}dk|$

$\leq{1\over(2\pi)^{1\over
2}}(|\int_{-n}^{n}|\mathcal{F}(f)(k)-\mathcal{F}(f_{m})(k)|dk|+\int_{|k|>n}|\mathcal{F}(f)(k)|dk+\int_{|k|>n}|\mathcal{F}(f_{m})(k)|dk)$

$\leq{1\over(2\pi)^{1\over 2}}(2D{n^{1\over 2}\over m^{1\over
2}}+\int_{|k|>n}{E\over|k|^{2}}dk+\int_{|k|>n}{Cm\over|k|^{3}}dk)$

$\leq{1\over(2\pi)^{1\over 2}}(2D{n^{1\over 2}\over m^{1\over 2}}+{2E\over
n}+{Cn^{3\over 2}\over n^{2}})$

$\leq{1\over(2\pi)^{1\over 2}}(2D({n\over[n^{3\over 2}]})^{1\over 2}+{2E\over
n}+{Cn^{3\over 2}\over n^{2}})$

$<\epsilon$

for sufficiently large $n$, so that, as $\epsilon>0$ was arbitrary, for
$x\in\mathcal{R}$;

$lim_{m\rightarrow\infty}\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)=\mathcal{F}^{-1}\mathcal{F}(f)(x)$,
$(****)$

and, by Definition 0.3, $(***)$, $(****)$;

$f(x)=lim_{m\rightarrow\infty}f_{m}(x)=lim_{m\rightarrow\infty}\mathcal{F}^{-1}(\mathcal{F}(f_{m}))(x)=\mathcal{F}^{-1}\mathcal{F}(f)(x)$

∎

## References

  * [1] Introduction to Real Analysis, R. Bartle and D. Sherbert, Wiley, (1982).

  * [2] On convergence and growth of partial sums of Fourier series, L. Carlsen, Acta Mathematica, 116 (1): 135–157, (1966).

  * [3] Lecture Notes on Fourier Transforms, W. Faris, Department of Mathematics, University of Arizona, (2008)

  * [4] Fourier Analysis, T. Korner, CUP, (1988)

  * [5] Maths Stackexchange, https://math.stackexchange.com/questions/138043/does-convergence-in-lp-imply-convergence-almost-everywhere, (2022)

  * [6] A Simple Proof of the Fourier Inversion Theorem Using Nonstandard Analysis, Tristram de Piro, available at http://www.curvalinea.net (19), (2013)

  * [7] Simple Proofs of the Riemann-Lebesgue Lemma using Nonstandard Analysis, Tristram de Piro, available at http://www.curvalinea.net (51), (2019)

  * [8] Some Arguments for the Wave Equation in Quantum Theory, Tristram de Piro, Open Journal of Mathematical Sciences, available at http://www.curvalinea.net (58), (2021)

  * [9] Some Arguments for the Wave Equation in Quantum Theory 4, Tristram de Piro, available at http://www.curvalinea.net (64), (2023), in progress.

  * [10] Fourier Analysis, An Introduction, E. Stein and R. Sakarchi, Princeton Lectures in Mathematics, (2003).