Integral exists if integrand decays faster than any polynomial.

Question

I am in the process of proving the Fourier transform is closed on the Schwartz class. And I have trouble showing Schwartz functions are $L^1(\mathbb{R}^n)$. So the simple question is suppose $f$ is real valued with it and its derivatives decay faster than any polynomial

for all $m,n\geq0$ $\sup_{x\in\mathbb{R}} |x|^n|f^{m}(x)|< \infty$ then $\int_{-\infty}^{\infty}f< \infty$

Answer 1

Use $$\begin{align} \int|f|\,dx &= \int (1+x^2)^{-2} (1+x^2)^{2}|f|\,dx \\ &\le\int (1+x^2)^{-2} \sup_{x\in\mathbb{R}}(1+x^2)^{2}|f|dx \\ &\le\sup_{x\in\mathbb{R}}(1+x^2)^{2}|f| \int (1+x^2)^{-2}\,dx\\ &<\infty. \end{align}$$

Answer 2

All numbered equations with no apostrophes are referenced from Baby Rudin Ch. 11.

Note that $\mu(\mathbb R_i \cap \mathbb R)=\mu(\mathbb R_i)\mu(\mathbb R)=|x|^{i-1}|f^{(i-1)}(x)|$ because $\cap$ translates into "multiplication" ($\cup$ translates into "addition"; see 11.2 Definition in Baby Rudin).

Also absolute value is used because the measure $\mu$ is a type of metric and absolute value is the most typically used metric and the one we need. See 11.9 Definition in Baby Rudin.

"11.16 Theorem: If $f$ is measurable, then $|f|$ is measurable."

Or rather

"11.16 Theorem: $|f|$ is measurable if $f$ is measurable."

Since theorems are restatements of definitions, and "if" in definitions is understood to be "if and only if," the next corollary follows:

11.16 Corollary: $f$ is measurable if and only if $|f|$ is measurable.

Assume for all $m,n\ge0$, $\underset{x\in\mathbb R}{\text{sup}}\ |x|^n|f^{(m)}(x)|<\infty$.

$I_\mathbb R(s)=\sum_{i=1}^kc_i\mu(\mathbb R\cap \mathbb R_i)\qquad\qquad(52)$

$\int_\mathbb R s \ d\mu=I_\mathbb R(s) \qquad\qquad\qquad\qquad\quad(54)$

$\Rightarrow s=\frac{\partial}{\partial\mu(\mathbb R)}I_\mathbb R(s)$
$\qquad=\frac{\partial}{\partial\mu(\mathbb R)}[\sum_{i=1}^kc_i\mu(\mathbb R\cap \mathbb R_i)]$
$\qquad=\sum_{i=1}^kc_i\frac{\partial}{\partial\mu(\mathbb R)}[\mu(\mathbb R\cap \mathbb R_i)]$
$\qquad=\sum_{i=1}^kc_i|x|^{i-1}\frac{d}{dx}|f^{(i-1)}(x)|$ $(\mu({\mathbb R_i})=|x|^{i-1}$ since ${\mathbb R_i}=\{x \ | \ s (x)=c_i\}$, i.e., every $x\in {\mathbb R_i}$)
$\qquad=\sum_{i=1}^kc_i|x|^{i-1}\frac{d}{dx}f^{(i-1)}(x)$ (since by definition $\mu(\mathbb R)$ is non-negative and hence $f^{(i-1)}$ is non-negative since it is measurable by 11.16 Corollary)
$\qquad=\sum_{i=1}^kc_i|x|^{i-1}f^{(i)}(x)$
$\qquad=\sum_{i=1}^kc_i|x|^{i-1}|f^{(i)}(x)|$ (since $f^{(i-1)}$ is non-negative, thus $f^{(i)}$ is non-negative) $\qquad(*)$

$s(x)=\sum_{i=1}^kc_iK_{\mathbb R_i}(x)\qquad(51)$

where

$K_{\mathbb R_i}= \begin{cases} 1 \quad (x\in {\mathbb R_i}), \\ 0 \quad (x\notin {\mathbb R_i}) \end{cases}\quad(48)$

since every $x\in \mathbb R\cap {\mathbb R_i}$, then every $x\in {\mathbb R_i}$. So $K_{\mathbb R_i}(x)=1$ and (51) becomes

$s(x)=\sum_{i=1}^kc_i$. $\qquad (51')$

Also since ${\mathbb R_i}=\{x \ | \ s (x)=c_i\}$ and every $x\in {\mathbb R_i}$, (51') becomes

$s(x)=c_i$. $\qquad\qquad\ (51'')$

Setting $(*)$ equal to (51) we have

$\sum_{i=1}^kc_i|x|^{i-1}|f^{(i)}(x)|=\sum_{i=1}^kc_iK_{\mathbb R_i}(x)$,

which implies

$|x|^{i-1}|f^{(i)}(x)|=K_{\mathbb R_i}(x)$
$\qquad\qquad\qquad=1$.
Integrating gives

$\int|x|^{i-1}|f^{(i)}(x)| \ d\mu(\mathbb R)=\int d\mu(\mathbb R)$

or

$|x|^{i-1}\int|f^{(i)}(x)| \ dx=\int dx$
$\qquad\qquad\qquad\qquad=x$

which implies

$x=|x|^{i-1}\int f^{(i)}(x) \ dx\qquad$ (since $f^{(i)}$ is non-negative as seen in $(*)$)
$\quad=|x|^{i-1} f^{(i-1)}(x) \qquad$

$\Rightarrow|x|^{i-1}|f^{(i-1)}(x)|=x\qquad$ (since $f^{(i-1)}$ is non-negative as seen in $(*)$)
$\qquad=|x|$ (since $\mu(\mathbb R \cap {\mathbb R_i})=|x|^{i-1}|f^{(i-1)}(x)|$ is non-negative by definition).$\quad(\star)$

Thus,

$\int_\mathbb R f \ d\mu=$ sup $I_\mathbb R(s)$ $\qquad\qquad\qquad\qquad(53)$
$\quad\qquad\ =$ sup $\sum_{i=1}^kc_i\mu(\mathbb R\cap \mathbb R_i)$
$\quad\qquad\ = \underset{x\in \mathbb R}{\text{sup}}\sum_{i=1}^kc_i|x|^{i-1}|f^{(i-1)}(x)|$
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}\{\sum_{i=1}^1c_i|x|^{i-1}|f^{(i-1)}(x)|\text{, ... ,}\sum_{i=1}^jc_i|x|^{i-1}|f^{(i-1)}(x)|\}$
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}\{\sum_{i=1}^1c_i|x|\text{, ... , }\sum_{i=1}^jc_i|x|\}\qquad\qquad$ (by $(\star)$)
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}\{|x|\sum_{i=1}^1c_i\text{, ... , }|x|\sum_{i=1}^jc_i\}$
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}\{|x|c_i\text{, ... , }|x|c_i\}\qquad\qquad$ (by $(51')$ and $(51'')$)
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}$ $c_i|x|$
$\quad\qquad\ =\underset{x\in\mathbb R}{\text{sup}}$ $c_i|x|^{i-1}|f^{(i-1)}(x)|$.$\qquad($by $(\star)$)

Since, by assumption, $m,n\ge0$, $\underset{x\in\mathbb R}{\text{sup}}\ |x|^n|f^{(m)}(x)|<\infty$ and $c_i$ is a constant, it follows that

$\int_{-\infty}^\infty f \ d\mu=\int_\mathbb R f \ d\mu$
$\quad\qquad\quad =\underset{x\in\mathbb R}{\text{sup}}$ $c_i|x|^{i-1}|f^{(i-1)}(x)|$
$\quad\qquad\quad<\infty$

References

Rudin, W. (2018). Chapter 11: The Lebesgue Theory. In W. Rudin, Principles of Mathematical Analysis (pp. 300-334). McGraw Hill Education.