2
$\begingroup$

The question is as follows:

Prove the linear boundedness of $f(x) = \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds$, for $x \in L_2(0 , + \infty)$. And find its norm.

$\textbf{some effort:}$

For to show it is bounded, we have $||fx||^2 = \int_{0}^{+\infty} \mid \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds \mid^2 dt$

$\hspace{7.6cm} \leq \int_{0}^{+\infty} \int_{0}^{+\infty} \mid \frac{x(s)}{\cosh(s)} \mid^2 ds dt$

$\hspace{7.6cm} = \int_{0}^{+\infty} \int_{0}^{+\infty} \frac{1}{\cosh(s)} \mid x(s) \mid^2 ds dt$

$\hspace{7.6cm} = \int_{0}^{+\infty} \frac{1}{\cosh(s)} \mid x(s) \mid^2 \int_{0}^{s} 1 dt ds $

$\hspace{7.6cm} = \int_{0}^{+\infty} \frac{s}{\cosh(s)} \mid x(s) \mid^2 ds $

$\hspace{7.6cm} \leq \bigg(\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds\bigg)^{\frac{1}{2}} \bigg(\int_{0}^{+\infty} \mid x(s) \mid^2 ds \bigg)^{\frac{1}{2}} $

Which will imply that $||f|| \leq \bigg(\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds\bigg)^{\frac{1}{4}} $.

If I am right until now, then we need to calculate $\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds$.

Can you please let me know if my calculation is far behind being correct?

And can you please let me know that how can I find its norm?

Thanks!

$\endgroup$
5
  • 2
    $\begingroup$ Your linear operator is $x \mapsto \langle x,y \rangle = \int_0^\infty x(s) y(s)ds $ where $y(s)=\frac{1}{\cosh(s)}$. For $\|x\|_2 < \infty$ then $|\langle x,y \rangle| \le \|x\|_2 \|y\|_2$. $\endgroup$ – reuns Dec 9 '17 at 21:54
  • 1
    $\begingroup$ There should be no double integral involved. $f(x)$ is a number. $\endgroup$ – user99914 Dec 9 '17 at 21:55
  • 1
    $\begingroup$ Your $f$ is a functional, not an operator. Therefore the norm is the $L^2$-norm of the function $g(x) = \frac{1}{\cosh x}$. You should be able to check that the norm therefore is 1. $\endgroup$ – Hans Engler Dec 9 '17 at 21:56
  • 2
    $\begingroup$ Using $x$ both as a variable on the argument of $f$ and as a function on the integrand is a nice and evil way to confuse students. $\endgroup$ – Pedro Tamaroff Dec 9 '17 at 22:08
  • 1
    $\begingroup$ $\frac{1}{\cosh x}\in\mathcal{S}(\mathbb{R})$ is a renowned almost-fixed point of the Fourier transform: $$ \int_{-\infty}^{+\infty}\frac{e^{-2\pi i \xi x}}{\cosh x}\,dx = \frac{\pi}{\cosh(\pi^2 \xi)} $$ hence if $g(s)\in L^2(\mathbb{R})$ we have $$ \left|\int_{-\infty}^{+\infty}\frac{g(s)}{\cosh s}\,ds\right|\leq \|g\|_2\cdot\sqrt{\int_{-\infty}^{+\infty}\frac{ds}{\cosh^2 s}}=\sqrt{2}\,\|g\|_2. $$ $\endgroup$ – Jack D'Aurizio Dec 9 '17 at 23:48
1
$\begingroup$

Ok, it seems that you are misunderstanding what you want to do. If $f$ is defined as you did then it is a linear operator $f:L^2(0,+\infty)\rightarrow \mathbb R$, so when you fix $x^* \in L^2(0,+\infty) \rightarrow f(x^*)=\int_{0}^{+\infty} \frac{x^*(s)}{cosh(s)}ds \in \mathbb R$! So it makes no sense write $\|fx\|^2=\int_{0}^{+\infty}|\int_{0}^{+\infty}\frac{x(s)}{cosh(s)}ds|^2dt$, in fact $f(x) \notin L^2$.

The natural norm of $f(x)$ is the natural norm in $\mathbb R$ which is the absolute value, now we want to show that: $$|f(x)|\le k \cdot \|x\|_{L^2(0,+\infty)}, \forall x \in L^2(0,+\infty).$$ Now notice that the function $y(t)=\frac{1}{cosh(t)} \in L^2(0,+\infty)$ so thanks to cauchy schwarz's inequality we have: $$|f(x)|=<x,y>_{L^2(0,+\infty)}\le \|x\|_{L^2(0,+\infty)} \|y\|_{L^2(0,+\infty)}$$ so $f $ is bounded, now take $x=y$:

$$|f(y)|=<y,y>=\|y\|^2_{L^2(0,+\infty)}=\|y\|_{L^2(0,+\infty)}\|y\|_{L^2(0,+\infty)}.$$ So we also have that $\|f\|=\|y\|_{L^2(0,+\infty)}$.

Leave aside the exercise, it's very important that you understand the first part because otherwise you won't be able to solve these kind of problems.

$\endgroup$
1
  • $\begingroup$ Many thanks for your nice and knowledgeable answer! I didn't know that! Thanks! $\endgroup$ – Nikita Dec 9 '17 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.