# $f(x) = \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds$ is bounded for $x \in L_2(0 , + \infty)$. And find its norm.

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!

• 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 $|\langle x,y \rangle| \le \|x\|_2 \|y\|_2$. Dec 9 '17 at 21:54
• There should be no double integral involved. $f(x)$ is a number.
– user99914
Dec 9 '17 at 21:55
• 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. Dec 9 '17 at 21:56
• 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. Dec 9 '17 at 22:08
• $\frac{1}{\cosh x}\in\mathcal{S}(\mathbb{R})$ 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 $$\left|\int_{-\infty}^{+\infty}\frac{g(s)}{\cosh s}\,ds\right|\leq \|g\|_2\cdot\sqrt{\int_{-\infty}^{+\infty}\frac{ds}{\cosh^2 s}}=\sqrt{2}\,\|g\|_2.$$ Dec 9 '17 at 23:48

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 $\|f\|=\|y\|_{L^2(0,+\infty)}$.