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?


  • 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, 2017 at 21:54
  • 1
    $\begingroup$ There should be no double integral involved. $f(x)$ is a number. $\endgroup$
    – user99914
    Dec 9, 2017 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. $\endgroup$ Dec 9, 2017 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
    Dec 9, 2017 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 $$ \left|\int_{-\infty}^{+\infty}\frac{g(s)}{\cosh s}\,ds\right|\leq \|g\|_2\cdot\sqrt{\int_{-\infty}^{+\infty}\frac{ds}{\cosh^2 s}}=\sqrt{2}\,\|g\|_2. $$ $\endgroup$ Dec 9, 2017 at 23:48

1 Answer 1


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.

  • $\begingroup$ Many thanks for your nice and knowledgeable answer! I didn't know that! Thanks! $\endgroup$
    – Nikita
    Dec 9, 2017 at 22:34

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