2
$\begingroup$

Let $A_f\colon L^2([0,1]) \rightarrow L^2([0,1]^2)$ be given by

$(Af)(x)=\displaystyle\sum_{k=1}^\infty \frac{1}{k\pi^2}\int_0^1 f(x)\sin(k\pi x)\sinh(k\pi y)\,\mathrm{d}x\sin(k\pi x)$

where $f\in C^1([0,1])$.

I want to show that $A$ is not bounded.

What I can tell for sure is, that

$\displaystyle \int_0^1 f(x)\sin(k\pi x)\sinh(k\pi y)\,\mathrm{d}x$ is bounded. This follows from a theorem which requires the absolute value of the kernel to be bounded in $L^1([0,1])$ for both $x$ and $y$.

Now I would have to show that

$\displaystyle (Bg)(x,y)=\sum_{k=1}^\infty \frac{1}{k\pi^2} g(y)\sin(k\pi x)$ is in general not bounded in $L^2([0,1]^2)$ for $g\in L^2([0,1])$

How can I do this? I have not much background in functional analysis, I only know how to deal with product spaces to be exact. I can only image that the $L^2$ norm would be integrating over two variables.

$\endgroup$
2
$\begingroup$

$$ \left\{\frac{1}{\sqrt{2}}\sin(k\pi x)\right\}_{n=1}^{\infty} $$ is an orthonormal basis of $L^2[0,1]$. The functions $\sinh(k\pi y)\sin(k\pi x)$ are mutually orthogonal in $L^2([0,1]\times[0,1])$ because of the $\sin$ terms, and $$ \|\sinh(k\pi y)\sin(k\pi x)\|^2 = 2\int_{0}^{1}\sinh(k\pi y)^2dy \\ = 2\int_0^1 \frac{e^{2k\pi y}+e^{-2k\pi y}-2}{4}dy \\ = \int_{0}^{1}(\cosh(2k\pi y)-1) dy \\ =\left.\frac{\sinh(2k\pi y)}{2k\pi}\right|_{0}^{1}-1 \\ = \frac{\sinh(2k\pi)}{2k\pi}-1. $$ Your map is unbounded because $\frac{1}{k}\|\sinh(k\pi y)\sin(k\pi x)\|$ is an unbounded sequence in $k$.

$\endgroup$
  • $\begingroup$ What argument did you use, to interchange infinite sum and integral? What norm did you use on the product space? $\endgroup$ – EpsilonDelta Oct 14 '18 at 19:13
  • $\begingroup$ @EpsilonDelta : Using $L^2[0,1]$ on the domain of the operator and $L^2([0,1]^2)$ on the range. The specific $\sin$ sequence, which is bounded, is mapped to the $\sin\sinh$ sequence which is unbounded. $\endgroup$ – DisintegratingByParts Oct 14 '18 at 20:14
  • $\begingroup$ I think you misunderstood me. I meant which norm you used on the space $L^2([0,1]^2)$. Also at some point you had to interchange the sum with the integral for making your estimates, how can you argument the interchange? $\endgroup$ – EpsilonDelta Oct 14 '18 at 21:08

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.