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.


$$ \left\{\frac{1}{\sqrt{2}}\sin(k\pi x)\right\}_{n=1}^{\infty} $$ 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 $k$.

  • $\begingroup$ 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

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