Show that Operator is not bounded

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 $$k$$.
• @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. – DisintegratingByParts Oct 14 '18 at 20:14
• 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? – EpsilonDelta Oct 14 '18 at 21:08