# What happens if we rotate the kernel of an integral operator?

Given an integral operator $K$ on $L^2(\mathbb R)$ with kernel $k(x, y)$, consider the integral operator $L$ on $L^2(\mathbb R)$, whose kernel has the form $k(\alpha x+\beta y, \gamma x+\delta y)$, where $\alpha\delta-\beta\gamma\ne 0$. It is well known that an integral operator belongs to the Hilbert-Schmidt class $\mathfrak S_2$ if and only if its kernel is square-summable; therefore, $K\in\mathfrak S_2$ if and only if $L\in\mathfrak S_2$. Is the same true if we replace the class $\mathfrak S_2$ by $\mathfrak S_1$, or by $\mathfrak S_p$ with $p\ge 1$? Can such an equivalence be proved for other properties? (References to known results would be appreciated.)

This question was not answered at SE. I was not able to use hints suggested in the discussion there.

In particular, this contains another question as a special case: does the integral operator on $L^2(\mathbb R)$, whose kernel is the indicator of the rhombus $\{|x|+|y|<1\}$, belong to the trace class?

• My impression is that rotations are not relevant to the theory. Most of the interesting kernels are singular on the diagonal $x=y$, but this locus is not invariant under rotations. Oct 9, 2014 at 11:37
• Integral operators discussed here belong to the Hilbert-schmidt class and thus their kernels are regular. Moreover, there are various serious questons about operators between two different $L^2$-spaces, and then there is no diagonal at all. My question stems from a concrete setting, and I do not pretend that it cover all interesting kernels. Oct 9, 2014 at 13:00
• I voted to close but this was based on an error in my thinking and on 2nd thought I don't see how the comments at MSE can be used to answer the question Oct 9, 2014 at 17:45

Consider the characteristic function of the rhombus $\{|x|+|y| <1\}$ as a kernel on $L^2(I)$, with $I:=[-1,1]$. Since this kernel is symmetric, the trace norm of the corresponding compact operator $T$ is just the $\ell^1$ norm of the sequence of its eigenvalues. Let's find them.
We have for any $x\in I$ $$Tu(x)=\int^{1-|x|}_{-1+|x|}u(y)dy.$$
Let $v(x):=\int_0^xu(y)dy$ and $\lambda\neq0$. The eigenvalue equation $Tu=\lambda u$ then writes $$v(1-x)-v(-1+x)=\lambda v'(x)\quad \mathrm{for\quad } x >0\; ,$$ $$v(1+x)-v(-1-x)=\lambda v'(x)\quad \mathrm{for\quad } x <0\; ,$$ with $v(0)=0$. Equivalently, $v(x)$ is an odd function satisfying $$2v(1-x)=\lambda v'(x).$$ If we derive both sides and use the equation again we find $$v''(x)=-(2/\lambda)^2v(x),$$ with boundary conditions $v(0)=0=v'(1).$ Hence the eigenfunctions are cosine functions, with eigenvalues $\lambda_n\sim C/n$, and $T$ is not in the trace class.
• Double check: from $\lambda_n=4/(2n+1)\pi$ one finds for the HS norm of $T$ $\sum_{n\ge1}\lambda_n^2=2$, as it has to be. Oct 9, 2014 at 22:32
• Does it help to point out (the obvious fact) that $Tu=0$ if $u$ is odd; this confused me for a moment. Oct 9, 2014 at 23:27
• Also note that $T$ is in $\mathfrak{S}_p$ for all $p>1$, so the question remains open for $2\neq p>1$. Oct 10, 2014 at 10:41
• Yes, thanks. The case $p=1$ was the most important one for me. Oct 10, 2014 at 10:49