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](https://math.stackexchange.com/questions/955944/what-happens-if-we-rotate-the-kernel-of-an-integral-operator). I was not able to use hints suggested in the discussion there.


In particular, this contains [another question](https://math.stackexchange.com/questions/956596/does-the-integral-operator-whose-kernel-is-the-indicator-of-the-rhombus-belong) 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?



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


