Suppose you have a kernel operator on a torus, with a kernel of a spatially varying width $\epsilon(x)$, which might be zero at certain points. That is to say, for some approximate identity $\psi_h(x)$ that makes periodic $h^{-1} \psi(x/h) \in C^\infty_c$, and some $C^2$ function $\epsilon(x) \geq 0$, you have $$ \mathcal{K}\phi(x) := \int_{\mathbb{T}} \psi_{\epsilon(x)}(x-y) \phi(y)\,\mathrm{d}y.$$
Clearly $\mathcal{K}$ is bounded in $L^1$.
With some effort one can likely show that it is bounded in $L^\infty$: the idea being that if you evaluate at $x$, and find that $\epsilon(x)$ is small, then $\epsilon'(x)$ is also small, and so $\epsilon$ varies slowly enough that $\psi_{\epsilon(x)}(x-\cdot)$ stays integrable.
What I don't know how to prove is whether it is necessarily bounded in $L^2$. This seems quite non-standard, because it is not quite either a bounded kernel (where you could compute the Frobenius norm), or a convolution (where you could go to Fourier space and get a diagonal operator).
However, it is useful to think about, e.g. if you are interested in getting uniform bounds on harmonic (or complex) functions along curves in the domain, where the edges start to cause trouble with your Green's function.
