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).