Convergence of integral operators defined by integral kernels

Question

Let $U \subset \mathbb{R}^{2}$ open and let $A_{\epsilon}: L^{2}(U,\mathbb{C}^{4}) \rightarrow L^{2}(U,\mathbb{C}^{4})$ be an integral operator defined by $A_{\epsilon}f(x) = \int_{U}k_{\epsilon}(x,y)f(y)dy$, where $k_{\epsilon}(x,y)$, for $\epsilon \geq 0$, is a smooth matrix valued function $U \times U \setminus \Delta \rightarrow M_{4\times4}(\mathbb{C})$. Around the diagonal $|k_{\epsilon}(x,y)|$ (meaning an arbitrary entry of the matrix) behaves like $\frac{1}{|x-y|}$. Furthermore let $k_{\epsilon}(x,y) \rightarrow k_{0}(x,y)$ pointwise. Can one in this case deduce that $A_{\epsilon} \rightarrow A_{0}$ in the strong sense for $\epsilon \rightarrow 0$, that means for any $f \in L^{2}(U,\mathbb{C}^{4})$ it holds $\lVert A_{\epsilon}f-A_{0}f \rVert_{L^{2}(U,\mathbb{C}^{4})} \rightarrow 0$ for $\epsilon \rightarrow 0$? Since the singular behavior of the integral kernel isn't too strong, I would think that something like this should be possible. But because I am new to the topic of such integral operators, I struggle to show this. Can anyone help me out here?

Answer 1

You get strong convergence if you have a uniform bound on the kernels away from the diagonal. Assume there is a constant $C$ such that for all $\epsilon \ge 0$ and $x \ne y$ you have

$$|k_\epsilon(x,y)| \le \frac{C}{|x-y|}.$$

Also assume that for every $\delta > 0$ there is $M_\delta < \infty$ such that
$$\sup_{\epsilon \ge 0} \sup_{|x-y| \ge \delta} |k_\epsilon(x,y)| \le M_\delta.$$

Split the operator difference into two parts. Choose $\delta > 0$ and write
$$A_\epsilon - A_0 = I_\epsilon^\delta + J_\epsilon^\delta$$
where $I_\epsilon^\delta$ integrates over $|x-y| < \delta$ and $J_\epsilon^\delta$ over $|x-y| \ge \delta$.

On the near part $|x-y| < \delta$, the Schur test with the bound $C/|x-y|$ in two dimensions gives
$$\sup_x \int_{|x-y| < \delta} \frac{C}{|x-y|} dy = 2\pi C \delta.$$
This shows $\|I_\epsilon^\delta\| \le 2\pi C \delta$ for all $\epsilon$. Choosing $\delta$ small makes this as small as needed.

On the far part $|x-y| \ge \delta$, the bound $M_\delta$ makes the kernel square integrable over $U \times U$. Since $k_\epsilon \to k_0$ pointwise and the bound is uniform, dominated convergence gives
$$\iint_{|x-y| \ge \delta} |k_\epsilon - k_0|^2 \, dx \, dy \to 0$$
as $\epsilon \to 0$. This gives $\|J_\epsilon^\delta\| \to 0$ in Hilbert–Schmidt norm and therefore in operator norm.

First pick $\delta$ so that $\|I_\epsilon^\delta\| < \eta/2$. Then take $\epsilon$ small enough so that $\|J_\epsilon^\delta\| < \eta/2$. This yields $\|A_\epsilon - A_0\| < \eta$ for small $\epsilon$. You get operator norm convergence and thus strong convergence.

Answer 2

Let now $U$ be a bounded, open two dimensional surface in $\mathbb{R}^{3}$ such that one can define a tubular neighborhood around $U$, lets say for parameter $|t|<1$, and denote by $\nu(x)$ the normal vector on $U$ in $x \in U$. In this case let $k_{\epsilon}:\mathbb{R}^{3} \times \mathbb{R}^{3} \setminus \Delta \rightarrow M_{4 \times 4}(\mathbb{C})$ be kernels with $|k_{\epsilon}(x,y)|$ behaving like $\frac{1}{|x-y|}$ near the diagonal. Again it should hold $k_{\epsilon} \rightarrow k_{0}$ pointwise. Using these kernels I want to define $k'_{\epsilon}\big((x,t),(y,t)\big) = k_{\epsilon}\big(x+t\nu(x),y+s\nu(y)\big)$ for $\epsilon \geq 0$. By means of these kernels define integral operators $A_{\epsilon}: L^{2}(U\times(-1,1),\mathbb{C}^{4}) \rightarrow A_{\epsilon}: L^{2}(U\times(-1,1),\mathbb{C}^{4})$ by $A_{\epsilon}f(x,t) = \int_{-1}^{1}\int_{u}k'_{\epsilon}\big((x,t),(y,s)\big)f(y,s)d\sigma(y)ds$, where $\sigma$ denotes the surface measure on $U$ (lets assume $\sigma(U)<\infty$). I have shown that $|k_{\epsilon}\big((x,t),(y,s) \big)| \leq \frac{C}{|x-y|}$. With this I can follow your arguments by splitting the inner integral over $U$ into an integral over $\{|x-y|>\delta\}$ and one over $\{|x-y|<\delta\}$. Although there also is the integral $\int_{-1}^{1}$ I would think that one can still deduce norm convergence and then consequently strong convergence. Is this correct or can something go wrong in this case?