What's the condition to prove the equicontinuity?

Question

Let $K: I\times I\rightarrow \mathbb{R}$ be a scalar kernel, where $I=[0,1]$, and $a: I \rightarrow (0,+\infty)$ an $L^1(I)$ function.

For $t_1,t_2\in I$, define $$I_{t_1,t_2}=\int_{0}^{1} \left |\big(K(t_2,s)-K(t_1,s) \right | \cdot a(s) \, d s + \int_{t_1}^{t_2} \left |K(t_2,s)\right | \cdot a(s) \, d s$$

Can you suggest (optimal, if possible) conditions on $K$ and $a$ so that the following holds?

For each $\epsilon>0$, there exists $\delta>0$ such that, for any $t_1,t_2\in I$, if $| t_1-t_2|<\delta$ then $I_{t_1,t_2}<\epsilon.$

Answer 1

Continuity of $K$ suffices.

Proof. Assume that $K$ is continuous and let $\varepsilon > 0$ be given. Since $I\times I$ is compact we know that $K$ is uniformly continuous. By uniform continuity of $K$ we may find a $\delta > 0$ such that for all $t_1,t_2,s\in I$ we have that $|K(t_2,s) - K(t_1,s)| < \frac{\varepsilon}{2\|a\|_{L_1}}$ if $|t_2-t_1| < \delta$. Now, it follows from dominated convergence that $(t_1,t_2)\mapsto \int_{t_1}^{t_2}a(s)ds$ is a continuous function. By using the compactness of the diagonal of $I\times I$, we may further assume that $\delta$ is small enough so that if $|t_2-t_1| < \delta$ then $\int_{t_1}^{t_2}a(s)ds < \frac{\varepsilon}{2\|K\|_{\infty}}$. Now, if $|t_2-t_1| < \delta$ we get: $$ I_{t_1,t_2} < \frac{\varepsilon}{2\|a\|_{L_1}} \|a\|_{L_1} + \frac{\varepsilon}{2\|K\|_{\infty}}\|K\|_{\infty} = \varepsilon. $$

Improved condition

Assume that $K$ is bounded and continuous in the first variable (i.e. $K(\cdot,s)$ is continuous for each $s\in I$). Then the conclusion holds.

Proof. Let $\varepsilon > 0$ be given and let $C$ be a constant such that $|K| \leq C$. Using the same argument as above we may find $\delta > 0$ such that if $|t_1-t_2| < \delta$ then $\int_{t_1}^{t_2}a(s)ds < \varepsilon / (2C)$. Further, the function $(t_1,t_2)\mapsto \int_0^1|K(t_2,s)-K(t_1,s)|a(s)ds$ is continuous by dominated convergence. By using the compactness of the diagonal of $I\times I$ we see that we may further assume that $\delta$ is small enough so that if $|t_1-t_2| < \delta$ then $\int_0^1|K(t_2,s)-K(t_1,s)|a(s)ds < \varepsilon / 2$. Now it follows that if $|t_1-t_2| < \delta$ then $I_{t_1,t_2} < \varepsilon$.