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I asked the following question on MathStackExchange, but I have not received the answer that I'm looking for. Although it may not be a research-level question, I thought I could ask it here.

I'm currently reading this paper (and working on a similar one). Specifically, I'm trying to improve the two hypotheses (2.5) and (2.6) of O'Regan's paper.

The main goal is to study the Hammerstein integral equation (in $\mathcal{C}(I,E))$:

$$x(t) = \int_{0}^{t} K(t,s)f\big(s,x(s)\big)ds,\quad t\in I;$$

where $I=[0,1]$, $K $ is a scalar kernel, $E$ a Banach space and $f:I \times E \rightarrow E$ is a given function.

Suppose that $s \mapsto K(t, s)$ is integrable and $t \mapsto K(t, s)$ is continuous.

My goal is to see if the following is true:

\begin{array}{l} \text { For each } \epsilon>0 \text { , there exists } \delta>0 \text { such that, for any } t_{1}, t_{2} \in I, \text { if }\left|t_{1}-t_{2}\right|<\delta \text { then }\\ \int_{t_1}^{t_2} K(t_2,s) d s< \epsilon \end{array}

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This does certainly not follow from your other hypotheses, as what you want to conclude is not much weaker than the equi-integrability of $\{K(t,\cdot):t\in I\}$ (sometimes also called absolute continuity in the $L_1$-norm), but what you assume is only a continuity in one variable which implies nothing about the $L_1$-norm, not even the boundedness.

A counterexample might look as follows: Around the line $t=s/2$ with $t\in(0,1)$, say, for $s\in[t/3,t/2]$, the function $K(t,s)$ assumes huge values (for instance, $1/t^2$). Then you extend the function to a nonnegative function preserving your condition such that $K(t,s)=0$ for $t\in\{0,1\}$ or $s\ge t$.

Then you have even $\sup_{\lvert t_1-t_2\rvert\le\delta}\int_{t_1}^{t_2}K(t_2,s)ds=\infty$ for every $\delta>0$.

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  • $\begingroup$ Great, thank you. I wasn't sure that those conditions are sufficient. May you suggest any improvements? (I have this idea of letting $K(t,s)\leq h(s), \;\forall t\in $I$$ with $h$ is a $L^1(I)$ function) $\endgroup$
    – Motaka
    Commented Aug 21, 2021 at 13:35
  • $\begingroup$ As I mentioned, a sufficient condition is the equi-integrability of the family $\{K(t,\cdot):t\in I\}$, that is $\sup_{t\in I}\lVert\chi_{E_n}(\cdot)K(t,\cdot)\rVert_{L_1(I)}\to0$ whenever $E_1\supseteq E_2\supseteq\ldots$ are measurable with $\bigcap_nE_n=\emptyset$. An integrable majorant $h$ (that is, $\lvert K(t,s)\rvert\le h(s)$ for almost all $s$), is sufficient for this (Lebesgue's dominated convergence theorem) but not necessary. $\endgroup$
    – Martin Väth
    Commented Aug 21, 2021 at 15:46
  • $\begingroup$ Vath: Can you explicit your "counterexample" more, I want to check why the two hypotheses fail. Thanks! $\endgroup$
    – Motaka
    Commented Sep 8, 2021 at 18:39
  • $\begingroup$ Slightly different, but easier to make explicit: For every $t\in[0,1]$ put $K(t,0)=0$, and if $s$ belongs to $I_n=(1/(n+1),1/n]$ put $K(t,s)=\max\{n^2(n+1)(1-\lvert nt-1\rvert),0\}$. Then on the one hand $\int_{1/(n+1)}^{1/n}K(1/n,s)ds=\int_{I_n}n^2(n+1)ds=n\to\infty$. On the other hand, for fixed $t$ the condition $K(t,s)>0$ holds only if $s\in I_n$ with $\lvert nt-1\rvert<1$. This holds only for finitely many $n$ so that $K(t,\cdot)$ is a simple function. $\endgroup$
    – Martin Väth
    Commented Sep 9, 2021 at 21:41

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