Let $(T_Kx)(t)=\int_0^1 K(t, s)x(s)ds$ be linear operator on $L_1[0,1]$. Here $K(t, s)$ is measurable function on $[0,1]\times[0,1]$ with $\sup\{\int_0^1|K(t,s)|dt ,s\in[0,1]\}<\infty$. Prove that $T_K:L_1[0,1]\to L_1[0,1]$ is compact.
My thoughts:
It is easy to show that $T$ is bounded. Next I want to show that $T$ is compact, i.e. for any bounded set $M\subset L_1[0,1]$ the image $T(M)$ is precompact. We have a criterion for precompactness in $L_1[0,1]:$ $A\subset L_1[0,1]$ is precompact $\Leftrightarrow$ $A$ is bounded and for any $\varepsilon>0$ there is $\delta>0$ such that for any $h\in[0,\delta]$ and $x\in A$ we have $\int_0^1|x(t+h)-x(t)|dt<\varepsilon$, where $x(t+h)=0$ when $t+h\notin[0,1]$. I want to check whether it is true for $T(M)$, where $M$ is a bounded set. We want
\begin{align}
\int^1_0|(Tx)(t+h)-(Tx)(t)|dt&=\int^1_0\Bigl|\int_0^1 \Bigl(K(t+h,s)-K(t,s)\Bigr)x(s)ds\Bigr|dt\\
&\leq \int^1_0\int_0^1 \Bigl|K(t+h,s)-K(t,s)\Bigr||x(s)|dsdt
\end{align}
to be small for all $x\in M$ and small $h$. To show that it is small I try to approximate $K(t, s)$ with continuous function. Using Luzin's theorem we can find continuous on $[0,1]\times[0,1]$ functions $g(t,s)$ so that $K(t,s)$ and $g(t, s)$ are not equal only on set of arbitrarily small measure. We have
\begin{align}
\int^1_0\int_0^1 \Bigl|K(t+h,s)-K(t,s)\Bigr||x(s)|dsdt&=\iint_A \Bigl|K(t+h,s)-K(t,s)\Bigr||x(s)|dsdt \,+\\
&\quad \iint_B \Bigl|g(t+h,s)-g(t,s)\Bigr||x(s)|dsdt=:I_1+I_2
\end{align}
where $B=\{(t,s)\in [0, 1]\times [0,1]: g(t+h,s)=K(t+h,s), g(t,s)=K(t,s)\}$, $A=[0,1]\times[0,1]-B$.
If $g(t,s)$ is fixed, second integral $I_2$ will be small for small $h$ due to uniform continuity of $g(t,s)$ and estimate can be made independent of $x$. My problems began when I tried to estimate first integral $I_1$. Due to absolute continuity of Lebesgue's integral, $I_1$ is small when measure of $A$ is small, but this estimate depends on $x$.
How can the problem be solved?
Best Answer
It seems that the statement as is, is not true in general. Additional uniform conditions on the kernel $K$ are required for $T_K$ to be compact.
Consider the kernel $$ K(t,s)=\frac{1}{\sqrt{st}}\mathbb{1}(t<s)\mathbb{1}(0\leq s,t\leq1)$$ It satisfies the conditions of the problem for $\int^1_0|K(t,s)|\,dt\equiv2$. We have that $$\int^1_0|K(t+h,s)-K(t,s)|\,dt = 2\mathbb{1}(0<s\leq h) + \frac{2\sqrt{h}}{\sqrt{s}}\mathbb{1}(h<s\leq1)$$ Consequently \begin{align} \int^1_0|T_Kx(t+h)-T_Kx(t)|\,dt &\leq \int^1_0|x(s)|\int^1_0 |K(t+h,s)-K(t,s)|\,dt\,ds \\ &=2\int^h_0|x(s)|\,ds +2\sqrt{h}\int^1_h\frac{|x(s)|}{\sqrt{s}}\,ds \end{align} From this estimate, one sees that the term $\int^h_0|x(s)|\,ds$ cannot be controlled uniformly on $x\in M$ (unless M is finite or uniformly integrable).
The problem, in my view, stems from the lack of uniformity with which one can approximate a kernel $K$ of the type described in the problem by nice functions (continuous functions for example). That is, a given $K$ that satisfies the condition $\|\int^1_0K(t,\cdot)\,dt\|_\infty<\infty$ may not by approximated nicely by continuous functions so that \begin{align} \int^1_0|x(s)|\int^1_0 |K(t+h,s)-K(t,s)|\,dt\,ds &\leq \int^1_0|x(s)|\int^1_0 |K(t+h,s)-H(t+h,s)|\,dt\,ds + \\ &\quad \int^1_0|x(s)|\int^1_0|H(t+h,s)-H(t,s)|\,dt\,ds +\\ &\quad\int^1_0|x(s)|\int^1_0 |H(t,s)-K(t,s)|\,dt\,ds\\ &\leq \int^1_0|x(s)|\int^1_0|H(t+h,s)-H(t,s)|\,dt\,ds + \\ &\quad 2\int^1_0|x(s)|\int^1_0 |H(t,s)-K(t,s)|\,dt\,ds. \end{align} with $H\in C([0,1]\times[0,1])$, becomes uniformly small in $M$.
Here is another example which shows that even weak compactness in $L_1[0,1]$ may not hold either. Define the kernel $$G(t,s)=\frac{1}{\sqrt{s}}\mathbb{1}(t\leq\sqrt{s})\mathbb{1}(0<s,t\leq1)$$ Then $\int^1_0|G(t,s)\,dt=1$, and $T_Gx(t):=\int^1_0x(s)G(t,s)\,ds=\int^1_{t^2}\frac{x(s)}{\sqrt{s}}\,ds$ The collection of maps $g_s:t\rightarrow G(t,s)$ is not uniformly integrable and in turn, this implies that for a general bounded set $M\subset L_1[0,1]$, $T(M)$ is not uniformly integrable.