Let $h\in $$L^2([0,1])$ and define $T:C([0,1])\to C([0,1])$ by $$(Tf)(x)=\int_0^x h(y)f(y) dy.$$ Is $T$ compact?
I am trying to show it is not the case. I need to find a bounded sequence $f_n$ of functions such that $Tf_n$ does not have a convergent subsequence. I have seen an example in $L^2$ where you can show that $||Te_i-Te_j||\geq \frac{\sqrt{2}}{2}$ and so $T(e_i)$ is not Cauchy and cannot have a convergent subsequence. But here I have no ON-basis so this does not work. I also tried to use the fact that $f$ is uniformly continuous on $[0,1]$. I know $x^n$ converges to a limit which is not continuous so it does not have a convergent subsequence. But my argument needs to work for all $h\in L^2$ so and I am not sure how to deal with that.
Best Answer
Just one observation following Dominik Kutek's solution. It suffices to assume that $h\in L_1[0,1]$ to see that the operator $$ Tf(x)=\int^x_0f(t)h(t)\,dt $$ is compact. The integrability of $h$ implies that for any $\varepsilon>0$ there is $\delta>0$ such that if $A\subset[0,1]$ is measurable and $\lambda_1(A)<\delta$, then $\int_A|h|\,d\lambda_1<\varepsilon$ (here $\lambda_1$ stands for Lebesgue measure on the line).
So, if $\mathscr{E}\subset\mathcal{C}[0,1]$ is bounded, $|x-y|<\delta$ implies that $$ |Tf(x)-Tf(y)|=\Big|\int^y_xf(s)h(s)\,ds\Big|\leq \|f\|_u\int^y_x|h|\leq \varepsilon\,\sup_{f\in\mathscr{E}}\|f\|_u $$ that is, $T(\mathscr{E})$ is uniformly equicontinuous. Since $\|Tf\|\leq \|h\|_1\|f\|_u$ for all $f\in\mathcal{C}[0,1]$, it follows from Ascolli--Arzela's theorem that $T(\mathscr{E})$ is relative compact.
The more regular $h$ is, that is the higher $p\geq1$ for which $h\in L_p$, the smoother the operator $Kf$ is. This can be seen from the following $$ \|T(x)-T(y)|\leq\|f\|_u\Big|\int^y_x|h(t)|\,dt\Big|\leq \|f\|_u\|h\|_p|x-y|^{1/q} $$ where $\frac1p +\frac1q=1$. If $p=\infty$, then $Tf$ is Lipchitz of order one; if $p=2$, $Tf$ is Lipschitz of order $1/2$.