Is the operator $T:C([0,1])\to C([0,1])$ defined by $(Tf)(x)=\int_0^x h(y)f(y) dy$ where $h\in L^2([0,1])$ compact? 
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.
 A: 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 $1/2$.
A: Take any $h \in L_2([0,1])$ and define $T = T_h: C[0,1] \to C[0,1]$ by $T(f)(x) = \int_0^x f(t)h(t) dt$.
Take any $f_n \in C[0,1]$ such that $||f_n|| < M$, where $||g|| = \sup\{ |g(t)| : t \in [0,1]\}$.
Looking at $g_n := T(f_n)$, we have $||g_n|| = \sup_{x \in [0,1]} | \int_0^x f_n(t)h(t) | \le \int_0^1 |f_n(t)||h(t)| \le M \cdot H$, where $H =\int_0^1 |h(t)|dt$
Note that $H$ is finite, since $L_2([0,1]) \subset L_1([0,1])$ (Lyapunov inequality, $[0,1]$ with Lebesgue Measure is a good probability space)
By that $||g_n|| < MH$ (some constant)
Moreover, looking at $(y>x)$, we have $g_n(y) - g_n(x) =  \int_x^y h(t)f_n(t) dt \le M\int_x^y |h(t)|dt$
If we could show that as $y-x \to 0$ then $\int_x^y |h(t)| dt$ tends to $0$ then it would mean that sequence $g_n$ is equicontinous. That + uniformly boundedness will give us (by Arzela-Ascoli) that $g_n$ has convergent subsequence, so that $T$ would be compact.
So let's look at $g(x) = \int_0^x |h(t)|dt$. We want to show that $g$ is uniformly continuous on $[0,1]$, but since $[0,1]$ is compact, we only need that $g$ is continuous. So let's fix $x$ and take sequence $x_n$ which converge to $x$. We have:
$|g(x_n) - g(x)| = \int_x^{x_n} |h(t)|dt  =  \int_0^1 \chi_{[x,x_n]}(t)|h(t)|dt$, now pointwiselly it converges to $0$ (under integral sign), whereas it is bounded by integrable function - $|h|$!. That means we can use dominated convergence theorem to ensure it tends to $0$. That shows (since $x_n$ was arbitrary tending to $x$), that $g$ is continuous at $x \in [0,1]$. Again since $x$ was arbitrary, it shows that $g$ is continuous on $[0,1]$, which by Cantor means it is uniformly continuous. So that we've showed sequence $g_n$ is equicontinuous. So by Arzela Ascolii we have subsequence $n_k$ such that $g_{n_k} = T(f_{n_k})$ converges
