I need some help with this exercise. Let $f\in C^0_b(R^2)$ and consider the operator $[T(v)](x)=\int_0^x f(x,y)v(y)dy$ for every $x\in R$. Is this a compact operator $T:C^0[0,1]\rightarrow C^1[0,1]$?
I think Ascoli Arzelà theorem might be useful. So, let $(v_n)$ be a bounded sequence in $C^0[0,1]$, $||v_n||<c$ for every $n$. I want to prove that the family of functions $\mathcal F=(Tv_n)\subset C^1[0,1]$ is equibounded and equicontinous. I have proved that it is equibounded, but I have some problems with equicontinuity.
Let $x,z\in[0,1]$, and suppose $z>x$, we also know that $|f|<M$. Now $|Tv_n(x)-Tv_n(z)|=|\int_0^x f(x,y)v_n(y)dy-\int_0^z f(z,y)v_n(y)dy|=|\int_0^xf(x,y)v_n(y)dy-\int_0^x f(z,y)v_n(y)dy-\int_x^zf(z,y)v_n(y)dy|<\int_0^x|f(x,y)-f(z,y)||v_n(y)|dy+\int_x^z |f(z,y)||v_n(y)|dy$. Now, from continuity and boundedness of $f$, from boundedness of $(v_n)$, we have that $|Tv_n(x)-Tv_n(z)|<c\epsilon x+Mc(z-x)<c(\epsilon+M)$, which is indipendent from $n$ but it is not proportional to $\epsilon$.
Any suggestions?
You can choose $0<\delta\leq\varepsilon$ for $x,z\in [0,1]$, $|x-z|<\delta \Rightarrow |f(x,y)-f(z,y)|\leq\varepsilon$, $y\in [0,1]$, because $f$ is uniformly continuous in $[0,1]\times [0,1]$. So in your case, $$|Tv_n(x)-Tv_n(z)|\leq c\varepsilon+Mc\delta\leq c(M+1)\varepsilon.$$