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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?

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  • $\begingroup$ Is it $T\colon C^0[0,1]\to C^1[0,1]$ or $T\colon C^0[0,1]\to C^0[0,1]$? $\endgroup$ – Julián Aguirre Nov 28 '13 at 10:57
  • $\begingroup$ The text of the exercise says $T:C^0[0,1]\rightarrow C^1[0,1]$. $\endgroup$ – batman Nov 28 '13 at 12:38
  • $\begingroup$ But if $f$ is assumed continuous, then $T\nu$ may not be differentiable. $\endgroup$ – Julián Aguirre Nov 28 '13 at 13:15
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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, $$|Tv_n(x)-Tv_n(z)|\leq c\varepsilon+Mc\delta\leq c(M+1)\varepsilon.$$

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  • $\begingroup$ I think you're right, but are you sure that we can choose $\delta$ such that $\delta<\epsilon$? $\endgroup$ – batman Nov 28 '13 at 14:37
  • $\begingroup$ Yes, because for each $\varepsilon$ >0, $\exists\delta'>0$ such that the inequality holds. So take $\delta = \min\{\delta',\varepsilon\}$. $\endgroup$ – Braun Nov 28 '13 at 14:58

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