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?

  • $\begingroup$ 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

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, $$|Tv_n(x)-Tv_n(z)|\leq c\varepsilon+Mc\delta\leq c(M+1)\varepsilon.$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.