I'm trying to figure out how to show compactness of an operator. This is an exercise on a past exam in functional analysis.
Define a linear operator $T: C([0,1]) \to C([0,1])$ by $$T(f)(x) = \frac{1}{x^2} \int_0^x t^2f(t)\, dt.$$ Show that $T$ is compact.
An operator is compact if the image of the closed unit ball is relatively compact.
I showed in a previous part that the same operator but with the integrand $tf(t)$ is not compact using the spectral theorem. I get the feeling that one can use Arzela-Ascoli some how, but I can't figure out how to show equicontinuity given a sequence $(f_n) \subset \overline{B(0,1)}$, because of the annoying factor $1/x^2$ in front of the expression. Doing the substitution $u=t/x$ almost works but not quite.
I'm grateful for any suggestions.
Best Answer
First we show $T(f)$ is differentiable everywhere, and deduce an upper bound for its derivative in terms of $\|f\|$. If $x\not=0$, then by the product rule and fundamental theorem of calculus, $$\frac{dT(f)}{dx}=-\frac{2}{x^3}\int_0^x t^2f(t)dt+\frac{1}{x^2} x^2f(x)$$ and $$|\frac{dT(f)}{dx}|\le \frac{2}{x^3}\int_0^x t^2\|f\|dt + \|f\|=(\frac{2}{3}+1)\|f\|=O(1)\|f\|$$
We can also take care of the special case $x=0$ (with l'Hospital's rule), but this is unnecessary for the purpose of mean value theorem, which implies for any $0\le x_1<x_2\le 1$, we have for some $x_0\in (x_1, x_2)$, $$T(f)(x_2)-T(f)(x_1)=\frac{dT(f)}{dx}(x_0)(x_2-x_1)$$ $$\|T(f)(x_2)-T(f)(x_1)|\le O(1)\|f\|(x_2-x_1)$$
That is, for any family of functions $f$ with $\|f\|\le M$ uniformly, we must have $T(f)$ are uniformly equicontinuous, and also uniformly bounded bounded by $\|T\|M$, hence by the Arzelà –Ascoli theorem, it contains a convergent subsequence, which implies the family of $T(f)$ is pre-compact.