Fall 2000 Problem 7

Exercise 12.1 (The Volterra Operator)   Write

$$Tf(x) = \int_0^x f(s)ds.$$ (12.1)

1. Show that $T$ defines a bounded linear operator on the Banach space $C([0,1])$, endowed with its usual norm.
2. Show that this operator on $C([0,1])$ is compact.

Proof. For part (a), linearity follows by the linearity of the integral. For boundedness, determine an upper bound as follows:

$$\Vert Tf\Vert _\infty = \sup_{x\in[0,1]}\left\vert\int_0^x f(s)ds... ...t_0^x\vert f(s)\vert ds = \int_0^1 \vert f(s)\vert ds \leq \Vert f\Vert _\infty$$ (12.2)

Therefore, $\Vert T\Vert\leq 1$, indicating $T$ is bounded.

For part (b), let $\{f_n\}\subseteq C([0,1])$ be a sequence in the unit ball. Then we verify $T(\{f_n\})$ is precompact. Because $T$ is bounded, so is its image. All we have to show is equicontinuity. Let $\epsilon>0$ be given. If $\vert x-y\vert<\epsilon$, then

$$\left\vert Tf_n(x)-Tf_n(y)\right\vert = \left\vert\int_0^x f_n(s)ds-\int_0^y f_n(s)ds\right\vert= \left\vert\int_y^x f_n(s)ds\right\vert \leq \int_y^x \vert f_n(s)\vert ds$$ (12.3)

$$\leq \vert x-y\vert\Vert f\Vert _{\infty}$$ (12.4)

$$\leq \vert x-y\vert$$ (12.5)

$$< \epsilon$$ (12.6)

Therefore the hypotheses of Arzelà-Ascoli apply, indicating a converging subsequence exists. $\qedsymbol$