Consider the integral operator $\Lambda$ on $C([a, b])$ given by $(\Lambda f)(x) = \int_a^x f(t) dt. \ \ $ Prove that for any bounded set $S$ in $C([a, b])$, the set $\Lambda(S)$ is relatively compact in $C([a, b])$.
$\text{My attempt:}$
- $\text{Ascoli thm:}$ Let $E$ be a compact metric space. Let $F \subset C(E)$ be a uniformly bounded, equicontinuous family of functions. Then $F$ is a relatively compact subset of $C(E)$.
So to apply the Ascoli thm I need to verify that the $\Lambda f$ is uniformly bounded and $\Lambda S$ is equicontinuous.
- To show the uniform boundedness, let $f_n\in S \subset C([a,b])$ , and since $S$ is bounded, let $M=\max_n{M_n}$, where $M_n = \sup_{x\in[a,b]}|f_n(x)|$ for each $n$.
$$|\Lambda f (x)|\le \int_a^x |f(t)| dt \le M(x-a)\le M(b-a)<\infty \ \ \ ,\forall f\in S$$
- To show the equicontinuity, Let $f,g \in S$ since $\Lambda$ is bounded hence it is continuous, i.e. if $\|f - g \|_\infty =\sup_x |f(x)-g(x)|<\delta \implies \| \Lambda f -\Lambda g \|_\infty = \sup|\Lambda f(x) -\Lambda g(x)|<\epsilon$
Is above a correct solution?
The second part is not correct. What you have to show is the following:
Given $\epsilon >0$ there exists $\delta >0$ such that $|x-y| <\delta$ implies $|\Lambda f(x)- \Lambda f(y)| <\epsilon$ fro every $f \in S$ [$\delta$ not depending on $f$].
To prove this not that $|\Lambda f(x)- \Lambda f(y)|\leq \int_x^{y} |f(t)| dt$ (for $x \leq y$, the case $y<x$ being similar). Since $S$ is bounded there exists $M$ such that all the functions in $S$ are bounded in absolute value by $M$. Hence $|\Lambda f(x)- \Lambda f(y)|\leq M|x-y|$. So the choice $\delta=\frac M {\epsilon}$ works.