# Application of Arzela-Ascoli Theorem, proving uniform boundedness and equicontinuity

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?

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 $$y being similar). Since $$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.