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Let $M$ be a bounded subset of the space $C_{[a,b]}$. Prove that the set of all functions $F(x)=\int^{x}_{a}f(t)dt$ with $f\in{M}$ compact.

Some helpful definitions:

bounded - A subset $S$ of a metric space $(X, d)$ is bounded if it is contained in a ball of finite radius, i.e. if there exists $x$ in $X$ and $\epsilon > 0$ such that for all $s\in{S}$, we have $d(x, s) < \epsilon$.

compact - A set $S$ of real numbers is called compact if every sequence in $S$ has a subsequence that converges to an element again contained in $S$.

Any help/clarification/direction/hints would be greatly appreciated.

mmh0015
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2 Answers2

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(If $M$ is merely bounded, the claim is false - see answer below.)

Define $\int M:=\{F:F=\int f, f\in M\}$. I asssume you mean to show $\int M\subset C_{[a,b]}$ and that $\int M$ is compact.

It is clear $\int M\subset C_{[a,b]}$ by continuity of $f\in M$ and absolute continuity of the integral.

Let $\{F_{n}\}$ be a sequence in $\int M$. This induces a sequence $\{f_{n}\}$ in $M$ such that $f_{n_{k}}\to f\in M$ for some subsequence $\{n_{k}\}$, since $M$ is assumed compact (the convergence is taken with respect to $||\cdot||_{\infty}$, the "sup norm").

This automatically implies $F_{n_{k}}\to F\in\int M$, thereby showing $M$ is compact. Indeed, for $\epsilon>0$ and $k(\epsilon)$ sufficiently large, we have $$\sup_{x\in[a,b]}|F_{n_{k}}(x)-F(x)|\leq\sup_{x\in[a,b]}\int_{a}^{x}|f_{n_{k}}(t)-f(t)|\;dt\leq\sup_{x\in[a,b]}|x-a|\epsilon\leq (b-a)\epsilon.$$

Sargera
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  • Wow, very helpful! I was confused as to what it was asking at first but now it makes more sense. Thank you so much. – mmh0015 Apr 17 '14 at 00:43
  • The proof is wrong. – user64066 Mar 14 '15 at 13:11
  • I saw "$M$ compact" in the title and made that assumption, which makes the proof $\int M$ is compact trivial because of the uniform convergence norm. But I see now the assumption is that $M$ is merely bounded. – Sargera Mar 14 '15 at 13:33
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The answer given above is wrong. Actually the claim itself is wrong. The set asked in the question is not even closed. To see that consider the unit interval $[0,1]$, and take $M$ as the unit ball in $C_{[0,1]}$. Now consider the function $F$

$$ F(x)=\begin{cases} x\quad 0\leq x\leq 1/2\\ 1-x\quad 1/2\leq x\leq 1 \end{cases} $$ This function is not continuously differentiable and is not a member of the set asked in the question, but we can approximate it with a sequence $\{F_n\}$ of functions in that set. In fact, $F_n(x)=\int_0^x F_n^\prime(t) dt$ with $\mid F_n^\prime(t)\mid\leq 1$ for all $n$ and for all $t$. So this set is not closed. If we replace the claim with relatively compact then we can prove the claim. You either show that the set is totally bounded (since $C_{[a,b]}$ is complete the claim follows) or use Arzela's theorem.

user64066
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