Arzelá-Ascoli to show that a given subset of $C([a,b])$ is compact

Question

I have been exposed to the following version of the Arzelá-Ascoli Theorem

Theorem (Arzelá-Ascoli): Let $K \subset \mathbb{F}$ be compact. A subset $S \subset C(K)$ is pre-compact iff $S$ is bounded and equicontinuous.

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Problem: Let $c>0$ be a constant. Let $M$ denote a set given as $$ M:= \left\lbrace f \in C^1([a,b]) : \int_a^b |f(x)|^2dx + \int_a^b |f'(x)|^2 dx \leq c \right\rbrace$$ Claim: $\overline{M}$ is compact in $C([a,b])$ , working with the sup-norm .

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My approach: $S \subset C(K)$ is pre-compact if and only if $\overline{S}$ is compact. Thus by the Arzeá-Ascoli Theorem it suffices to show that $M$ is pre-compact, i.e. bounded and equicontinuous.

Without loss of generality I can take $a=0, b=1$ because I can use a change of variables for the integrals otherwise.

For Equicontinuouity:

Since $f \in C^1([0,1])$ by the intermediate value theorem I can find for all $x,y \in [0,1]$ a $\xi \in [x,y]$ such that $|f(x)-f(y)| = |f'( \xi)| |x-y|$. Squaring this identity and than taking the integral with respect to $d \xi$ over $[0,1]$ then yields that $$ |f(x) -f(y)| \leq c |x-y| \tag{*} $$ Since this Lipschitz bound is true for all $f \in M$ we can conclude that $M$ is equicontinuous.

Remark: The squaring part I did above confuses me, it seems to be superfluous because thanks to Hölder's Inequality (we work on a space with finite measure Lebesgue measure) we have for $p=q=2$ that $\| 1 f \|_1 \leq \|1\|_2 \|f\|_2 \iff \int_0^1 |f(x)|dx \leq \sqrt{ \int_0^1|f(x)|^2dx} \leq \sqrt{c}$

For Boundedness:

Thanks to (*) and the remark above we can proceed similarly as in Showing a subset of $C([0,1])$ is compact. i.e. we have $$ |f(x)-f(0)| \leq c |x| \leq c $$ Or equivalently $$ f(0) -c \leq f(x) \leq c + f(0)$$ From the left inequality we obtain $f(0) \leq f(x) + c$ after integrating this we get $f(0) \leq c + \sqrt{c}$

Question: Is my approach correct? It seems a bit weird to me because I don't really exhaust the inequality from the set, but only build new more strict inequalities.

Answer 1

Your argument for the first part is incorrect. For any $x, y$ you used mean value theorem to find $\xi$ so that

$$|f(x)-f(y)| = |f'( \xi)| |x-y|.$$

And then you wish to integrate with respect to $\xi$: that does not quite make sense. $\xi$ is fixed once you are given $x, y$. It is not a variable.

Instead, you should use the fundamental theorem of calculus:

$$ f(y) - f(x) = \int_x^y f'(\xi) d\xi.$$

Then by Holder's inequality,

$$\tag{1} |f(y) - f(x)| \le \int_x^y |f'(\xi)| d\xi \le |x-y|^{1/2} \left(\int_0^1 |f'(\xi)|^2 d\xi\right)^{1/2}\le c|x-y|^{1/2}.$$

The above inequality is sufficient to show equicontinuity.

Indeed, it is not possible to get the ineqaulity

$$\tag{2} |f(y) - f(x)| \le N|x-y|$$

in the sense that there is no $N$ so that the above is satisfied by all $f\in M$: Take $g$ be a nonzero $C^1$ function on $\mathbb R$ with support in $[0,1]$ and consider $f_n(x) = \frac{1}{\sqrt n} g(nx)$. Then

$$\int_0^1 |f_n(x)|^2 dx \le \frac{1}{n} \int_0^1 g^2 (nx) dx \le \frac 1{n^2} \int_{-\infty}^\infty g^2(x) dx .$$

One the other hand, $f'_n (x) =\sqrt n g'(nx)$ and

$$\int_0^1 (f'_n(x))^2 dx = \int_0^1 n (g'(nx))^2 dx = \int_{-\infty}^\infty (g'(y))^2 dy$$

Thus $f_n \in M$ for all $n$ (if $c$ is large enough). However, since the derivative of $f_n$ are not uniformly bounded, they cannot be uniformly Lipschitz (that is, there is no $N$ satisfying (2) for all $f$).

For the boundedness, you can use (1) instead of (2) to argue similarly.

Remark For those who are familiar with Sobolev space, $M= W^{1,2}([0,1])\cap C^1([0,1])$ and the above is really the proof of Sobolev embedding in 1-d case: there is an embedding $W^{1,2}([0,1])$ into the Holder space $C^{0,1/2} ([0,1])$.