5

Consider the set $\mathcal{F}$ of continuous functions on $[0;1]$ with boundary values $$ f(0)=f(1)=0 \qquad \forall f \in \mathcal{F}. $$ Define the metric $d(f,g) = \lVert f-g \rVert_\infty = \sup_{0 \leq x \leq 1} |f(x)-g(x)|$ and the Lipschitz constant $$ L(f) = \sup_{0 \leq x_1 < x_2 \leq 1} \left|\frac{f(x_1)-f(x_2)}{x_1-x_2}\right|. $$ Define the subset of Lipschitz-continuous functions as $\mathcal{L}_M = \left\{ f \in \mathcal{F} \colon L(f) \leq M \right\}$.

I want to show that the metric space $(\mathcal{L}_1,d)$ is a compact subset of $(\mathcal{F},d)$. What are the steps to do this?

Here is my proof sketch:

  1. Prove that $(\mathcal{L}_1,d)$ is totally bounded (easy!)
  2. Prove that any sequence in $\mathcal{L}_1$ is equicontinuous.
  3. Infer from 1. and 2. by the Arzelà-Ascoli Theorem that $(\mathcal{L}_1,d)$ is relatively compact in the Banach space $(\mathcal{F},d)$
  4. Prove that any Cauchy sequence in $(\mathcal{L}_1,d)$ converges to an element from $\mathcal{L}_1$, i.e., show that $(\mathcal{L}_1,d)$ is complete.
  5. Since $(\mathcal{L}_1,d)$ is relatively compact and complete, it is therefore compact. QED.

Step 4. gives me headaches. That's because for any Cauchy sequence $(f_n)_n$, the sequence $(L(f_n))_n$ can be any sequence you like (in the unit interval). As a remedy, I could switch to a stronger metric $$ d(f,g) = \lVert f-g \rVert_\infty + \sup_{0 \leq x_1 < x_2 \leq 1} \left|\frac{f(x_1)-f(x_2)}{x_1-x_2}-\frac{g(x_1)-g(x_2)}{x_1-x_2}\right| $$ But then how do I eventually get back to my original metric?

jens
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    Instead of the 4. you have, how about proving 4': the pointwise limit of a sequence of functions from $\mathcal{L}_1$ belongs to $\mathcal{L}_1$? – Daniel Fischer Apr 21 '14 at 13:58
  • OK, I think I see where this is going. Seems much easier than I thought. Thanks! – jens Apr 21 '14 at 14:15

1 Answers1

4

(Step 4)

Since the space $(\mathcal{F}, d)$ is complete, it is equivalent to show that if $f_n \to f$ under the infinity norm, with $f_n \in \mathcal{L}_1$, and $f \in \mathcal{F}$, then $f \in \mathcal{L}_1$.

Take any $a, b$ with $0 < a < b < 1$. Then \begin{align*} \left|\frac{f(a) - f(b)}{a - b} \right| &\le \left|\frac{f(a) - f_n(a)}{a-b}\right| + \left|\frac{f_n(a) - f_n(b)}{a-b}\right| + \left|\frac{f_n(b) - f(b)}{a-b}\right| \\ &\le \left|\frac{f(a) - f_n(a)}{a-b}\right| + 1 + \left|\frac{f_n(b) - f(b)}{a-b}\right| \end{align*} With $a$ and $b$ fixed, take the limit of the above as $n \to \infty$ to get $$ \left|\frac{f(a) - f(b)}{a - b} \right| \le 1 $$ i.e. $f \in \mathcal{L}_1$ as required.

Caleb Stanford
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