\documentclass{amsart}
\usepackage{calculus}

\begin{document}
\title{H\"older spaces (Midterm problem 3)}
\maketitle

\setlength{\columnsep}{.5cm}
\begin{definition}
Let $\alpha \in (0,1)$. We define the H\"older space with exponent $\alpha$ by
$$C^\alpha([0,1]) = \left\{ f:[0,1] \to \R \st \sup_{x,y \in [0,1]} \tfrac{\abs{f(x) - f(y)}}{\abs{x - y}^\alpha} < \infty \right\}$$
and define $\holdernorm{f}{\alpha} = \Lpnorm{f} + \sup\limits_{x,y \in [0,1]} \frac{\abs{f(x) - f(y)}}{\abs{x-y}^\alpha}$
\end{definition}

\begin{remark}
When $\alpha = 0$, one generally insists $f$ is continuous, and then $C^0$ is just the set of all continuous functions with the $\Lpspace{\infty}$ norm. When $\alpha > 0$, any function in $C^\alpha$ is automatically continuous. The larger $\alpha$ is, the closer your function becomes to differentiable.

When $\alpha = 1$, one generally calls these functions \textit{Lipschitz continuous}. These are almost differentiable (e.g. $f(x) = \abs{x}$ is Lipschitz, but not differentiable). There is a (hard) theorem of Radamacher that says that Lipschitz functions are differentiable `almost everywhere' (whatever that means).

When $\alpha > 1$, the only such functions are constants. (Prove this: It's easy).
\end{remark}

It is easy to see that $C^\alpha$ is a Banach space. (Most of you got this correct on the midterm). We prove the `harder' question from your midterm:
\begin{theorem}\label{3}
Let $0 < \alpha < \beta < 1$. Let $S \subset C^\alpha$ be defined by $S = \{f \in C^\alpha \st \holdernorm{f}{\beta} \leq 1 \}$. Then $S$ is compact in $C^\alpha$.
\end{theorem}

The above is an illustration of a more `general' philosophy. Compact subsets of function spaces can generally be obtained by considering subsets which are bounded in a \textit{stronger} norm. By \textit{stronger}, we mean some norm which requires a more stringent `differentiability' condition on your function.

You've seen this in many guises by now: The first example you've seen is probably the Arzella Ascolli theorem: The `equicontinuity' assumption there is really a uniform bound on the oscillation of the function. If for example the derivatives were uniformly bounded, you would have equicontinuity.

The other example you've seen is from your homework: The set of functions which is bounded in a higher Sobolev norm is compact in a lower Sobolev space. Higher Sobolev norms, are of course a more stringent condition on differentiability.

Now to prove the Theorem. First a Lemma:
\begin{lemma}
If $(f_n) \to f$ in $C^0$, and $\forall n \in \N$, $\holdernorm{f_n}{\beta} \leq 1$, then for any $\alpha < \beta$, $(f_n) \to f$ in $C^\alpha$.
\end{lemma}

\begin{proof}
First note that $f \in C^\beta$, and $\holdernorm{f}{\beta} \leq 1$. We see this because
\begin{align*}
\frac{\abs{f(x) - f(y)}}{\abs{x - y}^\beta} = \lim_{n \to \infty} \frac{\abs{f_n(x) - f_n(y)}}{\abs{x - y}^\beta} &\leq \lim_{n \to \infty} 1 - \Lpnorm{f_n}{\infty} = 1 - \Lpnorm{f}{\infty}
\end{align*}
Now take $\epsilon > 0$. Choose $N$ so that $n > N \implies \Lpnorm{f_n - f}{\infty} < \epsilon$. If $0 < \abs{x-y} < \epsilon$, then
\begin{align*}
\frac{ \abs{(f - f_n)(x) - (f - f_n)(y)}}{\abs{x - y}^\alpha} &\leq \abs{x - y}^{\beta - \alpha} \frac{ \abs{(f - f_n)(x) - (f - f_n)(y)}}{\abs{x - y}^\beta}\\
&\leq \epsilon^{\beta - \alpha} \holdernorm{f_n - f}{\beta}\\
&\leq \epsilon^{\beta - \alpha} (\holdernorm{f_n}{\beta} + \holdernorm{f}{\beta}) \leq 2 \epsilon^{\beta - \alpha}
\end{align*}
Now when $\abs{x - y} \geq \epsilon$, we have
\begin{align*}
\frac{ \abs{(f - f_n)(x) - (f - f_n)(y)}}{\abs{x - y}^\alpha} &\leq \frac{ \abs{(f - f_n)(x) - (f - f_n)(y)}}{\epsilon^\alpha}\\
&\leq 2\epsilon^{-\alpha} \Lpnorm{f - f_n}{\infty} \leq 2\epsilon^{1 - \alpha}
\end{align*}
Thus
\begin{align*}
\holdernorm{f - f_n}{\alpha} &= \Lpnorm{f - f_n}{\infty} + \sup_{\abs{x-y}<\epsilon} (\cdots) + \sup_{\abs{x-y} \geq \epsilon} (\cdots)\\
&\leq \epsilon + 2\epsilon^{\beta - \alpha} + 2\epsilon^{1-\alpha}.
\end{align*}
Since the right hand side converges to $0$ as $\epsilon \to 0$ (as all the exponents are positive), we have $(\holdernorm{f - f_n}{\alpha}) \to 0$ finishing the proof.
\end{proof}

\begin{remark}
The lemma is \textit{false} when $\alpha = \beta$ (see your homework).
\end{remark}

The proof of Theorem \ref{3} is now immediate!
\begin{proof}[Proof of Theorem \ref{3}]
Pick a sequence $(f_n)$ in $S$. Choose $\delta = \epsilon^{1/\beta}$. When $\abs{x - y} < \delta$, $\abs{f_n(x) - f_n(y)} \leq \holdernorm{f_n}{\beta} \abs{x - y}^\beta \leq \epsilon$. Thus by the Arzella Ascolli theorem, this sequence has a uniformly convergent subsequence. By the above lemma this sub sequence converges in $C^\alpha$ (since by definition of $S$, $\holdernorm{f_n}{\beta} \leq 1$).
\end{proof}

\begin{remark}
If you don't know the Arzella Ascolli theorem, then we still have a short proof of the above theorem: The idea is to first find a uniformly convergent subsequence, and then use the lemma as above.

Finding a uniformly convergent subsequence can be done as follows: First follow the hint given on the midterm, to show that $S$ is totally bounded in $C^0$. Showing $S$ is totally bounded in $C^\alpha$ lead to some pretty horrific details. However showing it is totally bounded in $C^0$, is quite short (a lot of you did this on the midterm).

$S$ is clearly closed in $C^0$. Thus once you show $S$ is totally bounded in $C^0$, we automatically get compactness. Thus any sequence in $S$ has a uniformly convergent subsequence. That's good enough by the above lemma!
\end{remark}

\end{document}