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So this is a very basic question on Hölder spaces.

Let $0 < \beta < \alpha \leq 1$. Prove that the unit ball of $C^{0,\alpha}[0,1]$ is compact in $C^{0,\beta}[0,1]$.

For reference:

$\| f \|_\alpha = \| f \|_\infty + \sup_{x \neq y} \frac{\left| f(x)-f(y) \right|}{\left| x-y\right|^\alpha}$ is the Hölder norm and $C^{0,\alpha}[0,1]=\{ f:[0,1]\rightarrow \mathbb{R} \mid \| f \|_\alpha < \infty \}$ the space of Hölder-continuous functions.

Idea would be to use the Arzela-Ascoli theorem, but I seem to get confused.

Obviously the unit ball of $C^{0,\alpha}[0,1]$ bounded with respect to $\| \cdot \|_\infty$ and uniformly equicontinuous. The Arzela-Ascoli theorem then yields, that the unit ball of $C^{0,\alpha}[0,1]$ has compact closure with respect to $\| \cdot \|_\infty$. But $\| \cdot \|_\gamma$ has a coarser topology than $\| \cdot \|_\infty$. So if the unit ball is precompact with respect to $\| \cdot \|_\infty$ it certainly needs to be precompact with respect to $\| \cdot \|_\alpha$ and $\| \cdot \|_\beta$. However the closed unit ball in a infinite dimensional Banach space cannot be compact, so the above conclusion must be false.

Where did I go wrong? And equally important, how can I solve this exercise?

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  • $\begingroup$ The larger metric always yields a stronger topology, not a weaker one.So the Holder- norm topology is a finer topology than $\|\cdot\|$, not a coarser one. $\endgroup$ – DanielWainfleet Jan 5 '16 at 11:27
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Where you went wrong has been answered in a comment.

The result is not that hard. Your proof that a sequence in the unit ball of $C^{0,\alpha}$ must have a uniformly convergent subsequence is correct. So we may as well assume that $||f_n||_\alpha\le1$ and $f_n\to0$ uniformly, and we need to show that $||f_n||_\beta\to0$.

Let $\epsilon>0$. For every $n$ and every $x,y$ we have

$$|f_n(x)-f_n(y)|\le|x-y|^\alpha=|x-y|^{\alpha-\beta}|x-y|^\beta.$$Choose $\delta>0$ so that $\delta^{\alpha-\beta}=\epsilon$. Then the above shows that $$|f_n(x)-f_n(y)|\le\epsilon|x-y|^\beta\quad(|x-y|\le\delta).$$On the other hand, if $n$ is large enough then $2||f_n||_\infty/\delta^\beta<\epsilon$, so that $$|f_n(x)-f_n(y)|\le2||f_n||_\infty\le\frac{2||f_n||_\infty}{\delta^\beta}|x-y|^\beta<\epsilon|x-y|^\beta\quad(|x-y|>\delta).$$

(Similarly if $f_n\to g$ uniformly...)

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