# Subspaces of $C^\alpha [0,1]$ are finite dimensional if closed in $C[0,1]$

For $0 < \alpha < 1$, let $C^\alpha([0,1])$ be the subspace of $C[0,1]$ consisting of continuous functions with norm $$\| f\|_\alpha = \|f\| + \sup_{x\neq y} \frac{|f(x) - f(y)|}{|x-y|^\alpha},$$ where $\|\cdot\|$ is the ordinary sup norm on $C[0,1]$.

Problem: Let $X$ be a linear subspace of $C^\alpha[0,1]$. Suppose further $X$ is closed in $C[0,1]$. Then $X$ is finite dimensional.

My strategy is to show that the unit ball $B \subseteq X$ is compact w.r.t. the $\| \cdot \|_\alpha$ norm. By the Arzela-Ascoli theorem, I can prove that $B$ is compact w.r.t. the $\| \cdot \|$ norm. It is clear to me that $$\|\cdot\| \leq \|\cdot\|_{\alpha}.$$ However, how can I show that there is some constant $C$ so that $$\|\cdot\|_\alpha \leq C\|\cdot\|?$$

• Why don't you show compactness w.r.t. $\|\cdot\|$ instead of $\|\cdot\|_\alpha$? Also, by assumption, $X$ is complete when equipped with $\|\cdot\|$ and $\Phi : (X,\|\cdot\|_\alpha)\to(X,\|\cdot\|),f \mapsto f$ is bounded and bijective. Now use the inverse mapping theorem. – PhoemueX Jan 17 '16 at 8:55
• Same argument as here. Not quite sure whether to mark as a duplicate. – Daniel Fischer Jan 17 '16 at 15:16
• @PhoemueX I have shown compactness w.r.t. $\|.\|$. However I don't understand your last comment. Why is $(X, \|.\|_{\alpha})$ a Banach space? We are not given that is a closed subspace of $C^\alpha([0,1])$. – Ben Lim Jan 17 '16 at 21:32
• The injection $\iota\colon C^{\alpha}([0,1]) \hookrightarrow C([0,1])$ is continuous. Hence $\iota^{-1}(X)$ is closed if $X$ is closed in $C([0,1])$. – Daniel Fischer Jan 17 '16 at 21:35
• @DanielFischer Thanks. I have posted an answer below. – Ben Lim Jan 17 '16 at 23:29

## 1 Answer

I already know that the unit ball in $X$ (denoted $B$) is compact in the $\|.\|$ topology. So I just need to have the estimate $$\| . \|_{\alpha} \leq C \|.\|$$

for some $C$ to conclude that $B$ is compact in the $\|.\|_{\alpha}$ topology. Now $X$ is closed in $C[0,1]$, the inclusion $i : C^\alpha[0,1] \to C[0,1]$ is continuous so that $i^{-1}(X) = (X,\|.\|_{\alpha})$ is closed in $C^{\alpha}[0,1]$. Now we consider the "identity map" $$\Phi : (X,\|.\|_{\alpha}) \to (X,\|.\|)$$ which is bijective and continuous. The domain is a Banach space by the paragraph above (a closed subspace of a Banach space is a Banach space). It follows by the open mapping theorem that $\Phi^{-1}$ is continuous, so we have the estimate above as desired.