# Subspace of $\alpha$ Holder continuous functions is Closed

Let $$\Lambda_{\alpha}([0,1])$$ be the space of $$\alpha$$ Holder continuous functions on $$[0,1]$$ with the norm: $$\|f\|_{\Lambda_{\alpha}} = |f(0)| + \sup_{x,y \in [0,1], x\neq y} \frac{|f(x) - f(y)|}{|x-y|^{\alpha}}$$ and consider the subspace $$\lambda_{\alpha}$$ given by $$\frac{|f(x) - f(y)|}{|x-y|^{\alpha}} \rightarrow 0 \text{ as } x \rightarrow y \quad \forall \, y\in[0,1]$$

I'm having trouble showing that for $$\alpha < 1$$ this is an infinite dimensional closed subspace of $$\Lambda_{\alpha}([0,1])$$. I showed it was a subspace (as a vector space). Is there some theorem I should be using here? I started with a cauchy sequence in $$\lambda_{\alpha}$$ but I'm having trouble saying anything about its limit -- other than it lives in $$\Lambda_{\alpha}$$.

EDIT: Does this work?

Let $$(f_n)$$ be a Cauchy sequence in $$\lambda_{\alpha}$$. \begin{align} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} & = \frac{|\lim_{n\rightarrow \infty}f_{n}(x)-\lim_{n\rightarrow \infty}f_{n}(y)|}{|x-y|^{\alpha}} \\ & = \lim_{n \rightarrow \infty}\frac{|f_{n}(x)-f_{n}(y)|}{|x-y|^{\alpha}} \rightarrow 0 \text{ as } x \rightarrow y \end{align}

• I suspect that $\lambda_\alpha$ equals the intersection $\bigcap_{\beta>\alpha}\Lambda_\beta$. – Giuseppe Negro Nov 20 '18 at 16:13
• ah that's interesting I hadn't considered this – yoshi Nov 20 '18 at 16:26
• Yeah, that looks nice but beware. I am not that sure. Actually, now that I think a bit more about this, I think that it does not hold. – Giuseppe Negro Nov 20 '18 at 16:51

In your edit, you can insert a zero: \begin{align} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} & \le \frac{|f(x)-f_{n}(x)|}{|x-y|^{\alpha}}+\frac{|f_n(x)-f_{n}(y)|}{|x-y|^{\alpha}} +\frac{|f_n(y)-f(x)|}{|x-y|^{\alpha}} \end{align} Choose $$\epsilon>0$$. Then the first and last term are less than $$\epsilon/3$$ for all $$n$$ large enough. Fix such an $$n$$. Then the second term is less than $$\epsilon/3$$ for $$|x-y|$$ small enough. This shows that $$\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\le \epsilon$$ for all $$y$$ close to $$x$$. This is the claim.
• The last term should be $f(y)$ – badatmath yesterday