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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}

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    $\begingroup$ I suspect that $\lambda_\alpha$ equals the intersection $\bigcap_{\beta>\alpha}\Lambda_\beta$. $\endgroup$ – Giuseppe Negro Nov 20 '18 at 16:13
  • $\begingroup$ ah that's interesting I hadn't considered this $\endgroup$ – yoshi Nov 20 '18 at 16:26
  • $\begingroup$ 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. $\endgroup$ – Giuseppe Negro Nov 20 '18 at 16:51
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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.

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  • $\begingroup$ The last term should be $f(y)$ $\endgroup$ – badatmath yesterday

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