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I am studying myself some facts about $\alpha$-Hölder-continuous functions but I don't get any further by proving the following:

$(1)$ $\forall\alpha\in ]0,1]$ is $C^{0,\alpha}$ dense in $C^0(D)$ concerning the uniform norm and $D\subset\mathbb R^n$.

$(2)$ $\forall\alpha\in ]0,1]$ and compact set $K\subset\mathbb R^n$ is $(C^{0,\alpha}(K),||\cdot||_{C^{0,\alpha}(K)})$ a complete space (with $||u||_{C^{0,\alpha}(K)}:=||u||_{\sup}+\sup\limits_{{x,y\in K\space\&\space x\ne y}}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$ and $C^{0,\alpha}(K)$ the space of all $\alpha$-Hölder-continuous functions)

$(3)$ All bounded closed subsets of $(C^{0,\alpha}(K),||\cdot||_{C^{0,\alpha}(K)})$ are compact.

So how do you prove one $(1),(2),(3)$ ?

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(1) Assume $D$ compact then by Weiertrass approximation you get that polynomials are dense in $C(D)$, since $D\subset B_R(0)$ for some $R>0$ and polynomials are $\alpha$-Hölder in $B_R(0)$ we get the result.

If $D$ is not compact the result is not true: Take $D=(0,1)$ and $f(x)=\sin(1/x)$. Take $0<r$ and $g\in C^{0,\alpha}$ such that $\| g-f\|_\infty <r$, then for every $n\in \mathbb{N}$ there exists $x_n,y_n\in (0,1)$ with $|x_n-y_n|<r$ and $f(x_n)=1$, $f(y_n)=-1$. Then $|g(x_n)-g(y_n)| \geq 2(1-r)>r^{\alpha}>|x_n-y_n|^{\alpha}$ if $r$ is small enough, contradicting $g\in C^{0,\alpha}$.

If $D=\mathbb{R}$ a similar argument works if we use something like $f(x)= \sin(2n\pi(x-n))$ if $x\in [n,n+1/n]$, $n\in \mathbb{N}$ and zero otherwise.

(2) We know that $C(K)$ is complete so we only need to check that the limit function of a Cauchy sequence in $C^{0,\alpha}$ is in this space and that the seminorms converge. For this notice $$ |f(x)-f(y)|\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)| $$ So, picking $n$ so that $\| f-f_n\|_{\infty} <\epsilon$ we get, and remembering that $\frac{|f_n(z)-f_n(w)|}{|z-w|^\alpha} \leq [f_n]_\alpha \leq C$, $$ \frac{|f(x)-f(y)|}{|x-y|^\alpha} \leq C+2\epsilon/|x-y|^\alpha $$ which gives $[f]_\alpha <C<\infty$.

To see that the seminorms converge just notice $$ \frac{|f(x)-f_n(x)-f(y)+f_n(y)|}{|x-y|^\alpha} \leq\limsup_m [f_m-f_n]_\alpha $$ which gives $$ \limsup_n [f-f_n]_\alpha \leq \limsup_n \limsup_m [f_m-f_n]_\alpha \to 0. $$ (3) I don't know about this statement as is, but Arzela-Ascoli gives that bounded closed sets in $C^{0,\alpha}(K)$ are precompact in $C(K)$

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    $\begingroup$ what's $[f_n]_\alpha$ in (2) ? $\endgroup$ – sheldoor Jun 16 '12 at 20:10
  • $\begingroup$ @sheldoor: In general $$[g]_\alpha:= \sup_{x,y} \frac{|g(x)-g(y)|}{|x-y|^\alpha}$$ $\endgroup$ – Jose27 Jun 17 '12 at 5:13
  • $\begingroup$ Nice, but one should add the subtle observation that in the inequality that gives the $[f]_\alpha <C$, epsilon is independent of $x, y$ so, one FIRST lets epsilon go to zero, and then takes sup over $x\neq y$. $\endgroup$ – Behnam Esmayli Mar 30 '18 at 15:26
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(1) You should be more specific regarding $D$. Compact sets are fine. In general you want to make sure that the elements of your space of continuous functions are bounded and uniformly continuous; otherwise it's hard to approximate them uniformly... Here is one possible approach. Given a bounded uniformly continuous $f\colon D\to\mathbb R$, consider $f_n(x)=\inf_{y\in D}(f(y)+n|x-y|)$. Each $f_n$ is Lipschitz and $f_n\le f$ everywhere. Moreover, $f_n\to f$ uniformly, as you should be able to prove using the uniform continuity of $f$ (the idea is that only $y$ closed to $x$ affect the infimum).

Re (2): it's a good idea to begin by studying the proof that $C(K)$ is complete.

(3) is false: take $K=[0,1]$ and $f_k(x)=\sqrt{\max(x-1/k,0)}$. The sequence $f_k$ is bounded in $C^{0,1/2}$ but has no convergent subsequence. Indeed, the limit would have to be $f(x)=\sqrt{x}$, but $\|f-f_k\|_{C^{0,1/2}}=1$ for all $k$.

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