Let $\left\{\alpha_{n}\right\}_{n\in \mathbb{N}}$ a sequence of positive real numbers with $$\alpha_{n}\in (0,1)\quad \textrm{and}\quad \alpha_{n}>\alpha_{n+1}$$ Moreover suppose $$\lim_{n\rightarrow +\infty}\alpha_{n}=\bar{\alpha}>0$$ Now let $\left\{f_{n}\right\}_{n\in \mathbb{N}}$ a family of functions with $f_{n}\in C^{0,\alpha_{n}}(B_{1})$ and suppose there exist $C$ independent of $n$ s.t. for every $n\in \mathbb{N}$ $$\left\|f_{n}\right\|_{C^{0,\alpha_{n}}(B_{1})}\leq C$$ Does exist a $C^{0,\frac{\bar{\alpha}}{2}}$-convergent subsequence of $\left\{ f_{n} \right\}_{n}$ with a limit $\bar{f}\in C^{0,\frac{\bar{\alpha}}{2}}(B_{1})$? I'd say yes, but i wonder if there are subtleties i don't see...
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$\begingroup$ obviously homework $\endgroup$– Nik WeaverJun 14, 2013 at 14:28
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$\begingroup$ Why wouldn't you remove the "$4,$" in all the exponents? $\endgroup$– Luis SilvestreJun 17, 2013 at 0:49
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$\begingroup$ If this isn't homework, why do you need the solution? $\endgroup$– Yemon ChoiJun 17, 2013 at 9:25
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$\begingroup$ As i said in the post i'm quite sure that the limit exists at least continuous by Ascoli-Arzela theorem, for Holder continuity i'm also quite sure, i embed $C^{0,\alpha_{n}}$ in $C^{0,\frac{3\bar{\alpha}}{4}}$, since the embedding is bounded i have that the sequence of $f_{n}$ is bounded and so i can extract a convergent subsequence in $C^{0,\frac{\bar{\alpha}}{2}}$... I was wondering if it is correct or there is some subtlety i don't see... that's all. $\endgroup$– studentJun 17, 2013 at 9:42
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