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Suppose a sequence of function $f_n:[a,b]\rightarrow \mathbb{R}$ differentiable and they are Holder continuous

$|f_n(x)-f_n(y)|<C|x-y|^\alpha$,

where $C$ is independent of $n$.

Moreover $f'_n$ is known to exist and $\alpha$-Holder continuous, then what can we say about the Holder continuity constant?

Also, $f_n \uparrow f$, where $f$ is continuous, so the convergence is uniform by Dini. Does this tell us if $f'$ exists for every point in $(a,b)$?

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  • $\begingroup$ Which kind of limit/convergence is it? Only $f_n\ f$ pointwise? $\endgroup$ – user251257 Oct 5 '15 at 15:57
  • $\begingroup$ Yes, also i know the derivatives of fn are bounded. I know f has derivatives everywhere except at one point, where it still admits left and right derivative, but not sure if they agree. Also fn and f are all convex. $\endgroup$ – Lost1 Oct 5 '15 at 16:01
  • $\begingroup$ hm... hard to say. Could you give $f_n$ explicitly? As how it looks, you get at best $f_n \to f$ uniformly, thus, $f$ is $\alpha$-Holder with the same $C$. Nothing more could say. $\endgroup$ – user251257 Oct 5 '15 at 16:07
  • $\begingroup$ no, i have proved it before, but i forgot how it was done now... $\endgroup$ – Lost1 Oct 5 '15 at 16:10
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The answer in general is no. Let $$ f_n(x)=1-\sqrt{x^2+\frac{1}{n^2}}\ . $$ Each $f_n$ is $C^\infty$, $|f'_n(x)|\le1$, $\{f_n(x)\}$ is increasing and converges uniformly to $1-|x|$, which is not differentiable at $x=0$.

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  • $\begingroup$ I also know my fn are monotonically decreasing. Somehow my set of conditions allows to prove something, but it is hard to write down all of them $\endgroup$ – Lost1 Oct 5 '15 at 17:42
  • $\begingroup$ The example can be modified to produce monotonically decreasing $f_n$'s. $\endgroup$ – Julián Aguirre Oct 5 '15 at 17:49

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