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Throughout, we denote $C^\alpha$ for $\alpha \in (0,1]$ is the collection of real functions on the domain $[0,1]$ with Holder continuity of $\alpha$, i.e. if $f\in C^\alpha$, we have its norm $$|f|_\alpha = \sup_{[0,1]} |f| + \sup_{x\neq y \in [0,1]} \frac{|f(x) - f(y)|}{|x- y|^\alpha}$$ I want to know if the following claim is true, if not, a counter-example is desirable.

[Claim 1] Let $|f_n|_\alpha <1$, and $\lim_n f_n(x) = 0$ for all $x\in [0,1]$. Then $f_n\to 0$ in $C^\alpha$.

The following weaker result seems correct, please check.

[Claim 2] Let $|f_n|_\alpha <1$, and $\lim_n f_n(x) = 0$ for all $x\in [0,1]$. Then $f_n\to 0$ in $C^\beta$ for all $\beta\in (0, \alpha)$

Proof of [Claim 2]. A bounded set in $C^\alpha$ is precompact in $C^\beta$. Thus, any subsequence has its own subsequence $f_{n_k} \to g$ in $C^\beta$. By pointwise convergence, we must have $g= 0$. QED

user79963
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2 Answers2

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The proof of your second claim is correct. For the first one, consider the function $f_n$ whose graph is the linear interpolation of the points $(0,0$, $(2^{-n},2^{-n\alpha})$, $(2^{-n+1},0)$ and $(1,0)$. Then the $C^\alpha$-norm of $f_n$ is $1$, we have $\left\lvert f_n(x)\right\rvert\leqslant 2^{-n\alpha}$ for all $x$.

Davide Giraudo
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The second claim is correct, as well as the proof.

As for the first term consider the case $\alpha = 1$ and the sequence $$f^n(x) = \frac{1}{n}cos(nx)$$ Clearly $f^n \to 0$ in $\|\cdot \|_{\infty.}$ On the other side we also have that $$ \sup_{x \neq y} \frac{|f^n(x) -f^n(y)|}{|x-y|} = 1 $$ for every $n$. Indeed $$\frac{d}{dx}f^n \Big(\frac{\pi}{2n}\Big) = -1.$$ So the Holder seminorm does not converge to zero.

Kore-N
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