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I badly need a bound like $$[u]_\alpha = \sup_{S}\frac{|u(x) - u(y)|}{|x-y|^\alpha} \leq C\lVert u \rVert_{C^0(S)} = C\sup_S |u|$$ where $S$ is compact and $C$ is a constant not depending on $u$. Can it be done? Please let me know. Thank you.

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up vote 2 down vote accepted

No: If such a $C$ existed then the $C^{\alpha}$ norm would be equivalent to the supremum norm. But $C^{\alpha}$ bounded sets in $C(S)$ are precompact (in the supremum norm). So you can't have such a $C$ working for an infinite dimensional subspace of $C(S)$.

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We can construct continuous function such that $f_n(0)=1$, $f_n$ is linear on $[0,1/n]$, equal to $0$ on $(1/n,1)$. Then $\lVert f\rVert_{\infty}=1$ and $$\frac{|f(0)-f(1/n)|}{|0-1/n|^{\alpha}}=n^{\alpha}.$$ (the maps $f_n$ are Hölder continuous of any exponent, $[f_n]_{\alpha}=n^{\alpha}$).

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