How does pointwise Hölder continuous on compact subsets not imply locally Hölder continuous?

Question

When Gilbarg and Trudinger introduced the Hölder spaces, they mentioned on page 52 that

Furthermore note that local Hölder continuity is a stronger property than pointwise Hölder continuity in compact subsets.

without further elaboration.

Can someone kindly explain why?

For future readers here is a picture of the page in question — Calvin Khor, Jul 05 '22 at 05:41

score 5 · Accepted Answer · answered Jul 05 '22 at 05:29

5

It is easiest to illustrate this for pointwise $\alpha$-Holder functions with $\alpha=1$, i.e., pointwise Lipschitz functions. Consider $$f(x)=x^2\sin(1/x^2)$$ for $x$ in $[-1,1] \setminus\{0\}$, with $f(0)=0$. The pointwise Lipschitz constant in this case is simply $|f'(x)|$, which is well defined and finite everywhere in $[-1,1]$, but is not bounded in any neighborhood of $0$.

answered Jul 05 '22 at 05:29

Yuval Peres

22,782

1

This is a very helpful counter example. I think I understand Hölder continuity much better now. Thank you so much! – Cheuk Hwang Jul 27 '22 at 19:55

How does pointwise Hölder continuous on compact subsets not imply locally Hölder continuous?

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