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Assuming $\alpha\in(0,1]$, which function $f_\alpha:\mathbb{R}\rightarrow\mathbb{R}$ is $\alpha$-Hölder continuous at $0$, but not $\beta$-Hölder continuous for $\beta>\alpha$?

I've been thinking about a function that somehow relates to $f_\alpha(t)\sim |t|^\alpha$, though I haven't been able to confirm that. Anyone that could come up with one?

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    $\begingroup$ $|t|^\alpha$ will work yes. Just plug it into each definition. $\endgroup$
    – Ian
    May 1 '18 at 22:44
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For the ease we can just search for $f_\alpha$ with $f_\alpha(0)=0$ so $|f_\alpha(t)-f_\alpha(0)|\leq M|t|^\alpha$ becomes $|f_\alpha(t)|\leq M|t|^\alpha$ (for some $M,\delta>0$ with $|t|<\delta$).

Now $|t|^\alpha$ is indeed a correct candidate for $\alpha$-Hölder continuity by choosing $M,\delta=1$. Now we need to show that $f_\alpha$ is not $\beta$-Hölder continuous, that is, we have $|f_\alpha(t)|> M|t|^\beta$ for any $M,\delta>0$ and some $|t|<\delta$. So we find $$|t|^\alpha> M|t|^\beta\iff |t|^{\beta-\alpha}< \frac{1}{M}\iff |t|< \left(\frac{1}{M}\right)^\frac{1}{\beta-\alpha}$$ Therefore $t=\frac{1}{2}\min\{\delta,\left(\frac{1}{M}\right)^\frac{1}{\beta-\alpha}\}$ will do.

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  • $\begingroup$ where does the factor $\frac{1}{2}$ comes from? Can't I just choose $t$ without that factor? $\endgroup$
    – Vic Ryan
    Oct 6 '20 at 14:00

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