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The $\alpha$-Hölder norm of a function $f(x)\colon I \to X$ where $I=[0,T]$ and $X$ is some Banach space with norm $\|\cdot\|$ is:

$$\|f(t)\|_{\alpha}\colon=\sup_{s \neq t \in I}\frac{\|f(t)-f(s)\|}{|t-s|^{\alpha}}$$

If $\|f(x)\|_{\alpha}<\infty$ we say $f(x)$ is $\alpha$-Hölder continuous. What is interesting is that if $f(x)$ is $\alpha$-Hölder continuous, then it is $\beta$-Hölder continuous for $0<\beta<\alpha$. Hölder continuity is integral in understanding how rough a function is. If a function is $1$-Hölder continuous, it is Lipschitz and thus a.e. differentiable. If a function is $1+\epsilon$-Hölder continuous it is constant.

Because of these things and other technical considerations we are interested in the quantity:

$$\sup\{\alpha>0 \colon \|f(x)\|_{\alpha} <\infty\}$$

My question is simple, is there a name for this quantity? I often use "the Hölder norm is $\frac{1}{2}$" for example but this is not quite right even if it is understood in context.

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  • $\begingroup$ Do you mean the (Hölder) exponent, see wiki? $\endgroup$ – mickep Aug 20 '15 at 5:59
  • $\begingroup$ @mickep Hölder exponent just means $\alpha$, I want to know what the supremum is called. $\endgroup$ – user223391 Aug 20 '15 at 6:00
  • $\begingroup$ @mickep I want to find the supremum of Hölder exponents such that a function is that exponent Hölder continuous. $\endgroup$ – user223391 Aug 20 '15 at 6:02
  • $\begingroup$ Thank you for clearifying (for me). $\endgroup$ – mickep Aug 20 '15 at 6:24
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It is sometimes referred to as Hölder exponent, see for example the abstract in this paper.

This terminology is also used in Signal and Image Multiresolution Analysis.


I would use this terminology with caution since on the Wikipedia page, for example, it appears that if $f$ is Hölder continuous with parameter $\alpha$, then $\alpha$ is a Hölder exponent for $f$. (there is no mention of the maximality of $\alpha$.)

Also, all the problems I have encountered where phrased in the following way:

  • prove that $f$ is Hölder continuous for $\alpha < \beta$,

instead of

  • prove that $\beta$ is the XXX for $f$.

So even if there is an established name for it you are not the only one unaware of it.

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