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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy