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The function space $H^{\alpha} (\Omega)$ for $0 < \alpha \le 1$, is the set of functions:

$$\{ f \in C^0(\Omega) : \sup_{x \neq y} \dfrac{|f(x) - f(y)|}{|x-y|^{\alpha}} < \infty \}$$

with the metric $d_{H^{\alpha}} = || f - g ||_{H^{\alpha}}$, where $$||f||_{H^{\alpha}} = ||f||_{sup} + [f]_{H^{\alpha}} \text{ , } [f]_{H^{\alpha}} = \sup_{x \neq y} \dfrac{|f(x) - f(y)|}{|x-y|^{\alpha}} $$

Now, if $0 < \alpha < \beta \le 1$, then

$$[f]_{H^{\alpha}} \le 2 ||f||_{sup}^{1-\frac{\alpha}{\beta}} [f]_{H^{\beta}}^{\frac{\alpha}{\beta}} \space \forall f \in H^{\beta}$$

And also, there is some constant $M$ so that:

$$||f||_{H^{\alpha}} \le M ||f||_{sup}^{1-\frac{\alpha}{\beta}} ||f||_{H^{\beta}}^{\frac{\alpha}{\beta}} \space \forall f \in H^{\beta}$$

These were some questions on a problem set: I have checked that $d_{H^{\alpha}}$ is a metric, and proved the two properties (in the second I found that $M = 2$ is sufficient). However, rather blindly. It's easy to show from the first that if $0 < \alpha < \beta \le 1$, then $H^{\beta} \subset H^{\alpha}$.

What else do these formulas mean? Are they just some useful inequalities, or do they establish some connection between $H^{\beta}$ and $H^{\alpha}$?


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Are you sure the $\alpha$ is on both sides in the first inequality? In the second? – Glen Wheeler Feb 9 '11 at 23:47
Yes indeed! I'm so sorry, copy and paste with all the repetition... – milcak Feb 10 '11 at 1:21
this article might be of interest. It deals a lot of inclusion of one Holder space into another, and gives an application of such inclusions. The main point is the same as given by Willie Wong -- precompactness. – mpiktas Feb 10 '11 at 13:08
up vote 8 down vote accepted

Those two final inequalities are known as "interpolation inequalities". The point being the following: you can "extend" the Holder norms to $\alpha = 0$ with the formal expression

$$ [ f ]_{H^0} = \sup_{x\neq y} \frac{|f(x) - f(y)|}{|x-y|^0} = \sup_{x\neq y} \frac{|f(x) - f(y)|}{1} \leq 2 [f]_{sup} $$

Or, in other words, you identify $H^0$ with $C^0$ equipped with the sup norm. As you observed, it gives you that $H^\alpha \subset H^\beta$ if $\alpha > \beta$. What's more, however, is that now, using the sup-norm factor in the interpolation inequality, you can use Arzela-Ascoli to show that the inclusion of $H^\alpha\subset H^\beta$ is pre-compact! That is, any bounded sequence in $H^\alpha$ would have a converging subsequence in $H^\beta$, for $\beta < \alpha$.

I think you understand how, whenever something allows you to extract a converging subsequence, it is very useful in analysis indeed.

Lastly, the expression illustrates a phenomenon that happens with regularity in classical analysis, which is that good "scales" of function spaces are often log-convex in the exponent. Your family of Holder space norms $H^\alpha$, parametrized by $\alpha$, by your two inequalities, is log-convex.

There presumably are very nice applications of the log convexity in interpolation theory etc for the Hölder spaces, but unfortunately none comes to mind immediately at the moment.

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I wish I could comment...

Anyway, I don't think the second inequality holds at all, for instance all constant functions violate it, and even taking these out will not solve the problem; for example: Assume the inequality holds, take $f\in H^\beta$, $f\geq 0$ and $c>0$, then since the seminorm doesn't see constants we get $(\| f\| _{sup} +c)^{\alpha /\beta} = \| f+c\| _{sup}^{\alpha / \beta} \leq M[f]_{\beta } ^{\alpha / \beta }$, and this is for all $c>0$.

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What do you mean the seminorm doesn't see constants? If $f(x) = c$, then $[f]_{H^{\alpha}} = \sup_{x \neq y} \dfrac{|f(x) - f(y)|}{|x-y|^{\alpha}} = \sup_{x \neq y} \dfrac{0}{|x-y|^{\alpha}} = 0$. So if $f$ is constant, $||f||_{H^{\alpha}} = c + 0 = c$. – milcak Feb 10 '11 at 1:30
Then you get on the RHS $M \cdot c^{1- \frac{\alpha}{\beta}} \cdot c^{\frac{\alpha}{\beta}} = M \cdot c$, when $M = 2$, as far as I know, $c\le M\cdot c$. – milcak Feb 10 '11 at 1:35
If $f$ is constant the seminorm is zero, so the bound you get is $c\leq 0$. Why are you making one seminorm zero and the other one non-zero? – Jose27 Feb 10 '11 at 1:39
Oh Im so sorry! The very last $f$ is in a norm not seminorm. It's so hard to Latex this correctly! – milcak Feb 10 '11 at 2:45

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