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When is $C^{k,\alpha}(\bar{\Omega})$ a subset of $C^{k',\alpha'}(\bar{\Omega})$?

Gilbarg and Trudinger says that "for the domains of interest in this work the inclusion will hold whenever $k + \alpha < k' + \alpha'$". They then give one example of a cusped domain for which $C^{1}(\bar{\Omega})$ is not a subset of $C^{\alpha}(\bar{\Omega})$, but don't elaborate further. Have the inclusions been characterized precisely in the form of a necessary and sufficient condition?

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2 Answers 2

It should be clear from the definitions that the example given is roughly the only thing that can go wrong. In other words, it is when $C^1$ does not go into $C^{0,1}$. This difference is of course because $C^1$ only cares about the points that are, in some sense, infinitely close. Whereas $C^{0,1}$ can jump across boundaries, in some sense.

Perhaps an example would make it clear:

Let $\Omega_1$ be the half plane $x < 0$ Let $\Omega_2$ be the region $x > y^k$ for some positive integer $k$. Let $f(x,y) = y^2$ in $\Omega_2$ and $- y^2$ in $\Omega_1$. Then $f$ is clearly in $C^1(\bar{\Omega_1})$ and $C^1(\bar{\Omega_2})$. To check that it is in $C^1(\overline{\Omega_1\cup\Omega_2})$ it suffices to check that $\lim_{(x,y)\to (0,0)}f$ and $\nabla f$ agree with the limit is taken from inside $\Omega_1$ or $\Omega_2$. In this case, both functions vanish at the origin.

It is also very clear that $f\in C^{0,1}(\bar{\Omega_i})$ for $i = 1,2$ individually. But by the definition of the Holder norms, to compute the Holder constant of a point $z\in \Omega_2$ relative to $\Omega_1\cup \Omega2$, you have to take the sup over, say, all points within distance 1 of $z$, not just those points in $\Omega_2$. So if you take the sequence of points $(x_j,y_j) = (2j^{-k}, j^{-1})$. Note that $(x_j,y_j) \in \Omega_2$ and $(-x_j,y_j)\in \Omega^1$. The distance between them is $4j^{-k}$, while the difference of the value of the function $f$ is

$$ | f(x_j,y_j) - f(-x_j,y_j) | = 2 j^{-2} \not\leq 4 C j^{-k} $$

if $k > 2$. Just to re-iterate, the problem happens because "points between distinct regions in the set $\Omega_1\cup \Omega_2$ are allowed to get together too close too fast." And in fact this is the only possible way for things to break. Once you put a lower bound on the speed of approach of points in $\Omega_2$ to $\Omega_1$ along the boundary, you can use continuity of the first derivative implied by $C^1$ to control how fast the the difference between the function can grow in the two regions. For example, you can compute in the above example, that you have $C^{0,\alpha}$ for every $0 < \alpha < 2/k$ , while no inclusion is possible for $\alpha > 2/k$.

You may consider this to be a defect to the definitions. (Whether you put the defect with $C^1$ or $C^{0,1}$ is up to you.) It is clear that $\Omega_1\cup\Omega_2$ is not a connected domain. So it is a bit funny that the Holder regularity at a point $x\in\Omega^2$ should depend on the values of $f$ in $\Omega_1$. On the other hand, the two domains do become tangent very fast, so it is also a bit funny that the classical regularity should fail to factor that into the definition. Anyway, this naturally leads to the external cone condition for domains, which is what you should search for if you want to read more.

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(Sorry about all the typos. For some reason MO is really slowing down my browser when I type in the text field. It's gotten bad enough to the point that I cannot reliably edit the text.) –  Willie Wong Oct 23 '10 at 15:25
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The fact that in general $C^1_b(\Omega)\not\subset C^{0,1}(\Omega)$ is linked to the reason why hiking in mountain may be dangerous. –  Pietro Majer Oct 23 '10 at 15:49
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@Pietro: that's a wonderful quip. I have to remember that for when I teach a class on this. –  Willie Wong Oct 23 '10 at 16:42

Provided that $\Omega$ is nice enough, it's not hard to check that you have the following chain of inclusion: $C^0(\bar \Omega) \supset C^{0,\alpha}(\bar \Omega) \supset C^{0,1}(\bar \Omega) \supset C^{1,0}(\bar \Omega) \supset C^{1,\alpha}(\bar \Omega) \supset C^{2,0}(\bar \Omega) \supset...$

Hence, to compare $C^{k,\alpha}(\bar \Omega)$ to $C^{m,\beta}(\bar \Omega)$, you first compare $k$ to $m$.

If for example, $k>m$ then $C^{k,\alpha}(\bar \Omega)\subset C^{m,\beta}(\bar \Omega)$.

If $k=m$ then you compare $\alpha$ and $\beta$, if $\alpha \geq \beta$ then $C^{k,\alpha}(\bar \Omega)\subset C^{k,\beta}(\bar \Omega)$.

Obviously like Willie said, things can go wrong with crazy $\Omega$.

Furthermore, there exists a function in $C^{0,1}$ but not $C^{1,0}=C^1$. We can take $f(x)=|x|$ for $x\in [-1,1]$ to see that fact. You can also point out similar examples for other inclusions.

Eventually, you have the chain of strict inclusion, which gives you the the full characterization of $C^{k,\alpha}$.

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@Willie: After writing, I saw my typos and edited it. Thanks! –  Hung Tran Oct 23 '10 at 18:07
    
I'd like to add an – hopefully – interesting statement to give an example of how different the $C^k$ norms are from $C^{k,\alpha}$. While the first space is separable, the second space has incountable many elements $f_c$, that are pairwise more than 1 unit away from each other: $\|f_c - f_{c'}\| \geq 1$, forall $c \neq c'$. –  kostja Aug 29 '11 at 11:54
    
and I forgot to mention, that $C^{k,\alpha}$ balls are pre-compact in $C^k$. –  kostja Aug 29 '11 at 11:59

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