# Inclusion relation of Hölder space

Here is Bruce K. Driver's lecture notes on Hölder space. Theorem 24.14 (left as an exercise there) reads:

Let $\Omega$ be a precompact open subset of $\mathbb{R}^d$,$\alpha,\beta\in [0,1]$ and $k, j\in\mathbb{N}_0$. If $j + \beta> k + \alpha$, then $C^{j,\beta}(\bar{\Omega})$ is compactly contained in $C^{k,\alpha}(\bar{\Omega})$.

And Page 53 of Trudinger's Elliptic Differential Equations of Second Order says:

Such a relation will not be true in general. For example,consider the cusped domain $$\Omega=\{(x,y)\in\mathbb{R}^2\,|\,y<|x|^{1/2},x^2+y^2<1\}$$ and for some $\beta, 1<\beta<2$, let $u(x,y)=(\rm{sgn}\,x)|y|^{\beta}$ if $y>0$,=0 if $y\le 0$. Clearly $u\in C^1(\bar{\Omega)}$. However, if $1>\alpha>\beta/2$, it's easily seen that $u\notin C^{\alpha}(\bar{\Omega})$, and hence $C^1(\bar{\Omega})\not\subset C^{\alpha}(\bar{\Omega})$.

Now, my question is: *is there something wrong on the condition of Theorem 24.14 (and Lemma 24.3)or am I misunderstand something? *

Since $\Omega$ is bounded, surely that $\bar{\Omega}$ is compact in $\mathbb{R}^d$, and then the counterexample above must satisfy the condition of Theorem 24.14. Hence $C^1(\bar{\Omega})=C^{1,0}(\bar{\Omega})\subset C^{0,\alpha}(\bar{\Omega})=C^{\alpha}(\bar{\Omega})$, which is impossible.

If $\Omega$ is supposed to be convex, then Theorem 24.12 follows easily from Proposition 24.13 and Lemma 24.3. And the above counterexample shows that generally $C^{1,0}(\bar{\Omega})\not\subset C^{0,1}(\bar{\Omega})$ which contradicts with Lemma 24.3.

I'm not sure whether I misunderstand something or actually the notes missed something. If the Theorem 24.12 and Lemma 24.3 are correct, please help me on proving them, if the notes makes some mistakes, please tell me the right conditions.

Thanks!

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This is to clarify that Lemma 24.3 is valid if $\Omega$ is Lipschitz. More generally, we have the following.

Lemma. Let $\Omega\in\mathbb{R}^n$ be a bounded domain with $C^{0,\alpha}$ boundary for some $1<\alpha\leq1$. Let $u\in C^1(\Omega)$ with all its partial derivatives bounded. Then $u\in C^{0,\alpha}(\Omega)$, i.e., $$[u]_{\alpha}=\sup_{x,y\in\Omega}\frac{|u(x)-u(y)|}{|x-y|^\alpha}<\infty.$$

Proof. Let $\{U_k\}$ be a (sufficiently fine) finite open cover of $\Omega$, such that in each $U_k$, the boundary of $\Omega$ is a rotated graph of a $C^{0,\alpha}$ function. Then note that there exists $\varepsilon>0$ such that $\{|x-y|<\varepsilon\}\subset\bigcup_kU_k\times U_k$, so that it suffices to prove $$\sup_{x,y\in U_k}\frac{|u(x)-u(y)|}{|x-y|^\alpha}<\infty,$$ for each $k$. Pick $x,y\in U_k$, and suppose that there is a piecewise smooth curve of length $\ell$ joining $x$ and $y$. Then by using the bounded derivative assumption we have $|u(x)-u(y)|\leq C\ell$ with some constant $C$. Now we have to construct such a curve with a reasonable length. If we can connect $x$ and $y$ by a straight line in $\Omega$, we have $\ell=|x-y|$. So we assume this is not the case. Without loss of generality, also assume that $\Omega$ is given by $x_n<f(x_1,\ldots,x_{n-1})$, with $f\in C^{0,\alpha}$. Let us denote by $x'$ and $y'$ the projections of $x$ and $y$, respectively, onto the plane $\{x_n=0\}$. Then we look at the function $f$ along the line segment $[x'y']$, and denote its minimum by $h$. There are two cases: 1) $x_n>h$ and $y_n>h$, and 2) $x_n\leq h$ or $y_n\leq h$. We will only treat the case 1), the other case being easier. If we now redefine $x'$ and $y'$ to be the projections of $x$ and $y$, respectively, onto the plane $\{x_n=h-\delta\}$ with some small $\delta>0$, then the curve $L=xx'y'y$ will lie in $\Omega$. The length of this curve can be estimated as $$\begin{split} \ell &\leq|x-x'|+|x'-y'|+|y-y'|\leq f(x')-h+|x'-y'|+f(y')-h+2\delta\\ &\leq C_1|x'-y'|^\alpha+|x'-y'|+C_1|x'-y'|^\alpha+2\delta\\ &\leq C_2|x-y|^\alpha+2\delta, \end{split}$$ leading to $$|u(x)-u(y)|\leq CC_2|x-y|^\alpha + 2C\delta.$$ As $\delta>0$ was arbitrary, we conclude that $$|u(x)-u(y)|\leq CC_2|x-y|^\alpha.$$

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Could you briefly explain how you get $|x-x'| \leq f(x')-h$ and also $f(x')-h \leq C|x'-y'|^{\alpha}$? –  Euler....IS_ALIVE Oct 8 '12 at 20:43
@Euler....IS_ALIVE: We have $|x-x'|\leq f(x')-h+\delta$ and $f(x')-h=f(x')-f(z)\leq C|x'-z'|^\alpha\leq C|x'-y'|^\alpha$ for some $z'$ on the line segment $[x'y']$. –  timur Oct 9 '12 at 0:01
Ah ok I get that part now. But also, you say "using the bounded derivative assumption, we have $|u(x)-u(y)| \leq C\ell$" I'm also not sure where that comes from. $C^1$ would just imply that $|u(x)-u(y)| \leq C|x-y|$ How can you say $\leq C\ell$? Thanks for your time by the way. –  Euler....IS_ALIVE Oct 9 '12 at 0:34
@Euler....IS_ALIVE: If you have $|u(x)-u(y)|\leq C|x-y|$, then you would immediately have $|u(x)-u(y)|\leq C\ell$ because $|x-y|\leq\ell$. But the point is that $|u(x)-u(y)|\leq C|x-y|$ is not true, as the line connecting $x$ and $y$ can cross the boundary of $\Omega$. If you have a piecewise smooth curve joining $x$ and $y$ then you just integrate along the curve to bound $|u(x)-u(y)|$. –  timur Oct 9 '12 at 0:44
I guess that's what I was getting at then. So you are saying then that $|u(x)-u(y)| = |\int{\frac{d}{dt}}u(\sigma(t))| \leq C\ell$ where $\ell$ is the length of the curve and $C$ is finite because we have a bounded derivative? –  Euler....IS_ALIVE Oct 9 '12 at 2:15

The counter-example by Gilbarg and Trudinger shows that the domain $\Omega$ cannot have cusps, in general. Theorem 24.14, in my humble opinion, suffers from a very common illness: unless specified, a domain need not have a smooth boundary. This is something that people in PDEs often forget: I could mention a book in which regularity theory for elliptic PDEs is presented withour any smoothness assumption on the boundary of the domain :-)

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Could you please show me the proof of Lemma 24.3 when there is no cusp? I can just prove the case when $\Omega$ is convex, by the uses of mean value theorem. It seems that we can choose a smooth curve $\sigma(t):t\in[0,1]$ with $\sigma(0)=x,\sigma(1)=y$. Then $|u(x)-u(y)|=|\int_0^1\frac{d}{dt}u(\sigma(t))\rm{d}t|=|\int_0^1Du(\sigma(t)) \sigma '(t)\rm{d}t|\le C|x-y|$, but I'm not sure about that, and I don't know how this fails in the counterexample's case. –  Y.Z Aug 16 '12 at 16:16
I am skeptical about Lemma 24.3 if $\Omega$ is not convex: $\int_0^1 |\sigma'(t)|\, dt$ is the length of the curve $\sigma$, which might be longer than $|x-y|$. In PDE theory, a domain is usually a convex, open set, by definition. –  Siminore Aug 16 '12 at 16:29
@Siminore: I think your last sentence is a bit imprecise. Convexity is a very strong assumption. A domain is a nonempty, open, connected set. In a beginning graduate level PDE, usually one assumes the boundary of the domain is smooth. –  timur Aug 16 '12 at 16:44
@timur Well, without convexity you cannot expect any intermediate-value theorem in higher dimension. It is often assumed to make estimates by the trick of integrating-the-derivative. I agree that you can drop it when you do local estimates. –  Siminore Aug 16 '12 at 16:54
@Siminore I don't remember ever seeing convexity as a part of definition of a domain. And I do read some PDE now and then. When integration along paths is needed, one can assume that the domains is quasiconvex, or uniform, or NTA... but assuming everything is convex would be throwing out way too much (for example, any nontrivial topology of the domain). –  user31373 Aug 16 '12 at 17:48