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!
Best Answer
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 $$ |u(x)-u(y)|\leq CC_2|x-y|^\alpha. $$