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!