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, 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 $C^{k,\alpha}$.

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 $C^{k,\alpha}$.