In Gilbarg and Trudinger, they have an example where a function is in $C^1(\bar\Omega)$ but not in $C^\alpha(\bar\Omega)$ where $\alpha<1$. $\Omega$ is bounded and is defined as follows


$\Omega:= (x,y): y<\sqrt{|x|},x^2+y^2<1 $ and the function is given by $u(x,y)=(\text{sign} x)y^\beta$ where $1<\beta<2$ for y>0 and the function is zero everywhere else. This function is in $C^1(\bar\Omega)$ but not in $C^\alpha(\bar\Omega)$ with $1>\alpha>\beta/2$. For some reason, I don't see why this is true?



H\"older spaces
If you can show that $u \in C^1(\bar{\Omega})$, then look at points $(x,y)$, $(-x,y)$ along the top of your domain (approximately on {$y = \sqrt{x}, y>0$}) converging to $(0,0)$. Then 
$$\frac{|u(x,y)-u(-x,y)|}{|(x,y)-(-x,y)|^\alpha} = \frac{2|y|^\beta}{|2x|^\alpha} \sim \frac{2|x|^{\beta/2}}{2^\alpha |x|^\alpha}$$
which blows up as $(x,y) \to (0,0)$ if $\alpha > \beta/2$. 


