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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?

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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 $\alpha > \beta/2$.

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  • $\begingroup$ Thanks. But the same two points would indicate that u is not Lipschitz!! because $1>\beta/2$ $\endgroup$
    – AAAA
    Commented Mar 29, 2012 at 20:16
  • $\begingroup$ @AAAA: The function in question isn't Lipschitz. If it were, then it would also be $C^\alpha$. $\endgroup$
    – user11046
    Commented Mar 30, 2012 at 0:58
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    $\begingroup$ The whole point of the exercise is to see what can break down when the boundary is not so nice. This example is not satisfying. It seems that the way it was presented is just not right. $\endgroup$
    – AAAA
    Commented Mar 30, 2012 at 7:02
  • $\begingroup$ I do not agree. This is precisely the kind of thing that can go wrong in this type of embedding: an inward spike in the boundary, allowing for a large oscillation of the function between two very close points. If the bdry is Lipschitz the phenomenon does not occur $\endgroup$
    – Piero D'Ancona
    Commented Mar 30, 2012 at 11:59
  • $\begingroup$ @Piero. I guess what I mean is that this example is not comparing "like with like". It seems that in one case, you are allowed to jump at the "other boundary" while in the other case, you are not. I wonder what happens if one uses some kind of "intrinsic metric"?? $\endgroup$
    – AAAA
    Commented Mar 30, 2012 at 17:44

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