Let $\Omega := \{(x,y) \in \mathbb{R}^2 \ : \ 0<x<1, \ \sqrt x < y < 1\}$ and the function $u:\Omega \to \mathbb{R}$ defined as $u(x,y) := y^{-1/3}$.
Prove that $u$ belongs to $W^{1,p}(\Omega)$ for all $p \in [1,9/4)$
This is a problem that was gave to me and I had no problem in showing that $u \in L^p(\Omega)$ for all $p<9$. Then considering the derivatives of first order of $u$ the solution given by the text says that $u \in C^1(\Omega)$ so the distributional derivatives are the usual derivatives and the solution comes from the integral of $\frac{\partial u}{\partial y}(x,y)$ etc etc..
My question is: Why is $u\in C^1(\Omega)$? for $(0,y) \to (0,0)$ doesn't explode $u$'s common derivative? My thought was to consider the distributional derivative near $(0,0)$ using dominated convergence theorem as
$$ \langle\frac{\partial u}{\partial y} , \phi\rangle = \lim_{\varepsilon \to 0^+} - \int_\varepsilon^1 dy\int_0^{y^2} dx \ \ y^{-1/3} \frac{\partial \phi}{\partial y}$$
And then going on to establish the distributional derivative..
Let me know what you think about it. Thank you.
I suspect your text is using the convention that $C^1(\Omega)$ and $C^1(\overline\Omega)$ are not the same thing; the latter is a Banach space, but the first may not be. So since the domain avoids points where $y=0$, the function is positive and finite everywhere it is defined (without a uniform upper bound).
(Since you can check which $L^p$ $u$ belongs to, you should be able to finish the question)