I have to check if the function $f(x,y)=\sqrt{x+y}$ is in the Sobolev space $H^1(\Omega)$. I also have to check if it has a restriction to the boundary that is $L^2(\Gamma)$. Here $\Omega=(0,1)\times(0,1)$.
I almost have no idea how to answer the second question. For the first one, I tried to find if there was a weak derivative $g$ of $f$ such that $(f,\phi')=-(g,\phi)$ for $\phi\in C_0^\infty(\Omega)$. Here $(m,n)=\int_\Omega mndx$. To that end I had \begin{align*} (f,\phi')&=\int_0^1\int_0^1\sqrt{x+y}\phi'dxdy\\ &=\ldots=\\ &=-\int_0^1\int_0^1\frac{1}{2\sqrt{x+y}}\phi dxdy\\ &=-(g,\phi) \end{align*}
where I have used integration by parts and the fact that $\phi\in C_0^\infty(\Omega)$. Obviously I am extremely skeptical about my calculations here. How to show if $f(x,y)$ is/isn't in $H^1(\Omega)$? The other thing that came to my mind was to somehow use the first identity of Green. But then I am thinking $f$ would need to be in a vector field.
It is not exactly clear what you mean by $\phi'$. You have to do the compuation for $\partial_x \phi$ and $\partial_y \phi$. But okay, the respective derivative of $f$ is the same anyway. Okay, we have established that the weak derivative wrt x and y is $-\frac{1}{2\sqrt{x+y}}$. Now, what does $H^1$ mean anyway? That $f \in L^2$ and the derivatives are in $L^2$. One might think that this might be problematic since $\int_0^1 \frac{1}{x}dx$ does not convergence, but $\int_0^1 \int_0^1 \frac{1}{x+y} dxdy$ is fine, check this.
The second question is about the boundary behavior. The trace theorem ensures the existence of a restriction in $L^2(\Gamma)$. You can also check that by hand. On the left hand side of the domain we have $x=0$ and $y \in (0,1)$. Then $\int_0^1 f(0,y)^2 dy=\int_0^1 y dy$ which is fine. You can check the other boundaries also. (Actually, you do not know whether you are allowed to do this, since a function in $H^1$ might not be continuous. But here it is obvious)