Consider the square $S:=(0,1)\times (0,1)\subset \mathbb{R}^2$ and smooth functions $f,g:S\rightarrow (0,\infty)$, which can be extended up to the boundary by zero (i.e. $g(x),f(x):=0$ for all $x\in\partial S$ are continuous extensions). Hence $f,g\in H_{0}^1(S)$.
The diagonal $y=x$ separates the square into two triangles. If we glue $f$ and $g$ along that diagonal, would the resulting function be in $H_0^1(S)$? More precisely, do we have $h\in H_0^1(S)$, where
$$h(x,y):=\begin{cases} f(x,y), \,if\,\, x\geq y\\
g(x,y), \,if\,\, x<y \end{cases}$$
I do not know how to check if $h\in H_0^1(S)$. I tried to look for a sequence $\varphi_n\in C^{\infty}_c(S)$ which converges in $H^1_0(S)$ to $h$. I couldn't find a good candidate. Maybe, if $h$ is not continuous on the diagonal $\lbrace x=y\rbrace$, it is not possible (That's my feeling at least). Can someone help me?
Are there any conditions on $f,g$ which would guarantee that $h$ becomes a function in $H^1_0(S)$?
Best Answer
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^{1,p}$ spaces.