Hoi, consider $L_2(\Omega)$ with $\Omega = (0,1)\times (0,1)$ and let $u\in L_2(\Omega)$ be defined as
$u(x,y) = 1$ for $x>y$ and $u(x,y) = 0$ for $x\leq y$. Let $A = \partial_x^2 - \partial^2_y$ an operator densely defined on $C_0^{\infty}(\Omega)\subset L_2(\Omega)$
Let $(f,g) = \int_{\Omega}f\overline{g}d\omega$
I want to show that $A' = A $(the adjoint).
Also I want to show that there exists $f\in L_2(\Omega)$ such that $$(f,\phi)= (u,A'\phi) $$ with $Amax f=0$
for all $\phi \in C_0^{\infty}(\Omega)$ so that $f = Au$ in the weak sense. How do we do this? I have an idea of how to do the first part, but in the second one im clueless.
Let $s=x+y$ and $t=x-y$. Note that $x=\frac{s+t}{2}$ and $y=\frac{s-t}{2}$. Write $h(s,t)=\phi(\frac{s+t}{2},\frac{s-t}{2})$ and calculate $\partial^2_{st} h$ to find that $$4\partial^2_{st} h=\partial^2_x\phi-\partial^2_y\phi$$
Substitue it in the expression $\int_{x>y}(\partial^2_x-\partial^2_y)\overline{\phi}$. Now, draw a picture of the new domain of integration, and convince yourself that it is good to do integration by parts with relation to the variable $s$.