# Showing that an operator is self-adjoint

I have an inner product

$$\langle\, f,g\,\rangle = \int\int_{\Omega}f(x,y)g(x,y)r(x,y)dA$$

and I want to show that an operator, $$L$$, is self-adjoint with respect to the inner product by showing that $$\langle \,Lf,g\,\rangle =\langle \,f,Lg\,\rangle$$ whenever $$f$$ and $$g$$ vanish on the boundary, $$\delta\Omega$$.

The operator $$L$$ is defined by $$Lu = u_{xx}+u_{yy}$$.

I believe I'm supposed to use integration by parts, so that the integral on the boundary vanishes. For example, excluding the parameters $$(x,y)$$ and applying I.B.P.,

$$\langle \,Lf,g\,\rangle =\int\int_{\Omega}[(f_{xx}+f_{yy})gr]dA = \int_{\delta\Omega}[(f_{x}+f_{y})gr]\delta\Omega-\int\int_{\Omega}[(f_{x}+f_{y})\nabla(gr)]dA$$

Then, the boundary term would vanish and we'd have $$\langle \,Lf,g\,\rangle =-\int\int_{\Omega}[(f_{x}+f_{y})\nabla(gr)]dA$$

Applying the gradient, we get $$\langle \,Lf,g\,\rangle =-\int\int_{\Omega}[(g_xr+gr_x+g_yr+gr_y)f_x + (g_xr+gr_x+g_yr+gr_y)f_y]dA$$

However, not all the terms cancel with $$\langle \,f,Lg\,\rangle =-\int\int_{\Omega}[(f_xr+fr_x+f_yr+fr_y)g_x+(f_xr+fr_x+f_yr+fr_y)g_y]dA$$

Is this the right approach? Am I missing something? It's possible that my operator $$L$$ is incorrect, but I've double-checked and it seems good.