Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 13 / 3 / b / Solution

If $u_1=T{v}_1$ and $u_2=T{v}_2$, linearity of the weak equation makes $\alpha u_1+\beta u_2$ solve the equation with forcing $\alpha v_1+\beta v_2$. Uniqueness therefore gives $T(\alpha v_1+\beta v_2)=\alpha T{v}_1+\beta T{v}_2$.

Test the equation for $u=Tv$ with $u$ itself. The Poincaré inequality yieldsIf the gradient norm is nonzero, divide by it; if it is zero, Poincare gives $u=0$ and the estimate is automatic. In either caseHence $T:L^2(\Omega )\to W_0^{1,2}(\Omega )$ is a bounded linear operator. Notice that the energy identity has a minus sign, so $⟨Tv,v⟩=-\Vert D(Tv){\Vert }_2^2\le 0$. Testing the equations for $Tv$ and $Tz$ against each other gives $⟨Tv,z⟩=-\int D(Tv)\cdot D(Tz)=⟨v,Tz⟩$. Thus the bounded Dirichlet inverse is a negative self-adjoint operator on $L^2$.