= Solution
If $u_1=Tv_1$ and $u_2=Tv_2$, 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 $T(\alpha v_1+\beta v_2)=\alpha Tv_1+\beta Tv_2$.
Test the equation for $u=Tv$ with $u$ itself. The <Poincaré inequality> yields
$$
\|Du\|_2^2=-\int vu\leq\|v\|_2\|u\|_2\leq C_P\|v\|_2\|Du\|_2.
$$
If the <gradient> norm is nonzero, divide by it; if it is zero, Poincare gives $u=0$ and the estimate is automatic. In either case
$$
\boxed{\|Tv\|_{W^{1,2}}\leq C_P\sqrt{1+C_P^2}\,\|v\|_2.}
$$
Hence $T:L^2(\Omega)\to W_0^{1,2}(\Omega)$ is a <bounded linear operator>. Notice that the energy identity has a minus sign, so $\langle Tv,v\rangle=-\|D(Tv)\|_2^2\leq0$. Testing the equations for $Tv$ and $Tz$ against each other gives $\langle Tv,z\rangle=-\int D(Tv)\cdot D(Tz)=\langle v,Tz\rangle$. Thus the bounded Dirichlet inverse is a negative <self-adjoint operator> on $L^2$.
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