Proving that an operator is closed This is a question about the notes located at https://www.math.univ-toulouse.fr/~raymond/book-ficus.pdf
In section 6.2.2, theorem 6.2.3, the author would like to prove that $(A,D(A))$ is the infinitesimal generator of a semigroup of contractions on $Y = L^2(\Omega) \times H^{-1}(\Omega)$, $D(A) = H_0^1(\Omega) \times L^2(\Omega)$, where $A$ is defined by 
$$A(y_1,y_2) = (y_2, \tilde{A}y_1),$$
and where $\tilde{A}$ is defined by its action as 
$$(\tilde{A}y_1, \xi) = - \int \nabla y_1\cdot \nabla (-\Delta)^{-1}\xi,$$ in the usual sense for $(-\Delta)^{-1}$.  
It must be shown that the operator is dense and closed.  No proof is given for this.
Density seems to be immediate.  For closed, I tried to follow the proof from the previous page.  Let $(y_{1n},y_{2n})$ converge to $(y_1,y_2)$ in $H_0^1(\Omega)\times L^2(\Omega)$, and $A(y_{1n},y_{2n})$ converge to some $(f,g) \in L^2 \times H^{-1}$. Because $\nabla$ is bounded from $H_0^1 \mapsto L^2$, and bounded operators preserve (weak) convergence, by the definition, we have that $y_2 = f$, and $\tilde{A} y_1 = g$.  But then how can we show convergence of $y_{1n}$?  We would need a Cauchy estimate like $$||y_{1n}-y_{1m}||_{H_0^1} \leq C||\tilde{A}y_{1n} - \tilde{A}y_{1m}||_{H^{-1}}.$$
Is this somehow immediately true and I'm missing something?
Edit:  The second part of this doesn't make sense either.  The author tries to prove the resolvent operator bound by using the $\Delta$ operator, but of course we must use $\tilde{A}$.  Is there a reason we can use $\Delta$ instead of $\tilde{A}$?   
 A: The definition of $\tilde A$ is not so clearly written in this file: $\tilde A$ is defined as
$$
\tilde A:=-\Delta : H^1_0(\Omega) \subset H^{-1}(\Omega) \to H^{-1}(\Omega),
$$
while the integral you wrote is the chosen scalar product on $H^{-1}(\Omega)$:
$$
(\zeta_1,\zeta_2)_{H^{-1}(\Omega)} :=\int_\Omega \nabla w_1\cdot \nabla w_2
$$
with $w_i:=(-\Delta^{-1})\zeta_i$, $i=1,2$. Then it follows
$$
(\tilde Ay_1,\zeta)_{H^{-1}(\Omega)}  =\int_\Omega \nabla y_1\cdot \nabla w
$$
with $w=(-\Delta^{-1})\zeta$ and using the trivial identity $(\Delta^{-1})\tilde Ay_1=y_1$ for $y_1\in H^1_0(\Omega)$.

It is now easy to argue closedness: Assume $(y_{1,n},y_{2,n}) \to (y_1,y_2)$ in $L^2(\Omega) \times H^{-1}(\Omega)$ with $A(y_{1,n},y_{2,n})\to (f,g)$ in $L^2(\Omega) \times H^{-1}(\Omega)$. Since $A(y_{1,n},y_{2,n}) = (y_{2,n},-\Delta y_{1,n})$, it follows $y_{2,n}\to f$ and $-\Delta y_{1,n}\to g$.
It follows that $f=y_2 \in L^2(\Omega)$. 
Since $(-\Delta)^{-1}$ is continuous from $H^{-1}(\Omega)$ to $H^1_0(\Omega$ it follows $(-\Delta)^{-1}(-\Delta y_{1,n})=y_{1,n} \to (-\Delta^{-1})g$.
Hence $y_1\in H^1_0(\Omega)$ and $g=-\Delta y_1$. It follows $(y_1,y_2)\in D(A)$ and $A(y_1,y_2)=(f,g)$.
