Let $X$ and $Y$ Banach spaces. Let $A:D(A)\subset X\to Y$ with $D(A)$ dense in $X$ and $y\in Y$. Let $x\in X$ a weak solution of $Ax=y$, i.e. $(x,A^*y')=(y,y')$ forall $y'\in Y^*$ where $A^*:D(A^*)\subset Y^*\to X^*$ is the adjoint of $A$. The domain is $D(A^*)=\left\{y'\in Y^*:\exists x'\in X^*: y'(Ax)=x'(x)\forall x\in D(A)\right\}$
Question 1. If $x\in D(A)$ then $Ax=y$?
My attempt: $x\in X$ weak solution then $(x,A^*y')=(y,y')$ forall $y'\in Y^*$. The adjoint $A^*$ implies $(x,A^*y')=(Ax,y')$ for all $x\in D(A),\, y'\in D(A^*)$. Therefore, $(Ax,y')=(y,y')$ for all $x\in D(A),\, y'\in D(A^*)$. Equivalent, $y'(Ax-y)=0$ for all $x\in D(A),\, y'\in D(A^*)$. From here it can be concluded that $Ax=y$?
Question 2. The definition of $x$ is a weak solution of an equation $Ax=y$ requires that $x\in D(A)$? or not?
For the first question… From the definition of adjoint operator, the definition of weak solution becomes $(Ax,y’)=(y,y’)$ for all $y’\in Y^*$ if $x\in D(A)$. By Hahn Banach theorem (or, say, since the weak topology separates points), $Ax=y$.
For the second question, I imagine the point could be precisely to define a solution even in the case when $x\not\in D(A)$.