I'm reading Theorem 3.24 in Brezis's book of Functional Analysis.
Let $E$ and $F$ be two reflexive Banach spaces. Let $A: D(A) \subseteq E \rightarrow$ $F$ be an unbounded linear operator that is densely defined and closed. Then $D\left(A'\right)$ is dense in $F'$. Thus $A'': D\left(A''\right) \subseteq E'' \rightarrow F''$ is well defined and it may also be viewed as an unbounded operator from $E$ into $F$. Then we have $A''=A$.
Below I can only show that there is an isometric embedding from the graph of $A$ to that of $A''$. I could not prove that this embedding is indeed surjective. I have not used the fact that $A$ is closed. The proof of the author does use this fact.
Could you elaborate on how to finish my proof?
My attempt: Let $J_1, J_2$ be the canonical isometric isomorphisms from $E$ to $E''$ and from $F$ to $F''$ respectively. First, we show that $J_1 [D(A)] \subseteq D(A'')$. Let $x \in D(A)$. Then $\langle J_1 x, f \rangle = \langle f, x \rangle \le |x| \cdot \|f\|$ for all $f\in D(A')$. So $J_1 x \in D(A'')$. Next we show that $A'' (J_1x) = J_2(Ax)$ for all $x\in D(A)$. In fact, $$\langle A''(J_1x), f \rangle = \langle J_1 x, A'f \rangle = \langle A'f,x \rangle = \langle f, Ax \rangle, \quad \forall f \in D(A').$$
On the other hand, $\langle J_2(Ax), f \rangle = \langle f, Ax \rangle$ for all $f \in F^\star$. It follows that $\langle A''(J_1x) - J_2(Ax), f \rangle = 0$ for all $f \in D(A')$. Because $D(A')$ is dense in $F'$, we get $A''(J_1x) - J_2(Ax) = 0_{F''}$ and thus $A''(J_1x) = J_2(Ax)$.
Let $G(A)$ and $G(A'')$ be the graphs of $A,A''$ respectively. Clearly, $G(A)$ is a vector subspace of $E \times F$. Also, $G(A'')$ is a vector subspace of $E'' \times F''$. Consider $$J:E \times F \to E'' \times F'', (x,y) \mapsto (J_1 x, J_2y).$$
Then $J$ is an isometric isomorphism. We have $J[G(A)] \subseteq G(A'')$.