If $E,F$ are reflexive Banach spaces, then the graphs of unbounded linear operators $A$ and $A''$ are isometrically isomorphic I'm reading Theorem 3.24 in Brezis's book of Functional Analysis. The statement of the theorem is:

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 $D(A')$ is dense in $F'$. Thus $A'': D(A'')\subseteq E'' \rightarrow F''$ is well defined and it may also be viewed as an unbounded operator from E into F. Then we have $A''=A$.

At the end of the proof, in order to prove $A''=A$ it is stated:
$$\big [I'[ G(A')]=G(A)^{\bot} \, \text{ and } \, I''[G(A'')]=G(A')^{\bot} \big ]\implies G(A'')=G(A)^{\bot \bot}=G(A)$$ with $$\begin{aligned}
I':F'\times E' \rightarrow E' \times F', (g, h) \mapsto (-h, g) \\
I'' : E'' \times F'' \to F'' \times E'', (\varphi, \phi) \mapsto (-\phi, \varphi)
\end{aligned}$$
I have tried to make his implicit arguments clearer to gain better understanding. I posted my proof as an answer below. Could you have a check on my attempt?
PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, I will happily accept theirs.
 A: Notice that $(E \times F)'$ is isometrically isomorphic to $E' \times F'$. Consider $$I' : F' \times E' \to E' \times F', (g, h) \mapsto (-h, g).$$
Then $I'$ is an isometric isomorphism (i.i.). Let $G(A) \subseteq E \times F$ and $G(A') \subseteq F' \times E'$ be the graphs of $A$ and $A'$ respectively.
We denote by $G(A)^\bot$ the orthogonal complement (o.c.) of $G(A)$ in $E' \times F'$. Notice that $(g, h) \in G(A') \iff I'(g,h) \in G(A)^\bot$. It follows that $\color{blue}{I'[G(A')] = G(A)^\bot}$, and consequently $\color{blue}{\{I'[G(A')]\}^\bot = G(A)}$ because $A$ is closed.
Again, $(F' \times E')'$ is isometrically isomorphic to $F'' \times E''$. We define another i.i.  $$I'' : E'' \times F'' \to F'' \times E'', (\varphi, \phi) \mapsto (-\phi, \varphi).$$
We denote by $G(A')^{\bot}$ the o.c. of $G(A')$ in $F'' \times E''$. Let $G(A'') \subseteq E'' \times F''$ be the graph of $A''$. As in the case of $I'$, we have $\color{blue}{I''[G(A'')] = G(A')^{\bot}}$.
We denote by $G(A')^{\bot}_\mathrm b$ the o.c. of $G(A')$ in $F \times E$. Notice that $(y, x) \in G(A')^\bot_\mathrm b \iff (-x, y) \in \{I'[G(A')]\}^\bot$. Let $$I:  F \times E \to E \times F, (y, x) \mapsto (-x, y).$$ Then $I$ is an i.i. and $I [G(A')^\bot_\mathrm b] = \{I'[G(A')]\}^\bot$. It follows that $\color{blue}{G(A) = I[G(A')^\bot_\mathrm b ]}$. Let $J_1 : E \to E''$ and $J_2:F \to F''$ be the canonical i.i. We define an i.i. $$J:E \times F \to E'' \times F'', (x,y) \mapsto (J_1 x, J_2y).$$
We have $$\begin{aligned}
(\phi, \varphi) \in G(A')^{\bot} \iff &\langle\phi, g \rangle +\langle \varphi, A' g \rangle = 0, \forall g \in D(A') \\
\iff &\langle g, J_2^{-1} \phi \rangle +\langle A' g, J_1^{-1} \varphi \rangle = 0, \forall g \in D(A') \\
\iff &(J_2^{-1} \phi, J_1^{-1} \varphi) \in G(A')^\bot_\mathrm b \\
\iff &J^{-1}(\phi, \varphi) \in G(A')^\bot_\mathrm b.
\end{aligned}$$
This means $\color{blue}{G(A')^\bot_\mathrm b = J^{-1} [ G(A')^{\bot}]}$. Let's combine all the i.i.'s we have built so far, i.e., $$G(A) = I[G(A')^\bot_\mathrm b ] = I \circ J^{-1} [ G(A')^{\bot}] = I \circ J^{-1} \circ I''[G(A'')].$$
It follows that $G(A)$ and $G(A'')$ are isometrically isomorphic. This completes the proof.
