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I'm trying to do exercise 2.18 in Brezis' book of Functional Analysis. Could you have a check on my attempt?

Let $(E, |\cdot|_E), (F, |\cdot|_F)$ be Banach spaces and $A: \operatorname{dom} A \subseteq E \to F$ be a densely defined unbounded linear operators. Moreover, the graph of $A$, denoted by $\operatorname{graph} A:= \{(x, Ax) \mid x\in \operatorname{dom} A\}$, is closed in $E \times F$. Let $A^\star: \operatorname{dom} A^\star \subseteq F^\star \to E^\star$ be the adjoint of $A$. Then $$\operatorname{ker} A = (\operatorname{im} A^\star)^\perp.$$

Proof: By the construction of $A^\star$, we have the identity $$\langle A^\star f, x \rangle_{E^\star, E} = \langle f, Ax \rangle_{F^\star, F}, \quad \forall f\in \operatorname{dom} A^\star, x\in \operatorname{dom} A.$$

Fix $x \in \operatorname{ker} A$. Then $Ax=0$ and thus $\langle A^\star f, x \rangle_{E^\star, E} = 0$ for all $f\in \operatorname{dom} A^\star$. This in turn implies $\langle g, x \rangle_{E^\star, E} = 0$ for all $g\in \operatorname{im} A^\star$. As such, $x \in (\operatorname{im} A^\star)^\perp$.

Now we are going to prove the converse. Fix $x \in (\operatorname{im} A^\star)^\perp$, i.e., $\langle A^\star f, x \rangle_{E^\star, E} = 0$ for all $f\in \operatorname{dom} A^\star$. This implies $\langle f, Ax \rangle_{F^\star, F} = 0$ for all $f\in \operatorname{dom} A^\star$. We claim that $Ax=0$. Assume the contrary that $(x, 0) \notin \operatorname{graph} A$. By Hahn-Banach theorem, there is $(g_1, g_2) \in E^\star \times F^\star \cong (E \times F)^\star$ such that $$\langle g_1, z \rangle + \langle g_2, A z \rangle < \alpha < \langle g_1, x \rangle + \langle g_2, 0 \rangle = \langle g_1, x \rangle, \quad \forall z \in \operatorname{dom} A.$$

Because $\operatorname{dom} A$ is a linear subspace of $E$, we get $\langle g_1, z \rangle + \langle g_2, A z \rangle =0$ for all $z \in \operatorname{dom} A$. It follows that $|\langle g_2, A z \rangle| = |\langle g_1, z \rangle| \le \|g_1\| |z|_E$ for all $z \in \operatorname{dom} A$. Hence $g_2 \in \operatorname{dom} A^\star$ and thus $\langle g_2, A x \rangle = 0$. It follows that $\langle g_1, x \rangle = - \langle g_2, A x \rangle = 0$, which is a contradiction.


Update: Without $\operatorname{graph} A$ being closed, we still get $\operatorname{ker} A^\star = (\operatorname{im} A)^\perp$.

By the construction of $A^\star$, we have the identity $$\langle A^\star f, x \rangle_{E^\star, E} = \langle f, Ax \rangle_{F^\star, F}, \quad \forall f\in \operatorname{dom} A^\star, x\in \operatorname{dom} A.$$

Fix $f \in \operatorname{ker} A^\star$. Then $A^\star f=0$ and thus $\langle f, Ax \rangle_{F^\star, F} = 0$ for all $x\in \operatorname{dom} A$. This in turn implies $\langle f, y \rangle_{F^\star, F} = 0$ for all $y\in \operatorname{im} A$. As such, $x \in (\operatorname{im} A)^\perp$.

Now we are going to prove the converse. Fix $f \in (\operatorname{im} A)^\perp$, i.e., $\langle f, Ax \rangle_{F^\star, F} = 0$ for all $x \in \operatorname{dom} A$. Clearly, $f \in \operatorname{dom} A^\star$. It follows that $\langle A^\star f, x \rangle_{E^\star, E} = 0$ for all $x \in \operatorname{dom} A$. Because $A^\star f \in E^\star$ and $\operatorname{dom} A$ is dense in $E$, we get $A^\star f = 0$ and thus $f \in \operatorname{ker} A^\star$.

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