I'm trying to prove Theorem 2.21 in Brezis' book of Functional Analysis. The author leaves the proof as an exercise. Could you have a check on my attempt?
Let $E, F$ be Banach spaces. Let $A: D(A) \subseteq F$ be an unbounded linear operator that is densely defined. Let $A^\star: D(A^\star) \subseteq F^\star\to E^\star$ be the adjoint of $A$. If $A^\star$ is surjective, then there is $c>0$ such that $|u| \leq c |A u|$ for all $u \in D(A)$.
My attempt: First, we need the following lemma:
Let $(E, |\cdot |)$ be a normed linear space and $(E^\star, \| \cdot \|)$ its continuous dual. As usual, $$\|f\| := \sup_{x \in E \setminus \{0\}} \frac{\langle f, x \rangle}{|x|}, \quad \forall f\in E^\star.$$ Then $$|x| = \max_{f \in E^\star \setminus \{0\}} \frac{\langle f, x \rangle}{\|f\|}, \quad \forall x\in E.$$ [A proof is given here]
Let $B:= \{u \in D(A) \mid |Au| = 1\} \subseteq E$. For each $b \in B$, we define $J_b \in E^{\star \star} := (E^\star)^\star$ by $$\langle J_b, g \rangle_{E^{\star \star}, E^\star} := \langle g, b \rangle_{E^\star, E} \quad \forall g \in E^\star.$$
We have $$\|J_b\| := \sup_{g \in E^\star} \frac{|\langle J_b, g \rangle|}{\|g\|} = \sup_{g \in E^\star} \frac{\langle g, b \rangle}{\|g\|} = |b|, \quad \forall b\in B.$$
The last equality follows from our lemma. Clearly, $(J_b)_{b \in B}$ is a collection of bounded linear maps. Fix $g\in E^\star$ and pick $f\in F^\star$ such that $A^\star f = g$. Such $f$ exists because $A^\star$ is surjective. Then $$\begin{aligned} \sup_{b \in B} |\langle J_b, g \rangle| &= \sup_{b \in B} |\langle g, b \rangle| &&= \sup_{b \in B} |\langle A^\star f, b \rangle| \\ &= \sup_{b \in B} |\langle f, Ab \rangle| &&\le \sup_{b \in B} \|f\| \cdot |Ab| \\ &= \|f\| &&< \infty. \end{aligned}$$
It follows from uniform boundedness principle that $$\sup_{b\in B} \|J_b\| = \sup_{b\in B} |b| =:c < \infty.$$
For $u \in D(A)$ such that $|Au| > 0$, we have $u/|Au| \in B$. It follows that $\big | u/|Au| \big |\le c$ and thus $|u| \le c|Au|$. This completes the proof.