I'm reading Theorem 2.20. in Brezis' book of Functional Analysis.
Let $A: D(A) \subset E \rightarrow F$ be an unbounded linear operator that is densely defined and closed. The following properties are equivalent:
- (a) $A$ is surjective, i.e., $R(A)=F$,
- (b) there is a constant $C$ such that $$ \|v\| \leq C\left\|A^{\star} v\right\| \quad \forall v \in D\left(A^{\star}\right), $$
- (c) $N\left(A^{\star}\right)=\{0\}$ and $R\left(A^{\star}\right)$ is closed.
The proof of $(a) \Rightarrow(b)$ is
Set $$ B^{\star}=\left\{v \in D\left(A^{\star}\right) ;\left\|A^{\star} v\right\| \leq 1\right\} $$ By homogeneity it suffices to prove that $B^{\star}$ is bounded. For this purpose-in view of Corollary $2.5$ (uniform boundedness principle) -we have only to show that given any $f_{0} \in F$ the set $\left\langle B^{\star}, f_{0}\right\rangle$ is bounded (in $\mathbb{R}$ ). Since $A$ is surjective, there is some $u_{0} \in D(A)$ such that $A u_{0}=f_{0}$. For every $v \in B^{\star}$ we have $$ \left\langle v, f_{0}\right\rangle=\left\langle v, A u_{0}\right\rangle=\left\langle A^{\star} v, u_{0}\right\rangle $$ and thus $\left|\left\langle v, f_{0}\right\rangle\right| \leq\left\|u_{0}\right\|$.
Assume $B^{\star}$ is bounded by $K >0$. Then we have $$\left\|A^{\star} v\right\| \leq 1 \implies \|v\| \le K.$$ Could you explain how it allows us get such constant $C$?
Let $v\in D(A^\star)$ be any vector. Assume $\|A^\star v\|\ne 0$ and let $w = v/\|A^\star v\|$.
Then $\|A^\star w\|\le 1$, thus $\|w\|\le K$.
More detailed answer now the OP has solved his problem:
Let $v\in D(A^\star)$ be any vector.
First case: $\|A^\star v\|\ne 0\,.$
Let $w = v/\|A^\star v\|$. Then $\|A^\star w\|\le 1$, thus $\|w\|\le K$. But $\|w\| = \|v\|/\|A^\star v\|\le K$, thus $$\|v\|\le K\|A^\star v\|\,.$$
Second case: $\|A^\star v\|= 0\,.$
Let $\lambda\in \mathbb R_+^*$. Then $\|A^\star \lambda v\|=\lambda \|A^\star v\|=0\le 1$, thus $\|\lambda v\|\le K$, or equivalently $\|v\|\le K/\lambda$. Since this is true for any $\lambda$, we can let $\lambda\to \infty$, yielding $\|v\|=0$, i.e. $v=0$.
In particular, it is still true that $$\|v\|=0\le K\|A^\star v\|=0$$
Conclusion: Taking $C=K$ gives the desired result.