I'm studying Pazy's book on semigroups, and i came across with the following result (page 15 section 1.4)
Corollary 4.4: Let $A$ be a densely defined closed linear operator. If both $A$ and $A^*$ are dissipative, the $A$ is the infinitesimal generator of a $C_0 $ semigroup of contractions on $X$
The proof goes as follows
By Lumer Phillips Theorem it suffices to prove that $R(I-A)=X$. Since $A$ is dissipative and closed $R(I-A)$ is a closed subspace of $X$. If $R(I-A) \neq X$ then there exists $x^* \in X^*, x^*\neq 0$ such that $\langle x^*, x-Ax \rangle = 0$ for $x\in D(A)$. This implies $x^* - A^*x^* = 0$. Since $A^*$ is dissipative it follows that $x^*=0$, contradicting the construction of $x^*$.
I believe that i understand the main idea of the proof, there are two parts of the proof that i don't really see why, and they are
1. The first one is why If $R(I-A) \neq X$ then there exists $x^* \in X^*, x^*\neq 0$ such that $\langle x^*, x-Ax \rangle = 0$ for $x\in D(A)$?. Why do we go and take an element on the dual space, and such element why is necessary orthogonal to every element to the domain of $A$.
2. How do we got this This implies $x^* - A^*x^* = 0$.?
For the record a linear operator $A$ is dissipative if and only if $||(\lambda I-A)x|| \geq \lambda ||x||$ for all $x\in D(A)$ and $\lambda >0$.
Hope u guys can help me. And thanks in advance.
(1) is the Hahn-Banach theorem. I assume $R$ means "range". Since $R(I-A)$ is at this point a proper closed subspace of $X$, Hahn-Banach says there exists a nonzero continuous linear functional $x^*$ which annihilates $R(I-A)$. That is to say, $\langle x^*, (I-A)x \rangle = 0$ for all $x \in D(A)$. Note that $\langle \cdot, \cdot \rangle$ here is denoting the pairing between $X^*$ and $X $, so $\langle x^*, x \rangle$ just means $x^*(x)$, the result of evaluating the functional $x^*$ at $x$. It is not an inner product.
For (2), note that for all $x \in D(A)$ we have $|\langle x^*, Ax \rangle| = |\langle x^*, x \rangle| \le \|x^*\|_{X^*} \|x\|_X$. This implies $x^* \in D(A^*)$. Then, for any $x \in D(A)$ we have $0 = \langle x^*, x - Ax \rangle = \langle x^* - A^* x^* , x \rangle$. Since $D(A)$ is dense, $x^* -A^* x^*$ is a contininuous linear functional which vanishes on a dense subset, so it must be identically zero: $x^* - A^* x^* = 0$.