Showing that $A$ is NOT an infinitesimal generator

Question

As a state space, choose $X=L^{2}(0,1)$.

Let $A$ be defined as $\displaystyle Af=\frac{df}{d\zeta}$ with domain

$D(A)=\{f\in L^{2}(0,1)|f$ is absolutely continuous and $\frac{df}{d\zeta}\in L^{2}(0,1)\}$.

I need to show that $A$ is not an infinitesimal generator on $L^{2}(0,1)$.

I can do this by showing that $\alpha I-A$ is not injective for any $\alpha>0$.

How can I prove that it is not injective?

Answer 1

I am assuming all vector spaces are over $\mathbb{C}$.

To show that $\alpha I - A$ is not injective for any $\alpha \in \mathbb{C}$, it is sufficient to find (given $\alpha \in \mathbb{C}$) an $f \in L^2(0,1;\mathbb{C})$ such that $\alpha f - Af = 0$.

Set $f(x) = e^{\alpha x}$. Then $|f(x)| \le e^{\max(0,\Re \alpha)}$ for all $x \in (0,1)$ and hence $f \in L^2$. Also, $$ A f(x) = f'(x) = \alpha e^{\alpha x} = \alpha f(x) $$ and therefore $\alpha f - Af = 0$.