Let $X=C([0,1])$ and $$ X_1=\{f\in C^1([0,1])\colon f(0)=0\}\\ X_2=\{f\in C^1([0,1])\colon f(0)=f(1)=0\} $$ and $$ A_i\colon X_i\subset X\rightarrow X;\;\;\;A_if=f' $$ for $f\in D(A_i)$, where $i=1,2$;
I want to show that
- The spectrum of $A_1$ is empty
- The resolvent set of $A_2$ empty.
If we let $g\in X$ such that $$ (A_1-\lambda I))f=A_1f-\lambda f =g\;\;\;\;\;f(0)=0, $$ then $R(\lambda,A_1)g=f(x)$ (the solution) exists for all $\lambda \in \Bbb{C}$, hence the spectrum of $A_1$ is empty.
My question is about (2): How can I show that the resolvent set of $A_2$ is empty?
Let $g \in C[0,1]$. Suppose $(A_2-\lambda I)f=g$ has a solution, where $f(0)=f(1)=0$. Then $$ e^{-\lambda t}(f'-\lambda f)=e^{-\lambda t}g \\ \frac{d}{dt}(e^{-\lambda t}f)=e^{-\lambda t}g \\ 0= e^{-\lambda t}f(t)|_{t=0}^{1}=\int_{0}^{1}e^{-\lambda t}g(t)dt. $$ So, it is not possible for $(L-\lambda I)$ to be surjective, which is easily seen by noting that there exists $g \in C[0,1]$ such that $$ \int_{0}^{1}e^{-\lambda t}g(t)dt \ne 0. $$