Find the resolvent set of a differential operator

Question

Let $X=C[0,1]$ and define: $$Af=f''\ \text{for }\ f \in D(A)=\{f\in C^{2}[0,1]: f(0)=f(1)=0 \}.$$ I want to find $\rho(A)$ that is the resolvent set of operator $A.$

I'm not sure how to proceed with this. From the definition $\rho(A)$ is the set of all $z\in \mathbb{C}$ such that $(A-zT)^{-1}$ is $1-1$ and onto.

My take on this problem is following. First I find the solutions to $Af-zf=f''-zf=0$ to get that $$f(t) = c_{1}e^{\sqrt{z}t}+c_{2}e^{-\sqrt{z}t}$$ solves the equation. Now by applying the boundary conditions I get that there is no $z \in \mathbb{C}$ such that the BC are satisfied and thus $\sigma(A)$ is empty and since $\rho(A) =\mathbb{C}\setminus\sigma(A) = \mathbb{C}.$ Is this correct?

Answer 1

The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2\pi^2$ where $n=1,2,3,\cdots$. This is because $\sin(n\pi x)$ vanishes at $0,1$ and $$ (A+n^2I)\sin(n\pi x) = 0,\;\; n=1,2,3,\cdots. $$

For $z \ne -n^2\pi^2$, the resolvent is determined by solving $$ f''+\lambda f=g,\;\;\; f(0)=f(1)=0. $$ Variation of parameters can be used to solve this equation. The solution is \begin{align} f(x)&=\frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}\sin(\sqrt{\lambda})}\int_{0}^{x}\sin(\sqrt{\lambda}y)g(y)dy\\&+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}\sin(\sqrt{\lambda})}\int_{x}^{1}\sin(\sqrt{\lambda}(y-1))g(y)dy. \end{align}

Answer 2

It seems not correct.E.g.:The spectrum is not empty. Consider the function $$f(t)=\sin(\pi t).$$ Then $$f^{''}(t)=-\pi^2\sin(\pi t)$$ thus $Af+\pi^2f=0$ and $f(0)=f(1)=0$.