Let $\mathcal{B}$ denote the Banach space of all continuous functions $f: [0,2\pi] \to \mathbb{C}$ such that $f(0) = f(2\pi) =0$. Let $A$ denote the operator $$Af = -f'', \qquad \text{Dom}(A) = \{ f \in \mathcal{B} : f \in C^2[0,2\pi] \}.$$
I would like to find the spectrum of $A$.
Note that if $\sqrt{\lambda} \in \left\{\frac{n}{2} | n \in \mathbb{N} \right\}$, then $$f(x) = e^{i \sqrt{\lambda}x}- e^{-i \sqrt{\lambda}x} = i2\sin(\sqrt{\lambda}x) \in \text{Dom}(A)$$ is an eigenfunction of $A$ with eigenvalue $\lambda$. Therefore the spectrum of $A$ contains $\left\{\frac{n^2}{4} | n \in \mathbb{N}\right\}.$
Showing that this is all the spectrum where I am stuck. If $\lambda \notin \left\{\frac{n^2}{4} | n \in \mathbb{N}\right\}$, I need to conjecture a formula for the resolvent $u =(A - \lambda)^{-1}f$, $f \in \mathcal{B}$, and then show it is a bounded map, and that $u$ is smooth and obeys the boundary condition. My best guess is that the resolvent is described using Fourier series $$(A - \lambda)^{-1} f = \sum_{k = -\infty}^\infty \frac{e^{ikx}}{k^2 - \lambda}\hat{f}(k), \qquad \hat{f}(k) = \int^{2\pi}_0 e^{-ikt}f(t) dt,$$ or some variant thereof.
Hints or solutions are greatly appreciated!
You can use variation of parameters to find a solution of the form $$ (A-\lambda I)^{-1}f= \frac{1}{w(\lambda)}\left[\psi_{\lambda}(x)\int_0^xf(t)\varphi_{\lambda}(t)dt +\varphi_{\lambda}(x)\int_x^{2\pi}f(t)\psi_{\lambda}(t)dt\right], $$ where $w(\lambda)$ is the Wronskian $w(\lambda)=\psi_{\lambda}(x)\varphi_{\lambda}'(x)-\psi_{\lambda}'(x)\varphi_{\lambda}(x)$, which does not depend on $x$, and $\psi_{\lambda},\varphi_{\lambda}$ are the unique solutions of $f''+\lambda f=0$ that satisfy the endpoint conditions $$ \varphi_{\lambda}(0)=0,\;\; \varphi_{\lambda}'(0)=1,\\ \psi_{\lambda}(2\pi)=0,\;\; \psi_{\lambda}'(2\pi)=1. $$ You find that there are simple zeroes of the Wronskian at the points of the spectrum; everywhere else the above expression defines a bounded inverse of $(A-\lambda I)^{-1}$. The solutions $\varphi_{\lambda}$ and $\psi_{\lambda}$ are $$ \varphi_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}},\;\; \psi_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}(x-2\pi))}{\sqrt{\lambda}}. $$ The Wronskian of these solutions is \begin{align} w(\lambda)&=\frac{\sin(\sqrt{\lambda}(x-2\pi))}{\sqrt{\lambda}}\cos(\sqrt{\lambda}x)\\ &-\cos(\sqrt{\lambda}(x-2\pi))\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}} \\ &= \frac{\sin(2\pi\sqrt{\lambda})}{\sqrt{\lambda}}. \end{align} The Wronskian has simple zeros at $\sqrt{\lambda}=\pm 1,\pm 2,\pm 3,\cdots$, and a removable singularity at $\lambda=0$. As you would then expect, the spectrum is $$ \sigma(A)=\{ \pi^2, 4\pi^2, 9\pi^2,\cdots, n^2\pi^2,\cdots \}, $$ and the general element of the spectrum, $n^2\pi^2$ for $n=1,2,3,\cdots$ is an eigenvalue with corresponding eigenfunction $\sin(n\pi x)$.
NOTE: If you find a discrepancy in the algebraic sign of this resolvent expression, or a missing constant somewhere, I'll let you fix that on your own. It's a good exercise for you to work through this example on your own using the classical techniques of ODEs.