Let $\mathcal{B}$ denote the Banach space of all continuous functions $f: [0,2\pi] \to \mathbb{C}$ such that $f(0) = f(2\pi)$. 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 $A$.
So far, I've been trying to compute the spectrum by hand, and I conjecture that $\text{spec}(A)$ is $\{n^2 : n = 0,1, 2, \dots\}$. We get one inclusion simply by noticing that, for each $n = 0, 1, 2, \dots$, $n^2$ is an eigenvalue of $A$ with periodic eigenfunction $e^{inx}$.
Showing the other inclusion is where I have gotten stuck. By factoring $$-\frac{d^2}{dx^2}- \lambda^2 = (D_x - \lambda)(D_x + \lambda), \qquad D_x = \frac{1}{i} \frac{d}{dx}, \, \lambda \neq 1, 2, \dots $$ and using the integrating factor method twice, we find that the solution $u$ to the equation $$(A - \lambda^2)u = f \in \mathcal{B},$$ is given by $$u(x) = c_2 e^{-i\lambda t} + \frac{c_1}{i2\lambda}e^{i\lambda t} + F(x), $$ for some $c_1, c_2 \in \mathbb{C}$ and where $$F(x) = \int_0^x e^{i2\lambda t} \int^s_0 e^{-i\lambda t} f(t) dt ds.$$
So the above expression for $u$ is our proposed formula for $(A - \lambda^2)^{-1}f$, but it remains to show that $(A - \lambda^2)^{-1}$ is bounded and that $u$ can indeed be made periodic.
Requiring that $u$ be periodic gives rise to a linear system to solve for $c_1$ and $c_2$. It appears this system has a unique solution unless $\lambda$ is an integer or half-integer. So, it may be that the spectrum consists of more than eigenvalues above. But I'm not sure how to decide if the half-integers are indeed in the spectrum or not.
Hints or solutions are greatly appreciated!
As you note any eigenvector of $A$ must be a solution of $-f'' - \lambda f=0$, all of which are in the form $f(t) = c_1 e^{i\sqrt\lambda t}+ c_2 e^{-i\sqrt \lambda t}$. Additionally we must impose periodicity of this solution, which becomes $$f(0)=c_1+c_2 \overset!= f(2\pi) = c_1 e^{i2\pi\sqrt \lambda }+c_2e^{-i2\pi\sqrt\lambda}.$$ Re-arrange and this is the same condition as: $$c_1(1-e^{i2\pi\sqrt\lambda})= c_2(e^{-i2\pi\sqrt\lambda}-1)$$
In the event that $\sqrt\lambda$ is an integer any choice of $c_1, c_2$ yields a solution and there is a two dimensional eingenspace. However if $\sqrt\lambda$ is not an integer, then any choice of $c_1$ with $c_2= -c_1\,\frac{1-e{i2\pi\sqrt\lambda}}{1-e^{-i2\pi\sqrt\lambda}}$ also yields a solution and the eigenspace is one dimensional.
Thus $A-\lambda$ is never injective and the spectrum is all of $\Bbb C$. If you want to recover a discrete spectrum you need boundary condition $f(0)=0=f(2\pi)$ .