How to show that the eigenvalues of this Hermitian operator are non-negative?

$$V$$ is the space of infinitely differentiable functions $$f:\mathbb R \to \mathbb C$$ which are periodic of $$2\pi$$, with the inner product $$ = \int_0^{2\pi} \overline f(t)g(t)dt$$. Let $$L:V\to V$$ be defined as $$L=-d^2/dt^2$$. It is known that $$L$$ is self-adjoint, $$L=L^*$$. How to show that the eigenvalues of $$L$$ are non-negative? And how to write $$L=A^*A$$ for an operator $$A:V\to V$$?

• It is straightforward to show that if $A = i\frac{d}{dt}$, then $A^*A = L$. – whpowell96 Dec 13 '18 at 4:46
• @whpowell96 How did you come up with that? – user398843 Dec 13 '18 at 4:53
• I know that $\frac{d}{dt}\frac{d}{dt} = \frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $\alpha$ such that $\alpha^2=-1$ – whpowell96 Dec 13 '18 at 4:54
• @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$? – user398843 Dec 13 '18 at 4:57
• The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not – whpowell96 Dec 13 '18 at 5:02

The associated form for $$L$$ is \begin{align} \langle Lf,f\rangle &= \int_{0}^{2\pi} -f''(t)\overline{f(t)}dt\\ &= -f'\overline{f}|_{0}^{2\pi}+\int_{0}^{2\pi}f'(t)\overline{f'(t)}dt \\ &= \int_{0}^{2\pi}|f'(t)|^2dt. \end{align} So, if $$Lf=\lambda f$$, then $$\lambda\|f\|^2=\|f'\|^2$$ forces $$\lambda$$ to be real and non-negative. Furthermore, $$\lambda =0$$ iff $$f'=0$$ or $$f$$ is constant.