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$V$ is the space of infinitely differentiable functions $f:\mathbb R \to \mathbb C$ which are periodic of $2\pi$, with the inner product $<f|g> = \int_0^{2\pi} \overline f(t)g(t)dt$. 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 $A:V\to V$?

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  • $\begingroup$ It is straightforward to show that if $A = i\frac{d}{dt}$, then $A^*A = L$. $\endgroup$ – whpowell96 Dec 13 '18 at 4:46
  • $\begingroup$ @whpowell96 How did you come up with that? $\endgroup$ – user398843 Dec 13 '18 at 4:53
  • $\begingroup$ 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 $\alpha^2=-1$ $\endgroup$ – whpowell96 Dec 13 '18 at 4:54
  • $\begingroup$ @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$? $\endgroup$ – user398843 Dec 13 '18 at 4:57
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    $\begingroup$ 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 $\endgroup$ – whpowell96 Dec 13 '18 at 5:02
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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.

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