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## Homework Statement

Consider the space ##P_n = \text{Span}\{ e^{ik\theta};k=0,\pm 1, \dots , \pm n\}##,

with the hermitian ##L^2##-inner product

##\langle f,g\rangle = \int_{-\pi}^\pi f(\theta) \overline{g(\theta)}d\theta##.

Define operators ##A,B,C,D## as

##A = \frac{d}{d\theta}, \; \; B= i\frac{d}{d\theta}, \; \; C= \frac{d^2}{d\theta^2}, \; \; D: f\to D f(\theta) = f(\theta) + f(-\theta)##.

Which of the operators are self-adjoint? Find the eigenvalues and eigenfunctions for each operator.

## Homework Equations

The operator ##T## is self adjoint iff

##\langle Tf,g \rangle = \langle f,Tg\rangle## for all ##f,g## in the ##P_n##.

## The Attempt at a Solution

Mostly looking for some quick input if I'm getting this roughly right or not since there's no answer and I'm a bit unsure.

I first write ##f(\theta) = \sum_{-n}^n c_k e^{ik\theta}##. We're allowed to differentiate since ##f(\theta)## is in ##C^\infty##. Differentiating we have ##f'(\theta) = \sum_{-n}^n ikc_k e^{ik\theta}## and ##f''(\theta ) = \sum_{-n}^n -k^2c_k e^{ik\theta}##. every part of ##f(\theta )## is ##2\pi##-periodic so the only time we get a contribution to the integral is when the the exponential terms cancel each other exactly.

A) In this case when the index matches so we have##\langle Af, g\rangle = \int_{-\pi}^\pi \sum_{-n}^n \left( ika_kb_k \right) d\theta = 2\pi \sum_{-n}^n ika_k\overline{b_k}##.

and ##\langle f, Ag\rangle = \int_{-\pi}^\pi \left( \sum_{-n}^n -ika_k\overline{b_k} \right) d\theta = -2\pi \sum_{-n}^n ika_k\overline{b_k}## so not self-adjoint.

Similary we see that both B and C are self-adjoint.

The eigenvalues would be the functions ##e^{ikx}## for ##k = \pm 1, \pm 2,\dots , \pm n## and the eigenvalues for for ##A,B## and ##C## are ##ik, -k## and ##-k^2## respectively.

The case ##D## is slightly harder we have

##\langle Df,g \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(-\theta)\overline{g(\theta)}d\theta## and

##\langle f,Dg \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##

So we need to prove that ##\int_{-\pi}^\pi f(-\theta)\overline{g(\theta)} = \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##. Again matching the coefficients we have for the left side

##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}} \right) d\theta##

and the right side

##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}}\right) d\theta## which matches so self adjoint. The eigenvalues should be ##\frac{a_k+a_{n-k}}{a_k}## and the eigenfunctions the same as before. Did I understand this right?