Complex integration and bilinear operators

Question

Let $V$ be the space of differentiable complex-valued functions on the unit circle in the complex plane, and for $f,g \in V$, define $$\langle f,g \rangle= \int_0^{2\pi} \overline{f(\theta)}g(\theta)\,d\theta. $$

Show that $T=i\frac{d}{d\theta}$ is a Hermitian operator on $V$, and determine its eigenvalues on $W$.

---

See the picture on this link https://i.stack.imgur.com/TZnJJ.png to see my attempt to prove that $<f,T(g)>$ $-$ $<T(f),g>$ = $ 0$

If I can prove that it is zero, I will have shown that T is a Hermitian operator, but I am stumped. How do I do this?

About my notation, $f(θ)=f_a(θ)+if_b(θ)$ and similarly for $g$.

By the way, I have not taken complex analysis yet, so please take that into account when responding; don't give me something that will be completely over my head!

---

EDIT: I realized that I forgot to define what $W$ is

Let $W$ be the subspace of $V$ of functions $f(e^{i\theta})$, where $f$ is polynomial of degree $\leq n$. Find an orthonormal basis for $W$.

In light of this, is the set of eigenvalues still $\mathbb{Z}$?

Answer 1

To show that $T$ is Hermitian just use integration by parts and the fact that you are on a circle.

Let us find the eigenvalues of $T$, i.e. let us consider the equation \begin{align} T\psi =i\frac{d\psi}{d\theta} =\lambda\psi \end{align} such that $\psi(\theta+2\pi) = \psi(\theta)$ (since you are on a circle). But note that the above equation is just a linear ode with general solution given by \begin{align} \psi(\theta) = Ce^{-i\lambda \theta}. \end{align} However, since $\psi(\theta+2\pi) = Ce^{i\lambda(\theta+2\pi)}= Ce^{i\lambda\theta}e^{i2\pi \lambda} = Ce^{i\lambda\theta} = \psi(\theta) $, then it follows that $e^{i2\pi \lambda} = 1$ which is true if and only if $\lambda \in \mathbb{Z}$. So the eigenvalues of $T$ are precisely the integers.

Answer 2

The key fact that you need is that $f,g$ are periodic. Using integration by parts, you have \begin{align} \langle Tf,g\rangle &=\int_0^{2\pi}\overline{if'(\theta)}g(\theta)\,d\theta =-i\int_0^{2\pi}\overline{f'(\theta)}g(\theta)\,d\theta =-i\left.\overline{f(\theta)}g(\theta)\vphantom\int\right|_0^{2\pi}+i\int_0^{2\pi}\overline{f(\theta)}g'(\theta)\,d\theta\\ \ \\ &=\int_0^{2\pi}\overline{f(\theta)}\,ig'(\theta)\,d\theta =\langle f,Tg\rangle \end{align} As for eigenvalues, if $Tf=\lambda f$, this means that $$\tag1 f'(\theta)=-i\lambda f(\theta). $$ The solutions to the differential $(1)$ are the multiplies of $f(\theta)=e^{-i\lambda\theta}$. Now, there is a restriction, since we need $f$ to be differentiable on the circle, which means that it has period $2\pi$. So we need $\lambda$ real. Moreover, we need $$ e^{-i\lambda(\theta+2\pi)}=e^{-i\lambda\theta},\ \ \theta\in[0,2\pi). $$ That is, we need $e^{-2i\lambda\pi}=1$, which forces $\lambda\in\mathbb Z$.