Orthonormal basis for $\mathcal{L}^2([0,T] \times \mathbb{T})$

Question

Let $T>0$ and let $\mathbb{T}=\mathbb{R}\setminus\mathbb{Z}=[0,1)$ be the one dimensional continuous torus.

I know for a fact that if $\psi_k(u)=e^{2 \pi i k u}$, then $\{\psi_k(u): k \in \mathbb{N}_0\}$ is an orthonormal basis for $\mathcal{L}^2(\mathbb{T})$ with inner product $\langle f,g \rangle= \int_{\mathbb{T}} f(u) \overline{g(u)} \ du$, where $ \overline{g(u)} $ is the conjugate of $g(u)$.

I want to find an orthonormal basis for $\mathcal{L}^2([0,T] \times \mathbb{T})$.

I would say that the basis would be $\{\psi_k(t,u)=e^{2 \pi i k (t+u)}: t \in [0,T] \text{ and }k \in \mathbb{N}_0\}$. However, I don't know what would be the inner product and how I would prove it (I probably have to divide by $T$ and integrate twice, but I'm not sure)

Does anybody know how to do it? If possible, I would like you to refer me to a book where this is done.

Answer 1

Based on kahen's comment, I think I came up with the answer:

For $t \in [0,T]$ and $u \in \mathbb{T}$, let $\phi_k(t)=\tfrac{1}{\sqrt{T}}e^{2\pi i k t/T}$ and $\xi_k(u)=e^{2\pi i k u}$. Then

• $\{ \phi_n: n \in \mathbb{N}_0\}$ is an orthonormal basis of $L^2([0,T])$ for the innner product $\langle f,g \rangle= \int_0^T f(t) \overline{g(t)} \ dt $
• $\{ \xi_k: k \in \mathbb{N}_0\}$ is an orthonormal basis of $L^2(\mathbb{T})$ for the innner product $\langle f,g\rangle= \int_\mathbb{T} f(u) \overline{g(u)} \ du $,

where $ \overline{g} $ is the conjugate of $g$.

Thus, if $\psi_{n,k}(t,u) = \phi_{n}(t)\xi_k(u) = \tfrac{1}{\sqrt{T}}e^{2 \pi i (nt/T+ku)}$, then $\{ \psi_{n,k} : n,k \in \mathbb{N}_0\}$ is an orthonormal basis of $L^2([0,T] \times \mathbb{T})$ for the inner product $\langle f,g \rangle= \int_0^T \int_\mathbb{T} f(t,u) \overline{g(t,u)} \ du \ dt $. It is easy to verify that the functions are orthonormal \begin{equation*} \begin{split} \langle\psi_{n,k},\psi_{n',k'}\rangle & = \int_0^T \int_\mathbb{T} \tfrac{1}{\sqrt{T}} e^{2\pi i (nt/T+ku)} \tfrac{1}{\sqrt{T}} e^{- 2\pi i (n't/T+k'u)} \ du \ dt\\ & = \tfrac{1}{T} \int_0^T \int_\mathbb{T} e^{2\pi i ((n-n')t/T+(k-k')u)} \ du \ dt = \begin{cases} 1 & \text{ if } n = n' \text{ and } k = k', \\ 0 & \text{ if } n \neq n' \text{ or } k \neq k'. \end{cases} \end{split} \end{equation*}