Orthonormal basis for $C([0,1])$

Question

Consider the vector space $C([0,1])$ of continuous complex-valued function on the unit interval. Endow it with the inner product $\int_{0}^1f(x)\overline{g(x)}dx$.

I want to show that the set $\{e^{2\pi i kt}: k\in \mathbb Z\}$ is orthonormal. I'm aware that a complex integral can be written in the form $\int_{a}^{b} u(x)dx +i\int_{a}^b v(x)dx$, but I'm not sure how to carry out the computations here.

I'd appreciate it if someone could give an explanation or point me to a source that includes this proof, (which I think would be somewhat common, given the relation to Fourier analysis).

Answer 1

I think you can just do the direct computation. Let's denote $e_k := e^{2\pi i k t}, k\in \mathbb{Z}$ then for $k \neq l$ we have \begin{equation} <e_k,e_l> = \int_0^1 e_k \bar e_l dt = \int_0^1 e^{2\pi i k t} e^{-2\pi i l t}dt = \int_0^1 e^{2\pi i(k - l)t}dt = \frac{1}{2\pi i (k-l)}(e^{2\pi i(k - l)} - e^0) = 0. \end{equation} But for $k = l$ we have \begin{equation} <e_k,e_k> = \int_0^1 e_k \bar e_k dt = \int_0^1 e^{2\pi i k t} e^{-2\pi i k t}dt = \int_0^1 1 dt = 1. \end{equation}

Answer 2

Establish the relationship for integers $m$ and $n$ $$ \int_{0}^{1} e^{2\pi i m t} e^{-2\pi i n t} dt = \int_{0}^{1} e^{2\pi i (m - n ) t} dt = \begin{cases} 1 & m=n \\ 0 & m\ne n \end{cases} $$

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When $m\ne n$, $$ \int e^{2\pi i (m - n ) t} dt = \frac{i \left(1 - e^{2 i \pi (m-n)}\right)}{2 \pi (m-n)} $$ When $m=n$ $$ \int e^{2\pi i (m - n ) t} dt = \int 0 \,dt = 0 $$