Verifying $\{e^{in\theta}\}$ is orthonormal in $L^2(\mathbb{T})$

Question

Stated without proof in a text on Fourier Analysis:

It is easy to see that

$$ \int_{-\pi}^{\pi}e^{in\theta} \overline{e^{in\theta}} = 1 $$

since $|e^{in\theta}| = 1$. But how does one show that for $n \ne m$:

$$ \int_{-\pi}^{\pi}e^{in\theta} \overline{e^{im\theta}} = 0 $$

That's not what you want to show. Drop that squared modulus. You want to show that $$\int_{-\pi}^\pi e^{in\theta}\overline{e^{i m \theta}}, d\theta = 0$$ if $n\ne m$. — Giuseppe Negro, Nov 06 '17 at 12:39

score 2 · Accepted Answer · answered Nov 06 '17 at 13:44

2

For $m\neq n$ then let $k=n-m\neq0$ and \begin{align} \int_{-\pi}^{\pi}e^{in\theta} \overline{e^{im\theta}}d\theta &= \int_{-\pi}^{\pi}e^{in\theta} e^{-im\theta} d\theta \\ &= \int_{-\pi}^{\pi}e^{i(n-m)\theta}d\theta \\ &= \int_{-\pi}^{\pi}e^{ik\theta}d\theta \\ &= \dfrac{e^{ik\theta}}{ik}\Big|_{-\pi}^{\pi} \\ &= \dfrac{e^{ik\pi}-e^{-ik\pi}}{ik} \\ &= \dfrac{2}{k}\sin k\pi \\ &= 0 \end{align}

answered Nov 06 '17 at 13:44

Nosrati

29,995

On the 4th line I assume you're making use of the product rule? – user1770201 Nov 06 '17 at 18:08
The 5th line is substitution. But the 4th line is as well? – user1770201 Nov 06 '17 at 18:13
Yes -- that's the product rule, correct? – user1770201 Nov 06 '17 at 18:15
1

no integration $\int e^{ax}=\dfrac{e^{ax}}{a}$ – Nosrati Nov 06 '17 at 18:16
I see. Studying that lemma will clear up my confusion. Thanks. – user1770201 Nov 06 '17 at 18:22

score 1 · Answer 2 · answered Nov 06 '17 at 13:00

1

If $n$ is an integer $ \ne 0$ then

$\int_{-\pi}^\pi e^{int} dt= \frac{1}{i n}(e^{in \pi}-e^{-in \pi}) =0$,

since $e^{in \pi}=e^{-in \pi}= \cos( n \pi)=(-1)^n$.

answered Nov 06 '17 at 13:00

Fred

77,394

Verifying $\{e^{in\theta}\}$ is orthonormal in $L^2(\mathbb{T})$

2 Answers2