Orthogonality Relations and Inner Products

I’m always on the lookout for interesting problems, particularly ones that I might assign my students at some point. The content of this post comes from Exercise 19, Chapter 4 of Rudin’s Real and Complex Analysis.

Fix a positive integer N, and set \omega:= e^{\frac{2\pi i}{N}}. Then we have the orthogonality relations

\displaystyle\dfrac{1}{N}\sum_{n=1}^{N}\omega^{nk}=\begin{cases}1&{k=0}\\ 0&{1\leq k\leq N-1}\end{cases}

Proof. It is clear that \frac{1}{N}\sum_{n=1}^{N}\omega^{nk}=1, for k=0. For the second assertion, observe that \omega^{k}\neq 1 and by the geometric series formula,

\displaystyle\dfrac{1}{N}\sum_{n=1}^{N}\omega^{nk}=\dfrac{1}{N}\omega^{k}\dfrac{\omega^{N}-1}{\omega^{k}-1}=0

\Box

In the case N\geq 3, we can use the orthogonality relations to derive the following identity for inner product spaces:

\begin{array}{lcl}\displaystyle\dfrac{1}{N}\sum_{n=1}^{N}\left\|x+\omega^{n}y\right\|^{2}\omega^{n}&=&\displaystyle\dfrac{1}{N}\sum_{n=1}^{N}\langle{x+\omega^{n}y,x+\omega^{n}y\rangle}\omega^{n}\\[.7 em]&=&\displaystyle\dfrac{1}{N}\sum_{n=1}^{N}\langle{x,x}\rangle\omega^{n}+\omega^{-n}\omega^{n}\langle{x,y}\rangle+\omega^{2n}\langle{y,x}\rangle+\omega^{n}\langle{y,y}\rangle\\[.7 em]&=&\displaystyle\langle{x,y}+\langle{x,x}\rangle\dfrac{1}{N}\sum_{n=1}^{N}\omega^{n}+\langle{y,x}\rangle\dfrac{1}{N}\sum_{n=1}^{N}\omega^{2n}+\langle{y,y}\rangle\dfrac{1}{N}\sum_{n=1}^{N}\omega^{n}\\[.7 em]&=&\displaystyle\langle{x,y}\rangle\end{array}

Similarly, for any inner product space, we have the identity

\begin{array}{lcl}\displaystyle\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\left\|x+e^{i\theta}y\right\|^{2}e^{i\theta}d\theta&=&\displaystyle\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\langle{x+e^{i\theta}y,x+e^{i\theta}y}\rangle e^{i\theta}d\theta\\[.7 em]&=&\displaystyle\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\left[\langle{x,x}\rangle e^{i\theta}+\langle{y,x}\rangle e^{2i\theta}+\langle{x,y}\rangle+\langle{y,y}\rangle e^{i\theta}\right]d\theta\\[.7 em]&=&\displaystyle\dfrac{\langle{x,y}\rangle}{2\pi}\int_{-\pi}^{\pi}d\theta\\[.7 em]&=&\displaystyle\langle{x,y}\rangle\end{array}

We can also obtain the preceding identity from the preceding identity and the dominated convergence theorem. Indeed,

\begin{array}{lcl}\displaystyle\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\left\|x+e^{i\theta}y\right\|^{2}e^{i\theta}d\theta=\dfrac{1}{2\pi}\int_{0}^{2\pi}\left\|x+e^{i\theta}y\right\|^{2}e^{i\theta}d\theta &=&\displaystyle\lim_{N\rightarrow\infty}\dfrac{1}{2\pi}\sum_{n=1}^{N}\left\|x+e^{\frac{2\pi i n}{N}}y\right\|^{2}e^{\frac{2\pi i n}{N}}\dfrac{2\pi}{N}\\[.7 em]&=&\displaystyle\lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^{N}\left\|x+e^{\frac{2\pi i n}{N}}y\right\|^{2}e^{\frac{2\pi in}{N}}\\[.7 em]&=&\displaystyle\lim_{N\rightarrow\infty}\langle{x,y}\rangle\\&=&\displaystyle\langle{x,y}\rangle\end{array}

Advertisements
This entry was posted in math.CA, Textbook Solutions and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s