I have the following problem:
Show that the set $\displaystyle \{e^{2\pi i (mx+ny)}\}_{n,m=-\infty}^\infty$ is an orthonormal set in $L^2(D)$, where $D$ is any square whose sides have length one and are parallel to the coordinate axes.
This problem is little confusing to me, because so far I have been doing similar problems, where there has been only one index letter, namely $n$ (for example $\{\phi_n\}_{n=1}^\infty$).
I think that I'm supposed to do the following:
A) Showing that:
$$\int_{k-1}^k\int_{k-1}^ke^{2\pi i(mx+ny)}\overline{e^{2\pi i(ax+by)}}\;dx\:dy = 1,$$
when $m=a$ and $n=b$ and $0$ otherwise. Have I understood this problem correctly? Any guides/hints?
Thnx for any help! =)
You are right: if you prove that that integral is $1$ for $a=m$ and $b=n$, and $0$ otherwise, you prove that the set is orthonormal.
To prove that, note that $$\overline{e^{2\pi i(mx+ny)}}=e^{-2\pi i(mx+ny)}$$
If $a=m$, $b=n$, the product in the integral is $1$, since is the product of a complex number whose mudulus is $1$ and its conjugate.
If not, the product is $e^{2\pi i(cx+dy)}$ for some integers $c,d$, not both zero. You can split the integral: $$\int_0^1e^{2\pi icx}dx\int_0^1e^{2\pi idy}dy$$
Assume, for example, that $c\neq 0$. Then $$\int_0^1e^{2\pi icx}dx=\frac1{2\pi i c}(\exp(2\pi ic)-\exp(0))=0$$ so the pruduct is $0$.
Same reasoning if $d\neq0$.