Completeness of Exponentials Using Complex Analysis In order to show completness of $\{ e^{inx} \}_{n=-\infty}^{\infty}$ for $L^2[0,2\pi]$, it is enough to show that, if $f\in L^2[0,2\pi]$ is orthogonal to all exponentials, then $f=0$. I believe that Complex Analysis can be used to accomplish this by considering the function
$$
             F(\lambda)=\frac{1}{e^{-2\pi i\lambda}-1}\int_0^{2\pi}e^{-i\lambda t}f(t)dt.
$$
Outline: To do this, suppose $\int_0^{2\pi}e^{-i\lambda t}f(t)dt=0$ for $\lambda=0,\pm 1,\pm 2,\pm 3,\cdots$. Then $F$ has only removable singularities, and it extends to an entire function of $\lambda$. If $F$ can be shown to be uniformly bounded on a nested sequence of contours (squares or circles, for example,) then that should be enough to show that $F$ is uniformly bounded on $\mathbb{C}$ and, hence, a constant function. So there would be a constant $C$ such that
$$
              \int_0^{2\pi}e^{-i\lambda t}f(t)dt = C(e^{-2\pi i\lambda}-1)
$$
And that constant would have to be $0$. Then it would follow that all derivatives at $\lambda=0$ would be $0$, leading to the conclusion that $\int_0^{2\pi}t^n f(t)dt=0$. Now, by Weierstrass approximation, it would follow that $f\equiv 0$.
Can all of this be made rigorous?
 A: I'll use $z=x+iy$ instead of $\lambda$. One can show that

*

*$|F(z)|$ is uniformly bounded on circles of radius $n + 1/2$, $n = 1, 2, 3, \ldots$, and

*$\lim_{n \to \infty } F((n+\frac 12)y) = 0$.

(1) implies that $F$ is constant (using the maximum modulus principle and Liouville's theorem). (2) then implies that $F$ is identically zero.
Without loss of generality we can assume that $\Vert f \Vert_2 \le 1$. Then
$$
 \left| \int_0^{2 \pi}  e^{-izt} f(t)\, dt \right|^2 \le \int_0^{2 \pi} | e^{-izt}|^2 \, dt =  \int_0^{2 \pi} e^{2yt} \, dt = \frac{e^{4\pi y}-1}{2y}
$$
and
$$
 \left| \frac{1}{e^{-2 \pi i z}-1} \right|^2 
 = \left| \frac{e^{i \pi z}}{e^{- \pi i z}-e^{i \pi z}} \right|^2
 = \frac{4 e^{-2\pi y}}{|\sin(\pi z)|^2}
$$
so that
$$
 |F(z)|^2 \le \frac{ 4\sinh(2 \pi y)}{y |\sin(\pi z)|^2} \quad \text{for } z = x+iy , y \ne 0\, .
$$
From now on we assume that $|z|= n + 1/2$ with a positive integer $n$.
Case 1: $|y| \le 1$. Without loss of generality, $0 < y < 1$. The slope $\frac{\sinh(2 \pi y)}{y}$ is maximal for $y=1$ because of the convexity of the hyperbolic sine. And $|\sin(\pi z)| \ge 1$ for $|\pi z| = (n+1/2) \pi$ is shown here. It follows that
$$
 |F(z)|^2 \le 4 \sinh(2 \pi) \, .
$$
For continuity reasons this holds in the case $y=0$ as well.
Case 2: $|y| > 1$. Then $|\sin(\pi z)|^2 = \sin^2(\pi  x) + \sinh^2 (\pi y) \ge \sinh^2 (\pi y)$, so that
$$
|F(z)|^2 \le \frac{4 \sinh(2 \pi y)}{y \sinh^2 (\pi y)} = \frac{4}{y \tanh(\pi y)}
\le \frac{4}{y \tanh(\pi)} \, .
$$
These estimates together show that (1) and (2) are true.
A: This seems to be exactly what Jeff Rauch does in his notes “Fourier Analysis from Complex Analysis”, pag.7:
http://www.math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf
The only difference is that Rauch uses $\sin \pi \lambda$ in place of your $e^{-i2\pi \lambda}-1$. To obtain the key uniform boundedness, he uses the “Cauchy inequalities”, but to be honest I don’t understand this fully.
