Prove that orthonormalsystem is an orthonormalbasis

Question

We have an orthonormalsystem in $L^2(0, 2\pi)$: $\{e^{ikx} : k \in \mathbb{Z}\}$. Now I want to show that it's also an orthonormalbasis.

I thought the easiest way to do that would be to show that for every $v \in L^2(0, 2\pi)$ with $(v,\phi_k) = 0, \phi_k(x) = e^{ikx}$ for all $k \in \mathbb{Z}$, $v(x) = 0$ for all $x$. But I'm stuck showing that if $\int_0^{2\pi} v(x)e^{ikx} dx = 0$ for all $k \in \mathbb{Z}$, then $v=0$.

How do I proceed? Is there an easier way?

Answer 1

This must be a duplicate... A trigonometric polynomial is a linear combination of those exponentials. One of the theorems in the book, if not one of the definitions, shows that you only need to show that the trigonometric polynomials are dense in $L^2(-\pi,\pi)$. The notation $||.||_2$ will refer to that space:

Say $f\in L^2$ and $\epsilon>0$. Somewhere else in the book it shows that there is a function $g\in C([-\pi,\pi])$ with $||f-g||_2<\epsilon$. It's easy to modify $g$ to obtain a $2\pi$-periodic continuous function $h$ with $||g-h||_2<\epsilon$. Now Weierstrass says there is a trigonometric polynomial $P$ with $|h(t)-P(t)|<\epsilon$ for all $t$, hence $||h-P||_2\le\sqrt{2\pi}\epsilon$. Put it all together and the triangle inequality shows that $$||f-P||_2\le(1+1+\sqrt{2\pi})\epsilon.$$

Answer 2

Suppose $\int_{0}^{2\pi}f(x)e^{-i\lambda x}dx$ is $0$ for $n=0,\pm 1,\pm 2,\cdots$. Define a function $$ F(\lambda)=\frac{1}{1-e^{-2\pi i\lambda}}\int_{0}^{2\pi}f(x)e^{-i\lambda x}dx. $$ This function has removable singularities at $\lambda=0,\pm 1,\pm 2,\ldots$ because the integral vanishes for such $\lambda$. So $F$ extends to an entire function of $\lambda$. We show that $F$ must be identically $0$ by first showing that $F$ is uniformly bounded on $\mathbb{C}$, applying Liouville's Theorem to conclude that $F$ is a constant, and then taking a limit to show the constant is $0$.

Let $C_{N}$ be the positively oriented simple closed contour which is the boundary of the square centered at $0$ of width $2N+1$. I'll leave it to you to work out the details of how there is a constant $M$ independent of $N > 0$ such that $$ \left|\frac{e^{-i\lambda x}}{e^{-2\pi i\lambda}-1}\right| \le M, \;\;\; 0 \le x \le 2\pi,\;\; \lambda \in C_{N}. $$ Then, for any $\lambda' \in \mathbb{C}$, there is $N'$ such that $\lambda'$ is inside $C_N$ for $N\ge N'$, and $$ F(\lambda') = \frac{1}{2\pi}\oint_{C_{N}}\frac{F(\lambda)}{\lambda-\lambda'}d\lambda,\;\;\; N \ge N'. $$ Using the uniform boundedness of $F$ on all $C_{N}$, you can take a limit of the above as $N\rightarrow \infty$ in order to show that $F$ is uniformly bounded everywhere in the plane. Hence $F$ is a constant by Liouville's theorem. Therefore, there is a constant $C$ such that $$ \int_{0}^{2\pi}f(x)e^{-i\lambda x}dx = C(1-e^{-2\pi i\lambda}), \;\;\; \lambda \in \mathbb{C}. $$ Set $\lambda = -ir$ and let $r\rightarrow\infty$ to determine the constant: $$ \lim_{r\rightarrow\infty}\int_{0}^{2\pi}f(x)e^{-rx}dx = \lim_{r\rightarrow\infty}C(1-e^{-2\pi rx}) \\ 0 = C. $$ Therefore, $$ \int_{0}^{2\pi}f(x)e^{-i\lambda x}dx = 0,\;\;\;\lambda\in\mathbb{C}. $$ That means you can take all orders of derivatives at $\lambda=0$ and also get $0$: $$ 0= \left.\frac{d^{n}}{d\lambda^{n}}\int_{0}^{2\pi}f(x)e^{-i\lambda x}dx\right|_{\lambda = 0} = (-i)^{n}\int_{0}^{2\pi}f(x)x^{n}dx. $$ So $f$ must be $0$ a.e. because the polynomials are dense in $L^{2}$.