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?

Do you know, for example, that the polynomials are dense in $L^{2}$ ? Also, do you know any complex analysis? — Disintegrating By Parts, Jul 26 '15 at 20:14
Yes, I know that and a little bit. I still want to know why if $\int_0^{2\pi} v(x)e^{ikx} dx = 0$ for all $k \in \mathbb{Z}$, then $v=0$. — Cosmare, Jul 26 '15 at 20:17
If you know some complex analysis, then you can show that $\int_{0}^{2\pi}v(x)e^{ikx}dx=0$ for all $k$ forces $\int_{0}^{2\pi}v(x)e^{ikx}dx=0$ for all complex $k$, which allows you to take derivatives with respect to $k$ and show that $\int_{0}^{2\pi}v(x)x^{n}dx=0$ for all $n=0,1,2,3,\cdots$. — Disintegrating By Parts, Jul 26 '15 at 20:20
Alternatively, if you know enough Fourier Analysis to know $x^{n}$ can be written as a Fourier series, then you can also conclude that $\int_{0}^{2\pi}v(x)x^{n}dx=0$ for all $n=0,1,2,3,\cdots$. — Disintegrating By Parts, Jul 26 '15 at 20:22

score 0 · Answer 1 · answered Jul 26 '15 at 19:32

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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 $$||f-P||_2\le(1+1+\sqrt{2\pi})\epsilon.$$

answered Jul 26 '15 at 19:32

David C. Ullrich

92,839

Is there no way to prove it via the definition of an orthonormalbasis? – Cosmare Jul 26 '15 at 19:34
The definitions are not the same in every book - what definition do you have in mind? If I were giving the definition of "orthonormal basis" the definition would be "orthonormal set such that the linear combinations are dense". Regardless, it's a theorem that that's equivalent to whatever your definition is. And it must be a theorem right there in the book... – David C. Ullrich Jul 26 '15 at 19:37
I have a bunch of pairwise equivalent properties for a orthonormalbasis. "Linear combinations are dense" is one of them. Another one is what I mentioned in my post, and a third one is that you can write each element as an infinite linear combination (fourier series). – Cosmare Jul 26 '15 at 19:44
Well then what's the problem? I sketched a proof that the linear combinations are dense, so you're done. Linear combinations dense plus the theorem prove what you say you wanted. (If you don't want to use the theorem learn the proof of the theorem! Linear combinations dense plus the proof of the theorem also give what you say you want...) Theorems are there to be used. – David C. Ullrich Jul 26 '15 at 19:46
I just wanted to know how to prove that if for every $k \in \mathbb{Z} \int_0^{2\pi} v(x)e^{ikx} dx = 0$ then $v = 0$. – Cosmare Jul 26 '15 at 19:51
I understand what you wanted to prove! I've explained how to prove that. – David C. Ullrich Jul 26 '15 at 19:52

Disintegrating By Parts · Answer 2 · 2015-07-26T21:38:29.320

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 $L^{2}$.

Prove that orthonormalsystem is an orthonormalbasis

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