\[\begin{eqnarray}f(x)\stackrel{?}{=}\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi i n x}, \quad x \in \mathbb{R},\end{eqnarray}\]
where the Fourier coefficients for $n\in \mathbb{Z}$ are given by
\[\begin{eqnarray}\hat{f}(n)=\int_{0}^1 f(x)e^{-2\pi i x }dx, \quad n \in \mathbb{Z},\quad f\in L^1([0,1])\end{eqnarray}\]
(by $f\in L^1([0,1]),$ we mean that the restriction of $f$ to $[0,1]$ is Lebesgue integrable).
First of all, if $f$ is a trigonometric polynomial, say $f(x)=\sum_{n=-N}^Na_ne^{2\pi i n x}$, then
\[\begin{eqnarray}\int_ {0}^1 f(x)e^{-2\pi i m x}dx=\sum_{n=-N}^N\int_{0}^1 a_n e^{2\pi i n x}e^{-2\pi i m x}dx=a_m,\end{eqnarray}\]
and more generally whenever integration and summation can be interchanged, the coefficients $\hat{f}(n)$ offer the only possible way to represent $f$ (pointwise) as a trigonometric sum.
Some natural questions about Fourier series arise:
(1) When is a Fourier series convergent, and in what sense:
(1a) pointwise,
(1b) absolutely,
(1c) uniformly,
(1d) on average (in Cesàro sense),
(1e) in $L^p$ norms,
(1f) almost everywhere?
(2) How does smoothness or integrability of $f$ affect the size of the Fourier coefficients $\hat{f}(n)$?
(3) Do the Fourier coefficients $\hat{f}(n)$ characterize $f$ almost everywhere?
(4) Given a convergent trigonometric series, is it the Fourier series of some function?
(5) Does the Fourier series always converge to $f$ if it converges in the first place?
These have been important questions in Fourier analysis for centuries. Nowadays, quite general answers are known to these questions; especially between 1850's and 1960's, a huge amount of progress was made. We will discuss these questions in the following posts.
A simple convergence criterion
We can already show that $\hat{f}$ determines the function $f$, if we are assuming continuity (In a later post, we will see that (3) is generally true). From that we can deduce a simple but a very useful convergence criterion, whose assumptions give uniform and absolute convergence of the Fourier series to the right value.
Theorem 1. Let $f,g\in L^1([0,1])$ be $1$-periodic. If $\hat{f}(n)=\hat{g}(n)$ for all $n\in \mathbb{Z}$, then $f(x_0)=g(x_0)$ at every continuity point $x_0$ of $f(x)-g(x)$.
Proof. By linearity, we may assume $g=0$. By taking the real and imaginary parts, we may also assume that $f$ is real. We have
\[\begin{eqnarray}\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x)e^{-2\pi i n x}dx=0\end{eqnarray}\]
for all $n\in \mathbb{Z}$ (by periodicity, we may integrate over $[-\frac{1}{2},\frac{1}{2}]$). By linearity again, this means that
\[\begin{eqnarray}\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x)P(x)dx=0\end{eqnarray}\]
for any trigonometric polynomial $P(x)=\sum_{n=-N}^N a_n e^{2\pi i n x}$.
Now let $f$ be continuous at $x_0$. By considering $f_1(x)=f(x-x_0)$, we may assume $x_0=0$. Assume for the sake of contradiction that $f(0)\neq 0$. By symmetry, we may assume $f(0)>0$. Then there exists $a\neq 0,r>0$ such that $f(x)>a$ for $|x|\leq r$. For every $\varepsilon>0$, choose a trigonometric polynomial $P_{\varepsilon}$ such that $\|P_{\varepsilon}-L_r\|_{\infty}<\varepsilon$, where $L_r(x)=1$ for $x\in [-\frac{r}{2},\frac{r}{2}]$, $L_r(x)=0$ for $|x|\geq r$, and $L_r$ increases linearly from $0$ to $1$ on $[-r,-\frac{r}{2}]$ and similarly decreases linearly on $[\frac{r}{2},r]$. This can be done since trigonometric polynomials are dense in the set of continuous functions on $[-\frac{1}{2},\frac{1}{2}]$ (this theorem of Weierstrass is well-known, but will be proved in a later post when dealing with Fejér kernels).
Then
Then
\[\begin{eqnarray}\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x)P_{\varepsilon}(x)dx&\geq &\int_{-\frac{r}{2}}^{\frac{r}{2}}f(x)(1-\varepsilon)dx-\int_{\frac{r}{2}<|x|<\frac{1}{2}}f(x)\cdot\varepsilon dx.\end{eqnarray}\]
As $\varepsilon\to 0$, this approaches $\int_{-r/2}^{r/2}f(x)dx\geq ar>0$, which is a contradiction. ■
Now we present the mentioned simple criterion for uniform and absolute convergence of Fourier series. This theorem and the one following it show that Fourier series can represent a much wider class of functions than for example Taylor series.
Theorem 2. Let $f\in L^1([0,1])$ be a $1$-periodic function such that $\sum_{n=-\infty}^{\infty}|\hat{f}(n)|$ converges. Then the Fourier series of $f$ converges absolutely, and
\[\begin{eqnarray}f(x)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi i n x}\end{eqnarray}\]
holds at the continuity points of $f$. If $f$ is also everywhere continuous, then the Fourier series converges absolutely and uniformly to $f(x)$.
Proof. Let
\[\begin{eqnarray}g(x)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi i n x}.\end{eqnarray}\]
Our assumption gives immediately that this series converges absolutely. The series also defines a measurable function (as it is the limit of measurable functions), which is also bounded. Hence
\[\begin{eqnarray}\int_{0}^{1}(f(x)-g(x))e^{-2\pi i k x}dx&=&\hat{f}(k)-\int_{0}^{1}g(x)e^{-2\pi i k x}dx\\&=&\hat{f}(k)-\sum_{n=-\infty}^{\infty}\int_{0}^{1}\hat{f}(n)e^{2\pi i n x}e^{-2\pi i k x}dx\\&=&\hat{f}(k)-\hat{f}(k)=0,\end{eqnarray}\]
where Fubini's theorem was used to interchange integration and summation. Now the function $f(x)-g(x)$ has all of its Fourier coefficients equal to $0$. Notice that $g(x)$ is continuous, since by Weierstrass' M-test, the series defining it is uniformly convergent. Therefore, the previous theorem gives $f(x_0)=g(x_0)$ at the continuity points $x_0$ of $f$.
Finally, if $f$ is everywhere continuous, then $f(x)=g(x)$ everywhere, and so
\[\begin{eqnarray}\left|f(x)-\sum_{n=-N}^N\hat{f}(n)e^{2\pi i n x}\right|\leq \sum_{|n|>N}|\hat{f}(n)|,\end{eqnarray}\]
and the upper bound approaches $0$ and does not depend on $x$.■
We complement the previous theorem by showing that its conditions hold for $C^2$-functions. We formulate a more general result.
Theorem 3. Let $f\in C^k$ (as always, $f$ is $1$-periodic and $f(0)=f(1)$). Then $|\hat{f}(n)|\leq \frac{M_k}{n^{k}}$ for some constant $M_k$ (depending on $f$). In particular, the Fourier series of a $C^2$-function converges absolutely and uniformly to $f(x).$
Proof. Let $f\in C^k$. In the introductory post, we mentioned that smoothness of $f$ corresponds to decay of $\hat{f}$, and this is exactly what happens here. By integration by parts, for $n\neq 0$,
\[\begin{eqnarray}\hat{f}(n)=\frac{1}{2\pi i n}\int_{0}^1 f'(x)e^{-2\pi i n x}dx\end{eqnarray}\]
(the substitution term vanishes because $f(0)=f(1)$). Applying integration by parts $k$ times, we arrive at
\[\begin{eqnarray}\hat{f}(n)=\frac{1}{(2\pi i n)^k}\int_{0}^1 f'^{(k)}(x)e^{-2\pi i n x}dx.\end{eqnarray}\]
In particular $|\hat{f}(n)|\leq \frac{M_k}{n^K}$, where $M_k=\sup |f^{(k)}(x)|$.
In the case $f\in C^2$, this yields the convergence of $\sum_{n=-\infty}^{\infty}|\hat{f}(n)|$, and by the previous theorem, the Fourier series of $f$ converges absolutely and uniformly to the right vaue. ■
An interesting remark is that the convergence of Fourier series of $C^1$-functions is somewhat harder to prove than the case of $C^2$ (but not terribly difficult; it will be shown in a later post). Questions about convergence for $C$-functions are extremely difficult, and such results (in particular, Carleson's theorem) were some of the greatest achievements in Fourier analysis in the previous century.
Sum identities
There are a couple of very useful identities which are employed often, and some of them can also be used to derive sum identities.
Theorem 4. Let $f$ be $1$-periodic, $f\in L^1([0,1]).$ We have
(i) $\hat{f'}(n)=2\pi i n \hat{f}(n)$ if $f$ is also continuously differentiable
(ii) $\widehat{f* g}(n)=\hat{f}(n)\hat{g}(n)$, where $f*g(x):=\int_{0}^{1}f(x-y)g(y)dy$ is the convolution
(iii) $\sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2=\|f\|_2^2$ (Parseval's identity).
Proof. The first identity follows just from partial integration, as was done above.
Now we deal with the second identity. Since $f*g$ is integrable (and actually satisfies $\|f*g\|_1\leq \|f\|_1\|g\|_1$ by Fubini's theorem), we may substitute the definition of convolution to the formula for Fourier coefficients and change the order of integration, arriving at the result.
In the third identity, we may suppose $f\in L^2$, as otherwise both sides are infinite. At this point, we will only provee it for $C^2$-functions, as that simplifies the proof and that case is sufficient in what follows (for a complete proof, see this post). Since the Fourier series of $f\in C^2$ converges absolutely to $f$, we may compute
\[\begin{eqnarray}|f(x)|^2=\left|\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi in x}\right|^2=\sum_{m,n=-\infty}^{\infty}\hat{f}(m)\overline{\hat{f}(n)}e^{2\pi i (m-n)}.\end{eqnarray}\]
As this series is still absolutely and uniformly convergent, we may integrate both sides, change the places of the integral and the sum, and use orthogonality of the functions $e^{2\pi in x}$ to arrive at the identity.■
Parseval's identity above is quite convenient in evaluating sums, but even more useful is the Poisson summation formula
Theorem 5. Let $f:\mathbb{R}\to \mathbb{R}, f\in C^2(\mathbb{R})$ (no periodicity here) be such that $|f(x)|+|f''(x)|\leq \frac{C}{1+x^2}$ for some $C>0$, and let $a\in \mathbb{R}$. Then it holds that
\[\begin{eqnarray}\sum_{n=-\infty}^{\infty}f(n+a)=\sum_{n=-\infty}^{\infty}e^{2\pi i n a}\mathcal{F}(f)(n),\end{eqnarray}\]
where $\mathcal{F}(f)(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x \xi}dx$ is the Fourier transform (usually this would be denoted by a hat, but here it would be confusing).
Proof. Consider the function $F(x)=\sum_{n=-\infty}^{\infty}f(n+x),$ which is obviously $1$-periodic and $C^2$ because $f\in C^2$ and the differentiated series converges uniformly. The Fourier coefficients of $F$ are
\[\begin{eqnarray}\hat{F}(k)&=&\int_{0}^1 \sum_{n=-\infty}^{\infty}f(n+x)e^{-2\pi i k x}dx\\&=&\sum_{n=-\infty}^{\infty}\int_{0}^1 f(n+x)e^{-2\pi i k x}dx\\&=&\int_{\mathbb{R}}f(x)e^{-2\pi i kx}dx\\&=&\mathcal{F}(f)(k),\end{eqnarray}\]
where we used absolute convergence. We know that the Fourier series of a $C^2$-function converges to the function pointwise, so $F(a)=\sum_{k=-\infty}^{\infty}\hat{F}(k)e^{2\pi i n a}$. By writing this out, we see that this is exactly the Poisson summation formula. ■
The most important case is of course $a=0$, in which case the formula says that it does not matter whether we sum $f$ or its Fourier transform over the integers. This means that we may choose the one that is easier to sum, and in this way derive unexpected identities. Below is an example of an identity that is easy with Poisson summation but otherwise difficult.
Theorem 6. Define the Jacobi $\theta$-function by
\[\begin{eqnarray}\theta(x)=\sum_{n=-\infty}^{\infty}e^{-n^2\pi x}\end{eqnarray}\]
for $x$ in the right half-plane. Then $\theta(x)=\frac{1}{\sqrt{ x}}\theta(x^{-1})$.
Proof. This follows directly by applying Poisson summation to the function $g(z)=e^{-\pi x z^2}$ since its Fourier transform is $\frac{1}{\sqrt{x}}e^{-\frac{\pi \xi^2}{x}}$ (this is a well-known Fourier transform pair, and we will prove it later in a post on the Fourier transform).■
We remark that the Jacobi theta function is not just a random function; it is essentially a Modular form. This term is used of a function $f$ that is analytic in the upper half-plane, $1$-periodic, analytic at $\infty$ (meaning that $f(z^{-1})$ is analytic at the origin) and satisfies $f(z^{-1})=z^kf(z)$ for some integer $k$. Hence $f(z)=\theta^{12}$ is a modular form, and this has many implications to number theory, including the functional equation of the Riemann zeta function.
We remark that the Jacobi theta function is not just a random function; it is essentially a Modular form. This term is used of a function $f$ that is analytic in the upper half-plane, $1$-periodic, analytic at $\infty$ (meaning that $f(z^{-1})$ is analytic at the origin) and satisfies $f(z^{-1})=z^kf(z)$ for some integer $k$. Hence $f(z)=\theta^{12}$ is a modular form, and this has many implications to number theory, including the functional equation of the Riemann zeta function.
Two applications
We are going to derive a famous and quite useful product representation of the sine function, namely,
Theorem 7. We have
\[\begin{eqnarray}\sin x =x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)\end{eqnarray}\]
for all $x\in \mathbb{R}$.
This is a special case of Hadamard's product formula Hadamard's product formula for entire functions of finite order.
We will also derive a formula for $\zeta(2k)=\sum_{n=1}^{\infty}n^{-2k}$, that is, the Riemann zeta function at even integers. It is a bit surprising that there is a formula for these numbers, involving only powers of $\pi$ and rational numbers, given that no closed formula is known for the values of $\zeta$ at odd integers greater than $1$.
Theorem 8. We have $\zeta(2k)=(-1)^{k+1}\frac{B_{2k}(2\pi)^{2k}}{2\cdot (2k)!}$, where the Bernoulli numbers are given by the generating function
\[\begin{eqnarray}\frac{x}{e^x-1}=\sum_{n=0}^{\infty}B_n\frac{x^n}{n!}.\end{eqnarray}\]
It is quite remarkable that the proofs of both theorems are based on the same formula, a formula for $\cot x$, which is obtained by forming the Fourier series of $\cos(\alpha x)$. Let us start by determining the Fourier coefficients of the function $f$, which is equal to $f(x)=\cos(\alpha x)$ for $-\frac{1}{2}<x\leq \frac{1}{2}$ and extended to a $1$-periodic function ($\alpha$ is not an integer). In the definition of Fourier coefficients, we may as well integrate over $[-\frac{1}{2},\frac{1}{2}]$, so
\[\begin{eqnarray}\hat{f}(n)&=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\cos(\alpha x)e^{-2\pi in x}dx\\&=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\cos(\alpha x)\cos(2\pi n x)dx\\&=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\cos((\alpha+2\pi n)x)+\cos((\alpha-2\pi n)x)}{2}dx\\&=&\int_{0}^{\frac{1}{2}}(\cos((\alpha+2\pi n)x)+\cos((\alpha-2\pi n)x)dx\\&=&(-1)^n(\frac{\sin \frac{\alpha}{2}}{\alpha+2\pi n}+\frac{\sin \frac{\alpha}{2}}{\alpha-2\pi n})\\&=&(-1)^n\frac{2\alpha \sin \frac{\alpha}{2}}{\alpha^2-4n^2\pi^2}.\end{eqnarray}\]
The Fourier series of $f$ converges absolutely, and $f$ is everywhere continuous since $f(\frac{1}{2})=f(-\frac{1}{2})$. Therefore, the simple convergence criterion gives
\[\begin{eqnarray}f(x)&=&\cos(\alpha x)=\sum_{n=-\infty}^{\infty}(-1)^n\frac{2\alpha \sin \frac{\alpha}{2}}{\alpha^2-4n^2\pi^2}e^{-2\pi i n x}\\&=&\sum_{n=-\infty}^{\infty}(-1)^n\frac{2\alpha \sin \frac{\alpha}{2}}{\alpha^2-4n^2\pi^2}\cos(2\pi n x),\quad x\in \mathbb{R}\end{eqnarray}\]
Taking $x=\frac{1}{2}$ and replacing $\alpha=2\beta$, we get
\[\begin{eqnarray}\cot \beta=\frac{1}{\beta}+2\sum_{n=1}^{\infty}\frac{\beta}{\beta^2-n^2\pi^2}\end{eqnarray}\]
when $\beta\neq m\pi$ for $m\in \mathbb{Z}.$ This formula is the key to proving the theorems above.
Proof of the product formula for sine. Let $f(x)=x\prod_{n=1}^{\infty}(1-\frac{x^2}{n^2\pi^2})$. This product converges absolutely and uniformly on compact sets since the corresponding series $\sum_{n=1}^{\infty}\frac{x^2}{n^2\pi^2}$ does, and the same holds for its derivatives. Exploiting this, we may compute the logarithmic derivative
\[\begin{eqnarray}\frac{d}{dx}\log f(x)&=&\frac{1}{x}+\sum_{n=1}^{\infty}\frac{\frac{-2x}{n^2\pi^2}}{1-\frac{x^2}{n^2\pi^2}}\\&=&\frac{1}{x}+2\sum_{n=1}^{\infty}\frac{x}{x^2-n^2\pi^2}\\&=&\cot x,\end{eqnarray}\]
when $x\neq m\pi$. Since $\frac{d}{dx}(\log \sin x-\log f(x))=0$ for $n\pi<x<(n+1)\pi$, where $n$ is any integer, we get $f(x)=C_n\sin x$ for $x\in (n\pi,(n+1)\pi)$ for some constants $C_n$. We may determine $C_0$ by observing that
\[\begin{eqnarray}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^2}\right)=\lim_{N\to \infty}\prod_{n=2}^{N}\frac{n+1}{n}\cdot \frac{n-1}{n}=\lim_{N\to \infty}\frac{\frac{1}{2}(N+1)!(N-1)!}{N!^2}=\frac{1}{2},\end{eqnarray}\]
since the products are telescopic. This additional information gives
\[\begin{eqnarray}\lim_{x\to \pi-}\frac{f(x)}{x(1-\frac{x^2}{\pi^2})}=\frac{1}{2},\end{eqnarray}\]
while the assumption $f(x)=C_0\sin x$ for $x\in (0,\pi)$ together with L'Hospital's rule tells that the limit is $\frac{1}{2}C_0$. Thus $C_0=1$. Lastly, $f$ is continuously differentiable (even analytic) since the corresponding series is, so the function $f(x)-C_0\sin x$ is continuously differentiable. Therefore, $C_1=C_0=C_{-1}$, as otherwise the derivative would be discontinuous. Continuing similarly, we see that $C_n=1$ for all $n$, so $f(x)=x$ for all $x$.■
Proof of the formula for $\zeta(2k)$. We return to the fundamental cotangent formula. It can be written as
\[\begin{eqnarray}\cot x=\frac{1}{x}+2\sum_{n=1}^{\infty}\frac{xn^{-2}\pi^{-2}}{(\frac{x}{n\pi})^2-1},\quad x\neq m\pi\end{eqnarray}\]
Using the formula $\frac{1}{1-y^2}=1+y^2+y^4+...$ for $|y|<1$, this becomes
\[\begin{eqnarray}\cot x&=&\frac{1}{x}-2\sum_{n=1}^{\infty}n^{-2}\pi^{-2}\sum_{m=0}^{\infty}\frac{x^{2m+1}}{(n\pi)^{2m}}\\&=&\frac{1}{x}-2\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{x^{2m-1}}{(\pi n)^{2m}}\\&=&\frac{1}{x}-2\sum_{m=1}^{\infty}\frac{\zeta(2m)}{\pi^{2m}}x^{2m-1}\end{eqnarray}\]
by absolute convergence. Therefore, the function $x\cot x$ has the Taylor expansion
\[\begin{eqnarray}x\cot x=1-2\sum_{m=1}^{\infty}\frac{\zeta(2m)}{\pi^{2m}}x^{2m}\end{eqnarray}\]
for $|x|<\pi$. Using the Bernoulli numbers, we shall derive another Taylor series for this function and then compare coefficients. We easily compute that $B_0=1, B_1=-\frac{1}{2}, B_2=\frac{1}{6}$ and $B_3=0$. Actually, $B_{2r+1}=0$ for all positive $r$ since the function $\frac{x}{e^x-1}-1-\frac{x}{2}$ is even. Now we compute
\[\begin{eqnarray}x\cot x&=&\frac{2ix}{e^{2ix}-1}\cdot\frac{e^{2ix}+1}{2}=\left(\sum_{n=0}^{\infty}\frac{B_n}{n!}\cdot (2i)^nx^n\right) \left(1+\sum_{m=0}^{\infty}\frac{(2i)^n}{2n!}x^n\right)\\&=&\sum_{n=0}^{\infty}\frac{\gamma_n}{n!}x^n,\end{eqnarray}\]
where the coefficients are given by the discrete convolution as
\[\begin{eqnarray}\gamma_n&=&\frac{1}{2}\sum_{r=0}^{n-1}\binom{n}{r}B_r(2i)^r (2i)^{n-r}+B_n2^ni^{n}\\&=&(2i)^n\left(\frac{1}{2}\sum_{r=0}^{n-1}\binom{n}{r}B_r+B_n\right).\end{eqnarray}\]
We are lucky here; the sum is
\[\begin{eqnarray}\sum_{r=0}^{n-1}B_r\binom{n}{r}=0\end{eqnarray}\]
if $n\geq 2$ and $1$ if $n=1$. This follows just by writing the identity $x=\frac{x}{e^{x}-1}(e^x-1)$ as a power series and multiplying out. Therefore, $\gamma_n=(2i)^n B_n$ for $n>1$, so $\gamma_{2m}=(-1)^m2^{2m}B_{2m}$ and $\gamma_{2m+1}=0$ since every other Bernoulli number is zero. We conclude
\[\begin{eqnarray}x\cot x=1-\sum_{m=0}^{\infty}\frac{(-1)^m 2^{2m}B_{2m}}{(2m)!}x^{2m},\end{eqnarray}\]
and by comparing the coefficients with the other Taylor series of $x\cot x$, we immediately get the formula for $\zeta(2k).$■
In particular, we have evaluated $\zeta(2)=\frac{\pi^2}{6}$, which is the Basel problem, as well as $\zeta(4)=\frac{\pi^4}{90}$. From the formula for $\zeta(2k)$ one sees that it is equal to some rational number times $\pi^{2k}$. To see this, it is enough to show that the Bernoulli numbers are rational. From the recursive formula $\sum_{r=0}^{n-1}B_r\binom{n}{r}=0$ this is evident. The product formula for the sine function also has a beautiful special case; taking $x=\frac{\pi}{2}$, we easily arrive at the Wallis product
\[\begin{eqnarray}\frac{2}{\pi}=\frac{1\cdot 3}{2^2}\cdot\frac{3\cdot 5}{4^2}\cdot\frac{5\cdot 7}{6^2}\cdot...\end{eqnarray}\]
