Properties of Fourier Series

Examples and Formulation

Our problem is about coefficients $a_n=\frac{1}{L}{\int }_0^{L}f(x)e^{-2\pi inx/L}dx$, for $n\in \mathbb{Z}$ where $f$ is complex-valued on $[0,L]$

For this text, we consider functions that are at least Reimann integrable, including

• Everywhere continuous functions
• Piecewise continuous functions, which have finite discontinuities
• Reimann integrable functions, which are bounded but may have infinitely many discontinuities

A function is Reimann integrable if for every $ϵ>0$ there is a subdivision $0=x_0<x_1<\cdots <x_{N-1}<x_N=L$ of the interval $[0,L]$ where $\mathcal{U}-\mathcal{L}<ϵ$ where $\mathcal{U}={\sum }_{j=1}^N[{\text{sup}}_{x_{j-1}\le x\le x_j}f(x)\cdot (x_j-x_{j-1})]$ and $\mathcal{L}={\sum }_{j=1}^N[{\inf }_{x_{j-1}\le x\le x_j}f(x)\cdot (x_j-x_{j-1})]$, essentially saying that Reimann integration converges to a value

For example,
$f(x)\{\begin{array}{ll}1& \text{if }1/(n+1)<x\le 1/n\text{ and }n\text{ is odd}\\ 0& \text{if }1/(n+1)<x\le 1/n\text{ and }n\text{ is even}\\ 0& \text{if }x=0\end{array}$
has infinitely many discontinuities but is also Reimann integrable

While there are infinite discontinuities, they still have measure 0

When our functions are defined on an interval on the line we often use $x$, while functions on the circle are usually notated with $\theta$, however we can ultimately view these functions the same, related with $f(\theta )=F(e^{i\theta })$

If $f$ is is an integrable function on $[a,b]$ ($b-a=L$), then the $n$th Fourier coefficient of $f$ is defined by $\hat{f}(n)=\frac{1}{L}{\int }_a^{b}f(x)e^{-2\pi inx/L}dx$

The Fourier series of $f$ is given by ${\sum }_{n=-\infty }^{\infty }\hat{f}(n)e^{2\pi inx/L}$

These are part of a larger family called trigonometric series, of the form ${\sum }_{n=-\infty }^{\infty }{c}_n{e}^{2\pi inx/L}$ with $c_n\in \mathbb{C}$
A trigonometric series with finitely many non-zero terms is called a trigonometric polynomial with degree equal to the largest value of $|n|$ with non-zero $c_n$

The $N$th partial sum of the Fourier series of $f$ is a trigonometric polynomial given by $S_N(f)(x)={\sum }_{n=-N}^N\hat{f}(n)e^{2\pi inx/L}$

Example:
Let $f(\theta )=\theta$ for $-\pi \le 0\le \pi$

We use integration by parts ${\int }_a^{b}u(x)v^{\prime }(x)dx=[u(x)v(x){]}_a^b-{\int }_a^b{u}^{\prime }(x)v(x)dx$ or $\int udv=uv-\int vdu$

$\hat{f}(n)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\theta e^{-in\theta }d\theta$
$=\frac{1}{2\pi }{\left[-\frac{\theta }{in}{e}^{-in\theta }\right]}_{-\pi }^{\pi }+\frac{1}{2\pi in}{\int }_{-\pi }^{\pi }{e}^{-in\theta }d\theta$
$=\{\begin{array}{ll}\frac{(-1{)}^{n+1}}{in}& \text{if }n\ne 0\\ 0& \text{if }n=0\end{array}$

$f(\theta )\sim {\sum }_{n\ne 0}^{}\frac{(-1{)}^{n+1}}{in}{e}^{in\theta }=2{\sum }_{n=1}^{\infty }(-1{)}^{n+1}\frac{\sin n\theta }{n}$

Example:
$D_N(x)={\sum }_{n=-N}^N{e}^{inx}$ for $x\in [-\pi ,\pi ]$ is called the $N$th Dirichlet kernel
Its Fourier coefficients are $a_n=1$ if $|n|\le N$ and $a_n=0$ otherwise

We can get a closed form formula by summing the geometric progressions ${\sum }_{n=0}^N{\omega }^n$ and ${\sum }_{n=-N}^{-1}{\omega }^n$ (with $\omega =e^{ix}$) and some manipulation yields
$D_N(x)=\frac{\sin \left(\left(N+\frac{1}{2}\right)x\right)}{\sin (x/2)}$

Given our partial sums, we can reformulate our overarching problem as “In what sense does $S_N(f)$ converge to $f$ as $N\to \infty$?”

To start, do we have pointwise convergence ${\lim }_{N\to \infty }{S}_N(f)(\theta )=f(\theta )$ for every $\theta$?

We can’t generally expect this because we can change an integrable function at one point without affecting its Fourier coefficients, so we narrow our question to continuous and periodic functions

However, it turns out even continuous functions can have diverging Fourier series, so this is not an easy question at all (settled in 1966)

Our result is different if we interpret our question a bit differently, by either only looking at points of continuity or defining convergence in the mean square sense

Uniqueness

Theorem: Suppose that $f$ is an integrable function on the circle with $\hat{f}(n)=0$ for all $n\in Z$. Then $f({\theta }_0)=0$ whenever $f$ is continuous at the point ${\theta }_0$

Given what we know about finite discontinuities of integrable functions, this means $f$ vanishes for “most” values of $\theta$

We prove this by contradiction, constructing a family of trigonometric polynomials $\{p_k\}$ that peak at 0, such that ${\int }_{}^{}{p}_k(\theta )f(\theta )d\theta \to \infty$. This leads us to our result, because this is equivalent to a linear combination of Fourier coefficients (which we’ve said vanish)

Assume without loss of generality that $f$ is real-valued, defined on $[-\pi ,\pi ]$, that ${\theta }_0=0$, and $f(0)>0$

Since $f$ is continuous at $0$, we can choose $0<\delta \le \frac{\pi }{2}$, such that $f(\theta )>f(0)/2$ whenever $|\theta |<\delta$, which essentially defines a region around $\delta$ close to $f(0)$

Let $p(\theta )=ϵ+\cos \theta$ where $ϵ>0$ is chosen such that $|p(\theta )|<1-ϵ/2$ whenever $\delta \le |\theta |\le \pi$

Then choose $0<\eta <\delta$ such that $p(\theta )\ge 1+ϵ/2$ for $|\theta |<\eta$

Essentially, we’ve defined $ϵ$ to lower $p(\theta )$ under $1$ outside of the region defined by $\delta$, and then identified the region greater than $1$ with $\eta$

Finally, let $p_k(\theta )=[p(\theta ){]}^k$ and select $B$ such that $|f(\theta )|\le B$ for all $\theta$

By our original reasoning, ${\int }_{-\pi }^{\pi }f(\theta )p_k(\theta )d\theta =0$ for all $k$

However our construction shows,
$|{\int }_{\delta \le |\theta |}^{}f(\theta )p_k(\theta )d\theta |\le 2\pi B(1-ϵ/2{)}^k$ by our established bounds
${\int }_{\eta \le |\theta |<\delta }^{}f(\theta )p_k(\theta )d\theta \ge 0$ since the region defined by $\delta$ is positive, and
${\int }_{|\theta |<\eta }^{}f(\theta )p_k(\theta )d\theta \ge 2\eta \frac{f(0)}{2}(1+ϵ/2{)}^k$

So under the whole domain, ${\int }_{}^{}{p}_k(\theta )f(\theta )d\theta \to \infty$

To recap, we divided the domain of our polynomial into three regions; the center has exponential growth, the outside has exponential decay, and the middle we say little about. The reason we need an intermediate region is because establishing a bound on the integral requires uniform behavior (otherwise how would we even define a bound), and allowing points arbitrarily close to $1$ would violate this.

We’ve shown that the existence of a single positive point under continuity leads to a positive region, which breaks our constraints

We can use the same logic on a complex function, by breaking it into parts $f(\theta )=u(\theta )+iv(\theta )$

Corollary: If $f$ is continuous on the circle and $\hat{f}(n)=0$ for all $n\in \mathbb{Z}$, then $f=0$

Corollary: Suppose that $f$ is a continuous function on the circle and that the Fourier series of $f$ is absolutely convergent, ${\sum }_{n=-\infty }^{\infty }|\hat{f}(n)|<\infty$. Then, the Fourier series converges uniformly to $f$, that is, ${\lim }_{n\to \infty }{S}_N(f)(\theta )=f(\theta )$ uniformly in $\theta$

Note that $g(\theta )={\sum }_{n=-\infty }^{\infty }\hat{f}(n)e^{in\theta }={\lim }_{N\to \infty }{\sum }_{n=-N}^N\hat{f}(n)e^{in\theta }$, the limit of a sequence of continuous functions converges uniformly and therefore must also be continuous

We can then show pretty easily that the Fourier coefficients of $g$ are those of $\hat{f}(n)$ (using the uniform convergence)

Now to better understand this corollary, when is a Fourier series absolutely convergent?

Definition: $f(x)=O(g(x))$ as $x\to a$ means for some constant $C$, $|f(x)|\le C|g(x)|$ as $x$ approaches $a$

Corollary: Suppose that $f$ is a twice continuously differentiable function on the circle, then $\hat{f}(n)=O(1/|n{|}^2)$ as $|n|\to \infty$, which therefore means the Fourier series of $f$ converges absolutely and uniformly to $f$

Let’s prove this for $n\ne 0$

We find this result by integrating by parts twice,
$2\pi \hat{f}(n)={\int }_0^{2\pi }f(\theta )e^{-in\theta }d\theta$
$={\left[f(\theta )\cdot \frac{-e^{-in\theta }}{in}\right]}_0^{2\pi }+\frac{1}{in}{\int }_0^{2\pi }{f}^{\prime }(\theta )e^{-in\theta }d\theta$
$=\frac{1}{in}{\int }_0^{2\pi }{f}^{\prime }(\theta )e^{-in\theta }d\theta$
$=\frac{1}{in}{\left[f^{\prime }(\theta )\cdot \frac{-e^{in\theta }}{in}\right]}_0^{2\pi }+\frac{1}{(in{)}^2}{\int }_0^{2\pi }{f}^{\prime \prime }(\theta )e^{-in\theta }d\theta$
$=-\frac{1}{n^2}{\int }_0^{2\pi }{f}^{\prime \prime }(\theta )e^{-in\theta }d\theta$

We write $|n|$ to emphasize the symmetry,
$2\pi |n{|}^2|\hat{f}(n)|\le |{\int }_0^{2\pi }{f}^{\prime \prime }(\theta )e^{-in\theta }d\theta |\le {\int }_0^{2\pi }|f^{\prime \prime }(\theta )|d\theta \le C$

So we have $\hat{f}(n)=O(1/|n{|}^2)$

Incidentally, we’ve also shown $\widehat{f^{\prime }}(n)=in\hat{f}(n)$

Further smoothness conditions on $f$ imply better decay of the Fourier coefficients

We can also show that the Fourier series of $f$ converges absolutely assuming only one continuous derivative, and even more generally if $f$ satisfies a Hölder condition of order $\alpha >1/2$

We also describe smoothness in classes $C^k$ ($f$ is $k$ times continuously differentiable)

Conveniently, we can also apply these results to a Fourier series defined in terms of coefficients $c_n$ which decay at a rate of $O(1/|n{|}^k)$ (note this is subtly different than a function)

Discontinuities

What actually happens at discontinuities? We know that single discontinuous points do not affect integrals, so how does that work for convergence?

Example:
Take $f(\theta )=\theta$ on $[-\pi ,\pi ]$
We get $\hat{f}(\theta )=\frac{(-1{)}^{n+1}}{in}$

${\sum }_{n=-\infty }^{\infty }\frac{(-1{)}^{n+1}}{in}{e}^{in\theta }$ does not satisfy our decay rate tests, because $|\hat{f}(n)|\le \frac{C}{|n|}$ does not converge, but it might still converge

Convolutions

Definition: The convolution of $f$ and $g$ on $\mathbb{R}$ in $[-\pi ,\pi ]$ is $(f\ast g)(x)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(y)g(x-y)dy$

This operation comes up a lot in Fourier analysis

$S_N(f)(x)={\sum }_{n=-N}^N\hat{f}(n)e^{inx}$
$={\sum }_{n=-N}^N\left(\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(y)e^{-iny}dy\right)e^{inx}$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(y)\left({\sum }_{n=-N}^N{e}^{in(x-y)}\right)dy$
$=(f\ast D_N)(x)$

$D_N$ is called the $N$th Dirichlet kernel, $D_N(x)={\sum }_{n=-N}^N{e}^{inx}$

Proposition: Suppose that $f,g,h$ are $2\pi$-periodic integrable functions,

1. $f\ast (g+h)=(f\ast g)+(f\ast h)$
2. $(cf)\ast g=c(f\ast g)=f\ast (cg)$ for any $c\in \mathbb{C}$
3. $f\ast g=g\ast f$
4. $(f\ast g)\ast h=f\ast (g\ast h)$
5. $f\ast g$ is continuous
6. $\widehat{f\ast g}(n)=\hat{f}(n)\hat{g}(n)$

This operation behaves very nicely and the last two points are very interesting, especially since they work with merely integrable functions

Lemma: Suppose $f$ is integrable on the circle and bounded by $B$, then there exists a sequence $\{f_k{\}}_{k=1}^{\infty }$ of continuous functions on the circle such that ${\sup }_{x\in [-\pi ,\pi ]}|f_k(x)|\le B$ for all $k$ and ${\int }_{-\pi }^{\pi }|f(x)-f_k(x)|dx\to 0$ as $k\to \infty$

We use this lemma to prove our final two properties

For property 5,
$f\ast g-f_k\ast g_k=(f-f_k)\ast g+f_k\ast (g-g_k)$

$|(f-f_k)\ast g(x)|\le \frac{1}{2\pi }{\int }_{-\pi }^{\pi }|f(x-y)-f_k(x-y)||g(y)|dx$
$\le \frac{1}{2\pi }{\sup }_y|g(y)|{\int }_{-\pi }^{\pi }|f(y)-f_k(y)|dy$ $\to 0$ as $k\to \infty$

I.e. $(f-f_k)\ast g$ converges uniformly and the same logic applies to the other term

Therefore, we have a sequence of continuous functions converging uniformly to $f\ast g$, so our property holds

For property 6,
If $f$ and $g$ are continuous then we can exchange the order of integration to obtain our result

We know from the previous property that $\widehat{f_k\ast g_k}(n)\to \widehat{f\ast g}(n)$ as $k\to \infty$, so
$|\hat{f}(n)-{\hat{f}}_k(n)|=\frac{1}{2\pi }|{\int }_{-\pi }^{\pi }(f(x)-f_k(x))e^{{}^{-inx}}dx|$
$\le \frac{1}{2\pi }{\int }_{-\pi }^{\pi }|f(x)-f_k(x)|dx$

Hence, $\widehat{f_k}(n)\to \hat{f}(n)$ as $k\to \infty$

Kernels

Definition: A family of kernels $\{K_n(x){\}}_{n=1}^{\infty }$ on the circle is said to be a family of good kernels if,

1. For all $n\ge 1$, $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(x)dx=1$
2. There exists $M>0$ such that for all $n\ge 1$, ${\int }_{-\pi }^{\pi }|K_n(x)|dx\le M$
3. For every $\delta >0$, ${\int }_{\delta \le |x|\le \pi }^{}|K_n(x)|dx\to 0$ as $n\to \infty$

Note that if $K_n(x)\ge 0$ then the second condition follows from the first

We interpret a kernel as a weight distribution around the circle which concentrated near the origin as $n$ grows

Theorem: Let $\{K_n\}$ be a family of good kernels and $f$ an integrable function on the circle, then ${\lim }_{n\to \infty }(f\ast K_n)(x)=f(x)$ whenever $f$ is continuous at $x$ and if $f$ is continuous everywhere then the limit is uniform

Because of this, $\{K_n\}$ is sometimes referred to as an approximation to the identity

$(f\ast K_n)(x)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x-y)K_n(y)dy$ can be thought of as a weighted average, and since the weight distribution of $K_n$ concentrates at $y=0$, it effectively approaches an identity

Proof:
If $ϵ>0$ and $f$ is continuous at $x$ then we can choose $\delta$ such that if $|y|<\delta$ then $|f(x-y)-f(x)|<ϵ$

Then,
$(f\ast K_n)(x)-f(x)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(y)f(x-y)dy-f(x)$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(y)[f(x-y)-f(x)]dy$

$|(f\ast K_n)(x)-f(x)|=|\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(y)[f(x-y)-f(y)]dy|$
$\le \frac{1}{2\pi }{\int }_{|y|<\delta }^{}|K_n(y)||f(x-y)-f(x)|dy+\frac{1}{2\pi }{\int }_{\delta \le |y|\le \pi }^{}|K_n(y)||f(x-y)-f(x)|dy$
$\le \frac{ϵ}{2\pi }{\int }_{-\pi }^{\pi }|K_n(y)|dy+\frac{2B}{2\pi }{\int }_{\delta \le |y|\le \pi }^{}|K_n(y)|dy$ where $B$ bounds $f$

The first term is bounded by $\frac{ϵM}{2\pi }$ by our second good kernel property and the second term approaches $0$

Therefore, for all large $n$ we have $|(f\ast K_n)(x)-f(x)|\le Cϵ$ for some constant $C$

If $f$ is continuous everywhere, then it is uniformly continuous, so $\delta$ can be chosen independent of $x$, and convergence is therefore uniform

Therefore, our theorem is proven

The natural question to ask is whether $D_N$ is a good kernel, since then $S_N(f)(x)=(f\ast D_N)(x)$ would approximate $f$ under the theorem’s conditions

Unfortunately, life is not so nice, and we can obtain the approximation ${\int }_{-\pi }^{\pi }|D_N(x)|dx\ge c\log N$, despite the fact that $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{D}_N(x)dx=1$ is satisfied

This all suggests that pointwise convergence of Fourier series is complicated and could even fail at points of continuity

Cesàro and Abel Summability

Since a Fourier series may fail to converge at specific points, we investigate other ways of interpreting ${\lim }_{N\to \infty }{S}_N(f)=f$

Definition: ${\sigma }_N=\frac{s_0+s_1+\cdots +s_{N-1}}{N}$ is called the $N$th Cesaro mean of the sequence $\{s_k\}$

Definition: If ${\sigma }_N$ converges to a limit $\sigma$ as $N$ tends to infinity, $\sum c_n$ is called Cesaro summable to $\sigma$

This definition is interesting because it gives us a “limit” to a series like ${\sum }_{i=0}^{\infty }(-1{)}^i$ (which Cesaro sums to $1/2$, since the partial sums alternate between $0$ and $1$)

This is more inclusive than convergence, i.e. if a series converges to $s$ then it is also Cesaro summable to $s$

We take the $N$th Cesaro mean of the Fourier series,
${\sigma }_N(f)(x)=\frac{S_0(f)(x)+\cdots +S_{N-1}(f)(x)}{N}$
$=\frac{(f\ast D_0)(x)+\cdots +(f\ast D_{N-1})(x)}{N}$
$=(f\ast F_N)(x)$

We call $F_N(x)=\frac{D_0(x)+\cdots +D_{N-1}(x)}{N}$ the $N$th Fejér kernel

We showed here that $F_N(x)=\frac{1}{N}\frac{{\sin }^2(Nx/2)}{{\sin }^2(x/2)}$ and moreover, this is a good kernel

Theorem: If $f$ is integrable on the circle, then the Fourier series of $f$ is Cesaro summable to $f$ at every point of continuity of $f$ and moreover if $f$ is continuous on the circle then the Fourier series of $f$ is uniformly Cesaro summable to $f$

Corollary: If $f$ is integrable on the circle and $\hat{f}(n)=0$ for all $n$ then $f=0$ at all points of continuity of $f$

This follows immediately from our Theorem

Corollary: Continuous functions on the circle can be uniformly approximated by trigonometric polynomials, i.e. there exists a trigonometric polynomial $P$ such that $|f(x)-P(x)|<ϵ$ for all $-\pi \le x\le \pi$

Cesaro summability is not the only alternate way to look at convergence

Definition: A series of complex ${\sum }_{k=0}^{\infty }{c}_k$ is said to be Abel summable to $s$ if for every $0\le r<1$, the series $A(r)={\sum }_{k=0}^{\infty }{c}_k{r}^k$ and ${\lim }_{r\to 1}A(r)=s$

This is even more general than Cesaro summability and can cover sequences like ${\sum }_{k=0}^{\infty }(-1{)}^k(k+1)$