Convergence of Fourier Series

This continues on the ideas of Properties of Fourier Analysis

We are interested in both global convergence on the whole domain and local convergence near a single point

Mean-Square Convergence

Our goal here is to show when a Fourier series has mean-square convergence

Recall the basics of vector spaces (Linear Algebra Lecture 1, Lecture 2), along with three important formulas,

1. $X\perp Y⟺\Vert X+Y{\Vert }^2=\Vert X{\Vert }^2+\Vert Y{\Vert }^2$
2. $|(X,Y)|\le \Vert X\Vert \Vert Y\Vert$
3. $\Vert X+Y\Vert \le \Vert X\Vert +\Vert Y\Vert$

Notably, we can work with infinite-dimensional vector spaces

Example:
$l^2(\mathbb{Z})$ over $\mathbb{C}$ is the set of all infinite sequences of complex numbers such that ${\sum }_{n\in \mathbb{Z}}^{}|a_n{|}^2<\infty$ where addition and scalar multiplication are defined component-wise

We define $(A,B)={\sum }_{n\in \mathbb{Z}}^{}{a}_n\overline{b_n}$
$\Vert A\Vert =(A,A{)}^{1/2}={\left({\sum }_{n\in \mathbb{Z}}^{}|a_n{|}^2\right)}^{1/2}$

To see this is a vector space, we define $A_N=(\dots ,0,0,a_{-N},\dots ,a_{-1},a_0,a_1,\dots ,a_N,0,0,\dots )$

$\Vert A_N+B_N\Vert \le \Vert A_N\Vert +\Vert B_N\Vert \le \Vert A\Vert +\Vert B\Vert$
${\sum }_{|n|\le N}^{}|a_n+b_n{|}^2\le (\Vert A\Vert +\Vert B\Vert {)}^2$
${\sum }_{n\in \mathbb{Z}}^{}|a_n+b_n{|}^2<\infty$

We can derive the triangle inequality from this line of reasoning, and use similar reasoning to show that the Cauchy-Schwarz inequality holds

This space has a strictly positive-definite inner product ($\Vert X\Vert =0⟹X=0$) and is complete, which makes this a Hilbert space

Example:
Let $\mathcal{R}$ denote the set of complex-valued Reimann integrable functions on $[0,2\pi ]$

This is a vector space over $\mathbb{C}$ with $(f+g)(\theta )=f(\theta )+g(\theta )$ and $(\lambda f)(\theta )=\lambda \cdot f(\theta )$

We define $(f,g)=\frac{1}{2\pi }{\int }_0^{2\pi }f(\theta )\overline{g(\theta )}d\theta$
$\Vert f\Vert =(\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta ){|}^{2}d\theta {)}^{1/2}$

Both conditions for a Hilbert space fail on $\mathcal{R}$

For one, $\Vert f\Vert =0$ only says $f$ vanishes at points of continuity, but says nothing about measure zero discontinuities (although we can sort of get around this by still calling such functions the zero function)

More problematically, $\mathcal{R}$ is not complete

As an example,
$f(\theta )=\{\begin{array}{ll}0& \theta =0\\ \log (1/\theta )& 0<\theta \le 2\pi \end{array}$
$f_n(\theta )=\{\begin{array}{ll}0& 0\le \theta \le 1/n\\ f(\theta )& 1/n<\theta \le 2\pi \end{array}$

Consider that $f_n$ is a Cauchy sequence in $\mathcal{R}$, but if it converges it must converge to $f$, which is not in $\mathcal{R}$

In the sense that completing the rational numbers gives $\mathbb{R}$, one could say that completing the Reimann-integrable functions gives the Lebesgue-integrable class of functions

Back to the goal of this section, with the inner product operator defined, our goal in this section is to show $\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta )-S_n(f)(\theta ){|}^{2}d\theta \to 0$, or equivalently $\Vert f-S_N(f)\Vert \to 0$

We define the set $\{e_n{\}}_{n\in \mathbb{Z}}$ where $e_n(\theta )=e^{in\theta }$ and note that this set is orthonormal, that is $(e_n,e_m)=\{\begin{array}{ll}1& \text{if }n=m\\ 0& \text{if }n\ne m\end{array}$

Let $a_n$ denote the Fourier coefficients of an integrable function on the circle $f$, we conveniently find that $a_n=(f,e_n)$

This lets us write $S_N(f)={\sum }_{|n|\le N}^{}{a}_n{e}_n$

$(f-S_N(f),e_n)=(f,e_n)-a_n=0$ for any $|n|\le N$
Therefore, $(f-S_N(f))\perp {\sum }_{|n|\le N}^{}{b}_n{e}_n$ for any $b_n$ and we can choose $b_n=a_n$ to show $f-S_N(f)$ is orthogonal to $S_N(f)$
$f=f-S_N(f)+S_N(f)$
$\Vert f{\Vert }^2=\Vert f-S_N(f){\Vert }^2+\Vert S_N(f){\Vert }^2$
$=\Vert f-S_N(f){\Vert }^2+{\sum }_{|n|\le N}^{}|a_n{|}^2$

Lemma: If $f$ is integrable on the circle with Fourier coefficients $a_n$, then $\Vert f-S_N(f)\Vert \le \Vert f-{\sum }_{|n|\le N}^{}{c}_n{e}_n\Vert$ for any complex number $c_n$, and equality holds when $c_n=a_n$

Geometrically, this tells us $S_N(f)$ is the trigonometric polynomial of degree at most $N$ which is closest to $f$, the projection of $f$ onto the plane spanned by $\{e_{-N},\dots ,e_N\}$

Now assume $f$ is continuous on the circle
For $ϵ>0$ there exists a trigonometric polynomial $P$ (of degree $M$) such that $|f(\theta )-P(\theta )|<ϵ$ for all $\theta$
$\Vert f-P\Vert <ϵ$

By our lemma, $\Vert f-S_N(f)\Vert <ϵ$ whenever $N\ge M$

Furthermore, assume $f$ is merely integrable

By the approximation lemma, we can choose a continuous function $g$ which satisfies ${\sup }_{\theta \in [0,2\pi ]}|g(\theta )|\le {\sup }_{\theta \in [0,2\pi ]}|f(\theta )|=B$ and ${\int }_0^{2\pi }|f(\theta )-g(\theta )|d\theta <{ϵ}^2$

$\Vert f-g{\Vert }^2=\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta )-g(\theta ){|}^{2}d\theta \le \frac{2B}{2\pi }{\int }_0^{2\pi }|f(\theta )-g(\theta )|d\theta$
$\le C{ϵ}^2$

Now we can approximate $g$ with a trigonometric polynomial $P$, so that $\Vert g-P\Vert <ϵ$ and therefore $\Vert f-P\Vert <C^{\prime }ϵ$

Again, by our lemma we have proven the partial sums converge in the mean square norm

Theorem: Suppose $f$ is integrable on the circle, then $\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta )-S_N(f)(\theta ){|}^{2}d\theta \to 0$ as $N\to \infty$

As a corollary, we obtain Parseval’s identity, ${\sum }_{n=-\infty }^{\infty }|a_n{|}^2=\Vert f{\Vert }^2$ when $f$ is an integrable function

In general, if $\{e_n\}$ is any orthonormal family of functions on the circle and $a_n=(f,e_n)$ then ${\sum }_{n=-\infty }^{\infty }|a_n{|}^2\le \Vert f{\Vert }^2$

This also establishes an isomorphism between $L^2(\mathbb{R}{/}_{2\pi }\mathbb{Z})$ and $l^2(\mathbb{Z})$, specifically infinite sequences with finite norm

The failure of $\mathcal{R}$ to be complete can be understand as meaning there are $\{a_n\}$ with ${\sum }_n^{}|a_n{|}^2<\infty$ that do not correspond to a Reimann integrable function

To summarize,
Theorem: Let $f$ be an integrable function on the circle with $f\sim {\sum }_{n=-\infty }^{\infty }{a}_n{e}^{-in\theta }$,

1. $\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta )-S_N(f)(\theta ){|}^{2}d\theta \to 0$ as $N\to \infty$
2. ${\sum }_{n=-\infty }^{\infty }|a_n{|}^2=\frac{1}{2\pi }{\int }_0^{2\pi }|f(\theta ){|}^{2}d\theta$

Theorem (Reimann-Lebesgue Lemma): If $f$ is integrable on the circle, then $\hat{f}(n)\to 0$ as $|n|\to \infty$

Equivalently, ${\int }_0^{2\pi }f(\theta )\sin (N\theta )d\theta \to 0$ and ${\int }_0^{2\pi }f(\theta )\cos (N\theta )d\theta$ as $N\to \infty$

Lemma: Suppose $F$ and $G$ are integrable on the circle, then $(f,g)={\sum }_{n=-\infty }^{\infty }\hat{f}(n)\overline{\hat{g}(n)}$

Pointwise Convergence

Theorem: Let $f$ be an integrable function on the circle which is differentiable at a point ${\theta }_0$, then $S_N(f)({\theta }_0)\to f({\theta }_0)$ as $N$ tends to infinity

Define $F(t)=\{\begin{array}{ll}\frac{f({\theta }_0-t)-f({\theta }_0)}{t}& \text{if }t\ne 0\text{ and }|t|<\pi \\ -f^{\prime }({\theta }_0)& \text{if }t=0\end{array}$

By definition, $F$ is bounded near $0$ and integrable on $[-\pi ,-\delta ]\cup [\delta ,\pi ]$ because $|t|>\delta$, which implies it is integrable on all of $[-\pi ,\pi ]$

$S_N(f)({\theta }_0)=(f\ast D_N)({\theta }_0)$
$S_N(f)({\theta }_0)-f({\theta }_0)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f({\theta }_0-t)D_N(t)dt-f({\theta }_0)$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }F(t)t{D}_N(t)dt$

$t{D}_N(t)=\frac{t}{\sin (t/2)}\sin ((N+1/2)t)$
$\sin ((N+1/2)t)=\sin (Nt)\cos (t/2)+\cos (Nt)\sin (t/2)$

We can apply the Reimann-Lebesgue lemma to both parts to show that $S_N(f)({\theta }_0)=f({\theta }_0)$

Note that our condition that $F$ is bounded near $0$ amounts to saying $|f({\theta }_0-t)-f({\theta }_0)|\le M|t|$, or equivalently $|f(\theta )-f({\theta }_0)|\le M|\theta -{\theta }_0|$ for some $M\ge 0$, which is the same as saying that $f$ satisfies a Holder condition $\alpha =1$, and is actually sufficient for our theorem

Weirdly enough, this result means the convergence of our Fourier series is dependent on the local behavior of $f$, despite the fact that the Fourier series is derived from the entire domain of $f$

Theorem: Suppose $f$ and $g$ are two integrable functions defined on the circle, and for some ${\theta }_0$, there exists an open interval $I$ containing ${\theta }_0$ such that $f(\theta )=g(\theta )$ for all $\theta \in I$, then $S_N(f)({\theta }_0)-S_N(g)({\theta }_0)\to 0$ as $N$ tends to infinity

This happens because $f-g$ is $0$ in $I$ and therefore differentiable at ${\theta }_0$, so we can apply our convergence theorem

A Diverging Continuous Function

Continuity is not strong enough to guarantee convergence, but finding an actual counter example for this is quite difficult

We will use a technique called symmetry-breaking, where symmetry refers to the relationship between frequencies $e^{in\theta }$ and $e^{-in\theta }$

Consider the sawtooth function $f(\theta )\sim {\sum }_{n\ne 0}^{}\frac{e^{in\theta }}{n}$

If we “split” this series and only view the negative frequencies, $\tilde{f}={\sum }_{n=-\infty }^{n=-1}\frac{e^{in\theta }}{n}$, then we can see the Abel mean $|A_r(\tilde{f})(0)={\sum }_{n=1}^{\infty }\frac{r^n}{n}$ which tends to infinity, so this function diverges

We define $f_N(\theta )={\sum }_{1\le |n|\le N}^{}\frac{e^{in\theta }}{n}$ and ${\tilde{f}}_N(\theta )={\sum }_{-N\le n\le -1}^{}\frac{e^{in\theta }}{n}$

• Since ${\sum }_{n=1}^{N}1/n\ge \log N$, we have $|{\tilde{f}}_N(0)|\ge c\log N$
• $f_N(\theta )$ is uniformly bounded in $N$ and $\theta$ (more complicated to prove)

Note that these are both trigonometric polynomials of degree $N$
We can form trigonometric polynomials $P_N$ and ${\tilde{P}}_N$ by displacing the frequencies of $f_N$ and ${\tilde{f}}_N$ by $2N$ units

$P_N(\theta )=e^{i(2N)\theta }{f}_N(\theta )$ has coefficients for $N\le n\le 3N$, $n\ne 2N$
${\tilde{P}}_N(\theta )=e^{i(2N)\theta }{\tilde{f}}_N(\theta )$ has degree $N\le n\le 2N-1$

$S_M(P_N)=\{\begin{array}{ll}{P}_N& \text{if }M\ge 3N\\ {\tilde{P}}_N& \text{if }M=2N\\ 0& \text{if }M<N\end{array}$

This operator “breaks” symmetry when $M=2N$ but is otherwise benign

We now look for a convergent series ${\sum }_{}^{}{\alpha }_k$ and $\{N_k\}$ such that $N_{k+1}>3N_k$ and ${\alpha }_k\log N_k\to \infty$

${\alpha }_k=\frac{1}{k^2}$ and $N_k=3^{2^k}$ works

Our desired function is now $f(\theta )={\sum }_{k=1}^{\infty }{\alpha }_k{P}_{N_k}(\theta )$
Since $|P_N(\theta )|=|f_N(\theta )|$ is uniformly bounded, this function converges

However, $|S_{2N_m}(f)(0)|\ge c{\alpha }_m\log N_m+O(1)\to \infty$
So this Fourier series diverges at $\theta =0$