# Existence of a continuous function with divergent Fourier series

In that article, I provided an example of a continuous function with divergent Fourier series. We prove here the existence of such a function using Banach-Steinhaus theorem, also called uniform boundedness principle.

Theorem (Uniform Boundedness Theorem) Let $$(X, \Vert \cdot \Vert_X)$$ be a Banach space and $$(Y, \Vert \cdot \Vert_Y)$$ be a normed vector space. Suppose that $$F$$ is a set of continuous linear operators from $$X$$ to $$Y$$. If for all $$x \in X$$ one has $\sup\limits_{T \in F} \Vert T(x) \Vert_Y \lt \infty$ then $\sup\limits_{T \in F, \ \Vert x \Vert = 1} \Vert T(x) \Vert_Y \lt \infty$

Let’s take for $$X$$ the vector space $$\mathcal C_{2 \pi}$$ of continuous functions from $$\mathbb R$$ to $$\mathbb C$$ which are periodic with period $$2 \pi$$ endowed with the norm $$\Vert f \Vert_\infty = \sup\limits_{- \pi \le t \le \pi} \vert f(t) \vert$$. $$(\mathcal C_{2 \pi}, \Vert \cdot \Vert_\infty)$$ is a Banach space. For the vector space $$Y$$, we take the complex numbers $$\mathbb C$$ endowed with the modulus.

For $$n \in \mathbb N$$, the map $\begin{array}{l|rcl} \ell_n : & \mathcal C_{2 \pi} & \longrightarrow & \mathbb C \\ & f & \longmapsto & \displaystyle \sum_{p=-n}^n c_p(f) \end{array}$ is a linear operator, where for $$p \in \mathbb Z$$, $$c_p(f)$$ denotes the complex Fourier coefficient $c_p(f) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t) e^{-i p t} \ dt$

We now prove that
\begin{align*}
\Lambda_n &= \sup\limits_{f \in \mathcal C_{2 \pi}, \Vert f \Vert_\infty=1} \vert \ell_n(f) \vert\\
&= \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} \right\vert \ dt = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert h_n(t) \right\vert \ dt,
\end{align*} where one can notice that the function $\begin{array}{l|rcll} h_n : & [- \pi, \pi] & \longrightarrow & \mathbb C \\ & t & \longmapsto & \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} &\text{for } t \neq 0\\ & 0 & \longmapsto & 2n+1 \end{array}$ is continuous.
Continue reading Existence of a continuous function with divergent Fourier series

# A continuous function with divergent Fourier series

It is known that for a piecewise continuously differentiable function $$f$$, the Fourier series of $$f$$ converges at all $$x \in \mathbb R$$ to $$\frac{f(x^-)+f(x^+)}{2}$$.

We describe Fejér example of a continuous function with divergent Fourier series. Fejér example is the even, $$(2 \pi)$$-periodic function $$f$$ defined on $$[0,\pi]$$ by: $f(x) = \sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]$
According to Weierstrass M-test, $$f$$ is continuous. We denote $$f$$ Fourier series by $\frac{1}{2} a_0 + (a_1 \cos x + b_1 \sin x) + \dots + (a_n \cos nx + b_n \sin nx) + \dots.$

As $$f$$ is even, the $$b_n$$ are all vanishing. If we denote for all $$m \in \mathbb N$$:$\lambda_{n,m}=\int_0^{\pi} \sin \left[ (2m + 1) \frac{t}{2} \right] \cos nt \ dt \text{ and } \sigma_{n,m} = \sum_{k=0}^n \lambda_{k,m},$
we have:\begin{aligned} a_n &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \ dt= \frac{2}{\pi} \int_0^{\pi} f(t) \cos nt \ dt\\ &= \frac{2}{\pi} \int_0^{\pi} \left(\sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]\right) \cos nt \ dt\\ &=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \int_0^{\pi} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right] \cos nt \ dt\\ &=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \lambda_{n,2^{p^3-1}} \end{aligned} One can switch the $$\int$$ and $$\sum$$ signs as the series is normally convergent.

We now introduce for all $$n \in \mathbb N$$:$S_n = \frac{\pi}{2} \sum_{k=0}^n a_k = \sum_{p=1}^\infty \sum_{k=0}^n \frac{1}{p^2} \lambda_{k,2^{p^3-1}} =\sum_{p=1}^\infty \frac{1}{p^2} \sigma_{n,2^{p^3-1}}$

We will prove below that for all $$n,m \in \mathbb N$$ we have $$\sigma_{m,m} \ge \frac{1}{2} \ln m$$ and $$\sigma_{n,m} \ge 0$$. Assuming those inequalities for now, we get:$S_{2^{p^3-1}} \ge \frac{1}{p^2} \sigma_{2^{p^3-1},2^{p^3-1}} \ge \frac{1}{2p^2} \ln(2^{p^3-1}) = \frac{p^3-1}{2p^2} \ln 2$
As the right hand side diverges to $$\infty$$, we can conclude that $$(S_n)$$ diverges and consequently that the Fourier series of $$f$$ diverges at $$0$$. Continue reading A continuous function with divergent Fourier series

# A trigonometric series that is not a Fourier series (Lebesgue-integration)

We already provided here an example of a trigonometric series that is not the Fourier series of a Riemann-integrable function (namely the function $$\displaystyle x \mapsto \sum_{n=1}^\infty \frac{\sin nx}{\sqrt n}$$).

Applying an Abel-transformation (like mentioned in the link above), one can see that the function $f(x)=\sum_{n=2}^\infty \frac{\sin nx}{\ln n}$ is everywhere convergent. We now prove that $$f$$ cannot be the Fourier series of a Lebesgue-integrable function. The proof is based on the fact that for a $$2 \pi$$-periodic function $$g$$, Lebesgue-integrable on $$[0,2 \pi]$$, the sum $\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}$ is convergent where $$(c_n)_{n \in \mathbb Z}$$ are the complex Fourier coefficients of $$g$$: $c_n = \frac{1}{2 \pi} \int_0^{2 \pi} g(t)e^{-ikt} \ dt.$ As the series $$\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln n}$$ is divergent, we will be able to conclude that the sequence defined by $\gamma_0=\gamma_1=\gamma_{-1} = 0, \, \gamma_n=- \gamma_{-n} = \frac{1}{\ln n} \ (n \ge 2)$ cannot be the Fourier coefficients of a Lebesgue-integrable function, hence that $$f$$ is not the Fourier series of any Lebesgue-integrable function. Continue reading A trigonometric series that is not a Fourier series (Lebesgue-integration)

# A trigonometric series that is not a Fourier series (Riemann-integration)

We’re looking here at convergent trigonometric series like $f(x) = a_0 + \sum_{k=1}^\infty (a_n \cos nx + b_n \sin nx)$ which are convergent but are not Fourier series. Which means that the terms $$a_n$$ and $$b_n$$ cannot be written$\begin{array}{ll} a_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \cos nt \, dt & (n= 0, 1, \dots) \\ b_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \sin nt \, dt & (n= 1, 2, \dots) \end{array}$ where $$g$$ is any integrable function.

This raises the question of the type of integral used. We cover here an example based on Riemann integral. I’ll cover a Lebesgue integral example later on.

We prove here that the function $f(x)= \sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is a convergent trigonometric series but is not a Fourier series. Continue reading A trigonometric series that is not a Fourier series (Riemann-integration)