Tag Archives: Fourier

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 prove 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)