 # 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.

So we’re reduce to prove the following lemma:

LEMMA: 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.

We have for $$n \ge 1$$: $\sum_{k=1}^n \frac{c_k-c_{-k}}{k} = – \frac{i}{\pi} \int_0^{2 \pi} g(x) \sum_{k=1}^n \frac{\sin kx}{k} \ dx$ If we prove that $U_n: x \mapsto \sum_{k=1}^n \frac{\sin kx}{k}$ is a pointwise convergent sequence of functions bounded by a constant independent of $$n$$ and $$x$$, we’ll be done applying Lebesgue dominated convergence theorem.

The pointwise convergence of $$(U_n)$$ can again be obtained using an Abel transformation. Now, denoting $D_n(x) = \sum_{k=-n}^n e^{ikx} = \frac{\sin \left((n+\frac{1}{2})x \right )}{\sin \left( \frac{x}{2} \right)},$ we have
$U_n(x) = \frac{1}{2} \int_0^x (D_n(t) – 1) \ dt.$ Therefore, we’ll be able to conclude if we prove that $$\displaystyle \int_0^x D_n(t) \ dt$$ is bounded by a constant independent of $$n$$ and $$x$$ for $$x \in [0,2 \pi]$$. As $$D_n$$ is even for all $$n \ge 1$$, it is sufficient to consider $$x \in [0, \pi]$$.

The map $t \mapsto \frac{1}{\sin (t/2)} – \frac{1}{t/2}$ defined on $$(0,\pi]$$ can be extended by continuity to $$0$$, which gives a map $$\phi$$ defined on $$[0,\pi]$$ and for $$\alpha \in [0,\pi]$$:
$\left\vert \int_0^\alpha \frac{\sin \left((n+\frac{1}{2})t \right )}{\sin \left( \frac{t}{2} \right)} \ dt – \int_0^\alpha \frac{\sin \left((n+\frac{1}{2})t \right )}{\frac{t}{2}} \ dt\right\vert \le \int_0^\pi \vert \phi(t) \vert \ dt.$ We get the desired conclusion as
$\int_0^\alpha \frac{\sin \left((n+\frac{1}{2})t \right )}{\frac{t}{2}} \ dt = 2 \int_0^{(n+\frac{1}{2})\alpha} \frac{\sin t}{t} \ dt$ and the integral $$\displaystyle \int_0^\infty \frac{\sin t}{t} \ dt$$ is known to be convergent.