Let $f$ be a Schwartz class function. Let $F(x)=\sum_{n\in\mathbb{Z}}f(x-2\pi n)$. Then $F$ is periodic of period $2\pi$.

How can we show that the Fourier series of $F$ converges to $F$ pointwise everywhere, i.e.



The derivatives of rapidly decreasing (Schwartz class) functions are also rapidly decreasing. Hence we can sum the derivative $f'$ in a like manner,

$$G(x) = \sum_{n\in\mathbb{Z}} f'(x-2\pi n).$$

The sum converges locally uniformly, hence $G$ is continuous, and

$$\begin{align} \int_0^x G(t)\,dt &= \int_0^x \sum_{n\in\mathbb{Z}} f'(t-2\pi n)\,dt\\ &= \sum_{n\in\mathbb{Z}} \int_0^x f'(t-2\pi n)\,dt\\ &= \sum_{n\in \mathbb{Z}} f(x-2\pi n) - f(0-2\pi n)\\ &= F(x) - F(0), \end{align}$$

so $F$ is continuously differentiable (even $C^\infty$, but we don't need that).

For a continuously differentiable $2\pi$-periodic function $h$, the Fourier coefficients are summable(1), $\sum\limits_{k\in\mathbb{Z}}\lvert \hat{h}(k)\rvert < \infty$, hence the Fourier series converges uniformly to $h$.

(1) $h$ and $h'$ both belong to $L^2([0,2\pi])$, so for $k\neq 0$

$$\begin{align} 2\pi\hat{h}(k) &= \int_{0}^{2\pi} h(t)e^{-ikt}\,dt\\ &= \left[\frac{i}{k}h(t)e^{-ikt}\right]_0^{2\pi} - \frac{i}{k}\int_{0}^{2\pi} h'(t)e^{-ikt}\,dt\\ &= \frac{i}{k}\left[h(2\pi) - h(0)\right] + \frac{1}{ik} \int_0^{2\pi} h'(t)e^{-ikt}\,dt\\ &= \frac{2\pi}{ik} \hat{h'}(k). \end{align}$$

Since both, $(\hat{h'}(k))$ and $\left(\frac1k\right)$ belong to $\ell^2(\mathbb{Z}\setminus\{0\})$, it follows that $(\hat{h}(k)) \in \ell^1(\mathbb{Z})$.

  • $\begingroup$ Thanks, Daniel. In your expression for $\int_0^xG(t)dt$, what do you use to exchange the sum and the integral? $\endgroup$ – PJ Miller Dec 14 '13 at 7:07
  • $\begingroup$ The locally uniform convergence of the series. The interval of integration is bounded, hence $\sum f'(x-2\pi n)$ converges uniformly on $[0,x]$ (resp. $[x,0]$). $\endgroup$ – Daniel Fischer Dec 14 '13 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.