# Fourier series convergence for sum of Schwartz class functions

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.

$$F(x)=\sum_{k\in\mathbb{Z}}\hat{F}(k)e^{ikx}?$$

## 1 Answer

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 $(\hat{h}(k)) \in \ell^1(\mathbb{Z})$.

• Thanks, Daniel. In your expression for $\int_0^xG(t)dt$, what do you use to exchange the sum and the integral? – PJ Miller Dec 14 '13 at 7:07
• 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. $[x,0]$). – Daniel Fischer Dec 14 '13 at 11:12