Convergence of infinite sum in Schwartz class Suppose I have a function $g$ in the Schwartz class. Consider the sum $$\dfrac{1}{L}\sum_{n=-\infty}^{\infty}g\left(\dfrac{n\pi}{L}\right)$$ In other words, just evaluate the value of $g$ at intervals of length $\pi/L$. As $L\rightarrow\infty$, does this sum always converge to $$\dfrac{1}{\pi}\int_{-\infty}^\infty g(x)dx?$$
 A: To see the convergence, we split the integral in suitable parts,
$$\begin{align}
\left\lvert\frac{1}{\pi}\int_{-\infty}^\infty g(x)\,dx - \frac{1}{L}\sum_{n=-\infty}^\infty g\left(\frac{n\pi}{L}\right)\right\rvert &= \left\lvert \frac{1}{\pi}\sum_{n=-\infty}^\infty \int_{a_n}^{b_n} g(x) - g\left(\frac{n\pi}{L}\right)\,dx\right\rvert\\
&\leqslant \frac{1}{\pi}\sum_{n=-\infty}^\infty\left\lvert \int_{a_n}^{b_n} g'\left(\frac{n\pi}{L}\right)\left(x-\frac{n\pi}{L}\right) + \frac12 g''(\xi)\left(x-\frac{n\pi}{L}\right)^2\,dx\right\rvert\\
&= \frac{1}{2\pi}\sum_{n=-\infty}^\infty\left\lvert \int_{a_n}^{b_n}  g''(\xi)\left(x-\frac{n\pi}{L}\right)^2\,dx\right\rvert\\
&\leqslant \frac{1}{2\pi} \sum_{n=-\infty}^\infty K_n\int_{a_n}^{b_n}\left(x-\frac{n\pi}{L}\right)^2\,dx\\
&= \frac{1}{3\pi} \sum_{n=-\infty}^\infty K_n \left(\frac{\pi}{2L}\right)^3,
\end{align}$$
where we used
$$a_n = \frac{\left(n-\frac12\right)\pi}{L}\quad \text{and}\quad b_n = \frac{\left(n+\frac12\right)\pi}{L}$$
for brevity, and $K_n$ is the supremum of $\lvert g''\rvert$ on the interval $[a_n,b_n]$. Since $g$ belongs to the Schwartz class, we have
$$K_n \leqslant \frac{C}{1 + \left(\frac{n\pi}{L}\right)^2}$$
for some constant $C$, and hence
$$\begin{align}
\left\lvert\frac{1}{\pi}\int_{-\infty}^\infty g(x)\,dx - \frac{1}{L}\sum_{n=-\infty}^\infty g\left(\frac{n\pi}{L}\right)\right\rvert
&\leqslant \frac{C'}{L}\sum_{n=-\infty}^\infty \frac{1}{L^2 + (n\pi)^2} \leqslant \frac{M}{L}
\end{align}$$
for some constants $C'$ and $M$.
