Convergencia de una suma infinita en la clase de Schwartz

Question

Supongamos que tenemos una función $g$ en la clase de Schwartz. Considerar la suma $$\dfrac{1}{L}\sum_{n=-\infty}^{\infty}g\left(\dfrac{n\pi}{L}\right)$$ 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 $$\dfrac{1}{\pi}\int_{-\infty}^\infty g(x)dx?$$

Answer 1

Para ver la convergencia, divide la integral en partes convenientes,

$$\begin{align} \left\lvert\frac{1}{\pi}\int_{-\infty}^\infty g(x)\,dx - \frac{1}{L}\sum_{n=-\infty}^\infty g\left(\frac{n\pi}{L}\right)\right\rvert &= \left\lvert \frac{1}{\pi}\sum_{n=-\infty}^\infty \int_{a_n}^{b_n} g(x) - g\left(\frac{n\pi}{L}\right)\,dx\right\rvert\\ &\leqslant \frac{1}{\pi}\sum_{n=-\infty}^\infty\left\lvert \int_{a_n}^{b_n} g'\left(\frac{n\pi}{L}\right)\left(x-\frac{n\pi}{L}\right) + \frac12 g''(\xi)\left(x-\frac{n\pi}{L}\right)^2\,dx\right\rvert\\ &= \frac{1}{2\pi}\sum_{n=-\infty}^\infty\left\lvert \int_{a_n}^{b_n} g''(\xi)\left(x-\frac{n\pi}{L}\right)^2\,dx\right\rvert\\ &\leqslant \frac{1}{2\pi} \sum_{n=-\infty}^\infty K_n\int_{a_n}^{b_n}\left(x-\frac{n\pi}{L}\right)^2\,dx\\ &= \frac{1}{3\pi} \sum_{n=-\infty}^\infty K_n \left(\frac{\pi}{2L}\right)^3, \end {Alinee el} $$

donde hemos utilizado

$$a_n = \frac{\left(n-\frac12\right)\pi}{L}\quad \text{and}\quad b_n = \frac{\left(n+\frac12\right)\pi}{L}$$

brevedad y $K_n$ es el supremum de $\lvert g''\rvert$ en el intervalo $[a_n,b_n]$. $g$ Pertenece a la clase de Schwartz, tenemos

$$K_n \leqslant \frac{C}{1 + \left(\frac{n\pi}{L}\right)^2}$$

% constante $C$y por lo tanto

$$\begin{align} \left\lvert\frac{1}{\pi}\int_{-\infty}^\infty g(x)\,dx - \frac{1}{L}\sum_{n=-\infty}^\infty g\left(\frac{n\pi}{L}\right)\right\rvert &\leqslant \frac{C'}{L}\sum_{n=-\infty}^\infty \frac{1}{L^2 + (n\pi)^2} \leqslant \frac{M}{L} \end {Alinee el} $$

constantes $C'$ y $M$.