Uniform convergence and Poisson sum

Suppose $f:\mathbb{R}\to\mathbb{C} $ is continuously differentiable twice and there exists $C $ so that: $$ \forall x:\left|f\left(x\right)\right|,\left|\frac{df}{dx}\right|,\left|\frac{d^{2}f}{dx^{2}}\right|\le\frac{C}{\left|x\right|^{2}+1} $$ Prove that $ \sum\limits _{n=-\infty}^{\infty}f\left(x+2\pi n\right) $ converges uniformally in $[-\pi,\pi] $ and deduce Poisson's sum: $$ \sum_{n=-\infty}^{\infty}f\left(2\pi n\right)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\int\limits _{-\infty}^{\infty}f\left(x\right)e^{inx}dx $$ What I tried doing was : Because we know $|x|\le \pi $ we can approximate $\frac{C}{\left(x+2\pi n\right)^{2}+1} \approx\frac{C}{4\pi n^{2}}$ and we know that: $$ f\left(x+2\pi n\right)\le\left|\frac{C}{\left(x+2\pi n\right)^{2}+1}\right| $$ By Weirstrass M test we know the series on the right converges, and so the entire series converges .
What I don't understand: why did we need the bounds on $ f''(x)\,or\,f'(x) $?

As for the Poisson sum we know we can develop a Fourier series because the function is continuously differentiable once. So $$ f\left(x\right)=\sum_{n=-\infty}^{\infty}\left(\frac{1}{2\pi}\int\limits _{-\pi}^{\pi}f\left(t\right)e^{-int}dt\right)e^{inx} $$ No idea how to even approach this from here. Would appreciate guidance

Solution 1:

You got it a bit wrong: $f$ is not periodic, so you cannot express it as a Fourier series. But $$\sum_{n\in\mathbb Z}f(x+ 2\pi n)$$ is periodic (you already proved the series converges). Its $k$-th Fourier coefficient is $$c_k=\frac 1 {2\pi}\int_{-\pi}^\pi \sum_{n\in\mathbb Z}f(x+ 2\pi n)e^{-ikx}dx$$ By the same bound you already used, you can show that the series inside the integral converges uniformly, so you can exchange integral and sum: $$c_k=\sum_{n\in\mathbb Z}\frac 1 {2\pi}\int_{-\pi}^\pi f(x+ 2\pi n)e^{-ikx}dx$$ Now if you change the variable $t=x+2\pi n$, you get $$c_k=\sum_{n\in\mathbb Z}\frac 1 {2\pi}\int_{(2n-1)\pi}^{(2n+1)\pi} f(t)e^{-ikt}dx=\frac 1 {2\pi}\int_{\mathbb R}f(t)e^{-ikt}dt$$ By integrating by parts, you can verify that $$|c_k|\leq \frac 1 {k^2}\int_{\mathbb R}|f^{\prime\prime}(x)|dx\leq \frac 1 {k^2}\int_{\mathbb R}\frac {C}{x^2+1}dx$$ That's where you needed the bound on the first and second derivatives!

So the Fourier series is normally convergent (thus uniformly convergent), and you have for all $x\in\mathbb R$, $$\sum_{n\in\mathbb Z}f(x+2\pi n)= \sum_{k\in\mathbb Z}c_ke^{ikx}=\frac 1 {2\pi}\sum_{k\in\mathbb Z}\int_{\mathbb R}f(t)e^{-ikt}dt \cdot e^{ikx}$$ Evaluating at $x=0$ yields the desired Poisson formula.