Proving that $f(z;\sigma)=\sum_{k\in\Bbb Z}\frac{1}{\sqrt{2\pi}\, \sigma}{\rm e}^{-\frac{(z-k)^2}{2\sigma^2}}$ converges to $1$ as $\sigma\to\infty$

Question

I have an application where I get following function as a result: $$f(z;\sigma) = \sum_{k \in \mathbb{Z}} \frac{1}{\sqrt{2 \pi} \, \sigma} \textrm{e}^{-\frac{(z - k)^2}{2 {\sigma}^{2}}}$$

It appears that $$\lim_{\sigma \rightarrow \infty} f(z;\sigma) = 1$$ but I can't currently find a way to prove this.

Is this property of the sum true, and if it is, why? Any references would be greatly appreciated.

Answer 1

As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that $$ \sum_{k\in \mathbb{Z}} f(x+k) = \sum_{j\in\mathbb{Z}} \hat{f}(j)e^{2\pi ijx},\quad \forall x\in \mathbb{R} $$ for all Schwartz function $f$. Here, $\hat{f}$ is the Fourier transform of $f$ on $\mathbb{R}$. In this case, let $$f_\sigma(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}= D^1_\sigma f_1(x)$$ where $D^s_\alpha g(x) = \frac{1}{\alpha^{\frac{1}{s}}}g(\frac{x}{\alpha})$ is a dilation operator. Then, it holds that $$ \widehat{f_\sigma}(\xi) =\widehat{D^1_\sigma f_1}(\xi) = D^\infty_{1/\sigma}\widehat{f_1}(\xi)=e^{-2\pi^2\sigma^2\xi^2},\quad\forall \xi\in\mathbb{R}. $$ Hence the given sum is $$ \sum_{k\in \mathbb{Z}} f_\sigma(x+k) = \sum_{j\in\mathbb{Z}} \widehat{f_\sigma}(j)e^{2\pi ijx}=\sum_{j\in\mathbb{Z}} e^{-2\pi^2\sigma^2j^2}e^{2\pi ijx} = 1+\sum_{j\neq 0} e^{-2\pi^2\sigma^2j^2}e^{2\pi ijx}. $$For $\sigma>1$, we have $$ |e^{-2\pi^2\sigma^2j^2}e^{2\pi ijx}|\leq e^{-2\pi^2j^2} \in l^1(\mathbb{Z}). $$ Thus, by Lebesgue's dominated convergence theorem, as $\sigma \to\infty$, we get $$ \sum_{j\neq 0} e^{-2\pi^2\sigma^2j^2}e^{2\pi ijx} \to 0, $$ and as a result $$ \lim_{\sigma\to\infty}\sum_{k\in \mathbb{Z}} f_\sigma(x+k) = 1,\quad \forall x\in \mathbb{R}. $$

Answer 2

While other people gave you mathematically rigorous solution, here is a more intuitive one:

Let's go in the other limit, $\sigma\to 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$\int_{-0.5}^{0.5}f(z,0)dz=1$$ When you increase $\sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $\sigma\to\infty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$\int_{-0.5}^{0.5}f(z,\infty)dz=\int_{-0.5}^{0.5}Cdz=C=1$$

Answer 3

Hint. Consider the gaussian function $g(w)=\frac{e^{-\frac{w^2}{2 {\sigma}^{2}}}}{\sqrt{2 \pi} \, \sigma} $. Then $$\sum_{k \in \mathbb{Z}} \frac{e^{-\frac{(z - k)^2}{2 {\sigma}^{2}}}}{\sqrt{2 \pi} \, \sigma} -\frac{e^{-\frac{z^2}{2 {\sigma}^{2}}}}{\sqrt{2 \pi} \, \sigma} =\sum_{k \in \mathbb{Z}\setminus \{0\}} g(z-k)\leq \int_{-\infty}^{\infty}g(w)\,dw=1$$ where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k \in \mathbb{Z}$ under the graph of $g$.

In a similar way we have that $$\sum_{k \in \mathbb{Z}} \frac{e^{-\frac{(z - k)^2}{2 {\sigma}^{2}}}}{\sqrt{2 \pi} \, \sigma} +\frac{e^{-\frac{z^2}{2 {\sigma}^{2}}}}{\sqrt{2 \pi} \, \sigma} =2g(0)+\sum_{k \in \mathbb{Z}\setminus \{0\}} g(z-k)\\\geq \int_{-\infty}^{\infty}g(w)\,dw=1$$ where this time the union of the rectangles contains the area under the graph of $g$.