Qfyilyi

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.

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

Let's go in the other limit, $sigmato 0$. 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 $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$

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 $g$.

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