Schwartz Class Functions on Integers On $\mathbb{R}$, we define the Schwartz class functions as infinitely differentiable functions such that
$$ \lim\limits_{|x|\to \infty} | x^{m}f^{(n)}(x) | = 0 $$
for all $m, n \in \mathbb{N}$ and $f^{(n)}$ denotes the $n^{th}$ derivative of $f$.
However, I was asked to define Schwartz class on integers, and I came up with the following:
$$ \mathcal{S}(\mathbb{Z}) = \{ f : \mathbb{Z} \to \mathbb{C} \;\; | \;\; \lim\limits_{|n| \to \infty} n^{k}f(n) = 0 \;\; \forall\,\, k \in \mathbb{N} \}$$
The first examples that come to my mind are the restrictions of Schwartz functions on $\mathbb{R}$ to $\mathbb{Z}$. For example, $f(n) = e^{-n^{2}}$ being the restriction of the Gaussian function.

I was wondering if the converse is true. That is, given a function in $\mathcal{S}(\mathbb{Z})$, can it be extended to a function in $\mathcal{S}(\mathbb{R})$?
I was thinking that we could smoothly interpolate the function in between the integers, where it is already defined. The question remains about the rapid decay of this function. However, since original function goes to zero rather faster, I expect this to go to zero fast enough as well, since the slopes cannot change erratically in this case.

So, is the statement true? And if so, what is the justification?

 A: What GiuseppeNegro says is true. Indeed, let $f_0$ be in $\mathcal{S}(\mathbb Z)$ as defined above, and extend it to a function $f:\mathbb R\to\mathbb R$ as follows: between $n$ and $n+1$, let $f$ be equal to $f_0(n)$ on $\left[n,n+\frac{1}{3}\right]$, and equal to $f_0(n+1)$ on $\left[n+\frac{2}{3},n+1\right]$. In $\left[n+\frac{1}{3},n+\frac{2}{3}\right]$, extend $f$ linearly. Then, for all $x\in\mathbb R$, you have that $|f(x)|\leq\max\{|f([x])|,|f([x]+1)|\}$, where $[x]$ is the floor function of $x$.
Now, let $\phi$ be an infinitely differentiable function, with support in $\left[-\frac{1}{3},\frac{1}{3}\right]$ and integral equal to $1$, and define $g$ to be the convolution of $f$ and $\phi$: $g=f\ast\phi$. You can check that, if $k\in\mathbb N$, then the $k$-th derivative of $g$ at $x$ is $g^{(k)}(x)=\left(f\ast\phi^{(k)}\right)(x)$. Also, if $n\in\mathbb Z$, then $$g(n)=(f\ast\phi)(n)=\int_{\mathbb R}\phi(y)f(n-y)\,dy=\int_{-\frac{1}{3}}^{\frac{1}{3}}\phi(y)f(n-y)\,dy=\int_{-\frac{1}{3}}^{\frac{1}{3}}\phi(y)f(n)\,dy=f(n)=f_0(n),$$ so $g$ is equal to $f_0$ on $\mathbb Z$.
Let $k,l\in\mathbb N$. $|\phi^{(k)}|$ is bounded above by some $C_k>0$, so you have that $$\left|x^lg^{(k)}(x)\right|\leq|x|^l\int_{-\frac{1}{3}}^{\frac{1}{3}}\left|\phi^{(k)}(y)f(x-y)\right|\,dy\leq|x|^lC_k\int_{x-\frac{1}{3}}^{x+\frac{1}{3}}|f(y)|,$$ which is bounded by $$\frac{2}{3}C_k|x|^l\sup\left\{|f(y)|:x-\frac{1}{3}\leq y\leq x+\frac{1}{3}\right\}.$$ Then, using the property $|f(x)|\leq\max\{|f([x])|,|f([x]+1)|\}$ and the fact that $f_0$ is in $\mathcal{S}(\mathbb Z)$, you have that $$\lim_{|x|\to\infty}\left|x^lg^{(k)}(x)\right|=0,$$ so $g$ is in $\mathcal{S}(\mathbb R)$ and extends $f_0$.
