Schwartz Class Functions on Integers

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 $f_0$.