Discrete analogous to du Bois-Reymmond's lemma? We say a function $\varphi:\mathbb Z^n\longrightarrow \mathbb C$ is rapidily decreasing if for for every $M\in\mathbb N\cup\{0\}$ there exists a constant $C_{\varphi, M}>0$ such that $$|\varphi(\xi)|\leq C_{\varphi, M}(1+|\xi|)^{-M},$$ for every $\xi\in\mathbb Z^n$. The constant $C_{\varphi, M}$ must not depend on $\xi$. 
Let $\psi\in\mathscr{S}(\mathbb Z^n)$. Is it true that if $$\sum_{\xi\in\mathbb Z^n} \varphi(\xi)\psi(\xi)=0,$$ for every $\varphi\in\mathscr{S}(\mathbb Z^n)$ then $\psi=0$? This would be a discrete analogous to du Bois-Reymond's lemma.
Thanks

I'm adding an alternative proof. We say $\varphi:\mathbb Z^n\longrightarrow \mathbb C$ is slowling increasing if there are $M\in\mathbb N_0$ and $C_M>0$ such that $$|\varphi(\xi)|\leq C_M(1+|\xi|)^M,$$ for every $\xi\in\mathbb Z^n$. We will denote by $\mathscr{S}^\prime(\mathbb Z^n)$ the vector space of these maps.
$\mathscr{S}(\mathbb Z^n)$ and $\mathscr{S}^\prime(\mathbb Z^n)$ are topological vector spaces and $\mathscr{S}^\prime(\mathbb Z^n)$ is isomorphic to the topological dual of $\mathscr{S}(\mathbb Z^n)$. The isomorphism is given by $T:\mathscr{S}^\prime(\mathbb Z^n)\longrightarrow (\mathscr{S}(\mathbb Z^n))^\prime$, $\psi\longmapsto T_\psi$, where $$\langle T_\psi, \varphi\rangle:=\sum_{\xi\in\mathbb Z^n} \varphi(\xi)\psi(\xi).$$
So the condition $$\sum_{\xi\in\mathbb Z^n} \varphi(\xi)\psi(\xi)=0,$$ for every $\varphi\in\mathscr{S}(\mathbb Z^n)$ means precisely $T(\psi)=T_\psi=0$. Since $T$ is injective $\psi=0$. 
