Continuity of translation from $\mathbb{R}$ to Schwartz space Given $f \in \mathcal{S}(\mathbb{R})$ where $\mathcal{S}(\mathbb{R})$ is the Schwartz space, the space of $C^\infty(\mathbb{R})$ such that: $\forall i, j\in \mathbb{N}$
$$
\rho_{i,j}(f) = \sup_{x\in \mathbb{R}} |x|^i|\frac{d^jf}{dx^j}(x)| < \infty
$$
I want to show that the map from $\mathbb{R} \to \mathcal{S}(\mathbb{R})$, $h \mapsto \tau_h(f)$ where $\tau_h(f)(x) = f(x - h)$ is continuous.
I know that is enough to take a sequence in $\mathbb{R}$ $\{h_n\}$ $h_n\to h$  as $n \to \infty$ and show that $\forall i, j \in \mathbb{N}$
$$
\rho_{i, j}(\tau_h(f) - \tau_{h_n}(f)) \to 0 \text{ as } n\to \infty
$$
But I'm stuck I've been trying to use the uniform continuity of the derivatives as I know that continuous functions that vanish at infinity are uniformly continuous but it seems to be of no help.
I'm supposed to solve this without using fourier transform as I haven't studied it yet in my course. Thanks in advance!.
 A: Here is one approach. Try to fill-in the gaps and/or adapting to your particular choice of generating norms.
Notice that the topology on $\mathcal{S}(\mathbb{R})$ is generated by the norms
$$\rho_m(\phi)=\sup_{\substack{x\in\mathbb{R}\\ 0\leq k\leq m}} |(1+|x|^2)^m D^k\phi(x)|,\qquad m\in\mathbb{Z}_+$$
The following inequality will be useful for our purposes:
\begin{align}
(1+|x|^2)\leq 4(1+|x-y|^2)(1+|y|^2)\tag{1}\label{one}
\end{align}
Fix $m\in\mathbb{N}$ and $h_0\in\mathbb{R}$. For any $k\in\mathbb{Z}_+$ and $h\in\mathbb{R}$
\begin{align}
(1+|x|^2)^m|D^k(\tau_h\phi-\tau_{h_0}\phi)(x)|&=(1+|x|^2)^m|\phi^{(k)}(x-h)-\phi^{(k)}(x-h_0)|\\
&=(1+|x|^2)^m|h-h_0|\phi^{(k+1)}(x+\theta(h-h_0)|
\end{align}
for some $\theta=\theta(k,x,h)\in(0,1)$ (here we apply the mean value theorem).
Dividing and multiplying by $(1+|x+\theta(h-h_0)|^2)^{m+1}$ yields
\begin{align}
(1+|x|^2)^m|D^k(\tau_h\phi-\tau_{h_0})(x)|&\leq|h-h_0|\frac{(1+|x|^2)^m}{(1+|x+\theta(h-h_0)|^2)^{m+1}}\rho_{m+1}(\phi)\\
&\leq4|h-h_0|(1+|h-h_0|^2)^{m+1}\rho_{m+1}(\phi)
\end{align}
Here we use \eqref{one} in the last inequality.
Therefore
$$\rho_m(\tau_h\phi-\tau_{h_0}\phi)\leq 4|h-h_0|(1+|h-h_0|^2)^{m+1}\rho_{m+1}(\phi)\xrightarrow{h\rightarrow h_0}0$$
The conclusion follows.
