When is this map continuous? Consider the map $h:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega}$
$$h((x_1,x_2,...))=(a_1x_1+b_1,...),a_i,b_i\in\mathbb{R},a_i>0$$ give $\mathbb{R}^{\omega}$ the uniform topology , under what conditions on $ a_i,b_i$ is $ h$ continuous ? 
 A: The sup metric is defined as $D(x,y) = \sup_n d(x_n, y_n)$, where $d(x,y) = \min(|x-y|,1)$, the truncated standard metric on the reals.
$h$ can be written as the composition of two mappings: $m((x_1,x_2,\ldots) = (a_1x_1, a_2x_2, \ldots)$ and $t(x_1,x_2,x_3,\ldots) = (x_1 + b_1, x_2 + b_2, x_3 + b_3, \ldots)$, so that $h = t \circ m$. 
It's easy to check that $t$ is an isometry on $(\mathbb{R}^\omega, D)$, so in particular a homeomorphism. So if $m$ is continuous, so is $h$ and vice versa.
This shows that the values of the $b_i$ are irrelevant. We'll just look at the continuity of $m$ from now on. And we only have to look at continuity at $0$ as $m$ is linear and the metric is translation invariant.
Let $U = \{x \in \mathbb{R}^\omega: D(x,0) < 1 \}$, which is open, as it is an open ball. 
Clearly $0 \in m^{-1}[U]$. Suppose that there exists $r>0$ such that $B(0,2r) \subseteq m^{-1}[U]$. So the sequence $(r,r,r,\ldots$ lies in that ball so its image under $m$, $(ra_1,ra_2,\ldots,) \in U$. This implies that the supremum of the $ra_n$ never exceeds 1 (otherwise the distance to $0$ would have been $1$). So the sup of the $a_n$ is bounded, and this is thus a necessary condition to be continuous. 
On the other hand, if the $a_i$ are bounded above, say by $A$, then $m$ is continuous at $0$: take for $\varepsilon > 0$ (and $<1$, wlog) $\delta = \frac{\varepsilon}{2A}$.
So $h$ is continuous iff $\sup_n a_n < \infty$. 
