$f: \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{R}}: x\mapsto (e^{t\sin(x)})_{ t\in \mathbb{R}}$

Question

I have the following function:

$f: \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{R}}: x\mapsto (e^{t\sin(x)})_{ t\in \mathbb{R}}$

I have to investigate if this function is continuous. I am intimidated of the space $ \mathbb{R}^{\mathbb{R}}$. Is there a general rule how to approach such problems?

I suppose I need to use topological spaces/metric spaces in order to do anything at all here. My naive attemp was at first this:

$f_1:x \mapsto \sin(x)$ is continuous. $f_2:x\mapsto e^x$ is continuous. Therefore, $f_1\circ f_2$ is also continuous, since I know this from general topology. But I have a set of Functions here and I need somehow to use the product topology...

Any help on this?

Answer 1

For spaces $X, Y$, there are many possible topologies for $Y^X$ (as interpreted as $Y^X = \mathcal C (X,Y)$, the space of continuous maps $X → Y$), the two most natural being

• the compact-open topology, which is the coarsest topology such that all sets of functions are open that map some fixed compact set of $X$ to some fixed open set of $Y$, and
• the product topology, which is the finest topology making all projections – in this case interpretable as evaluations – $Y^X → Y,~h ↦ h(x)$ for $x ∈ X$ continuous.

For the compact-open topology. For topological spaces $X$, $Y$ and $Z$ with $Y$ being locally compact, there turns out to be a bijection called the exponential law, namely $$Z^{X×Y} \to (Z^Y)^X$$ This bijection is also called currying.

In this case, the uncurried version of your $f$ is the map $$ℝ × ℝ → ℝ,~(x,t) ↦ \mathrm{e}^{t \sin x}.$$ Now the exponential law implies that $f$ is continuous if and only if this map is – so is it?

For the product topology, a map $f \colon X → Z^Y$ is continuous if and only if all its components, given by $X → Z,~x ↦ f(x)(y)$ for $y ∈ Y$ are.

In your case, the components are for all $t ∈ ℝ$ given by $$ℝ → ℝ,~x ↦ \mathrm{e}^{t\sin x}.$$ Are these continuous?

Caveat. As Asaf Karagila points out, $Y^X$ is ordinarily just interpreted as the set of all maps $X → Y$, in which case the single most natural topology is again the product topology, once more defined as the finest topology making all projections $Y^X → Y,~h ↦ h(x)$ for $x ∈ X$ continuous. In this case, the discussion about the product topology still holds and you can the check the continuity of $f$ component-wise.