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A function $f : \mathbb{R} \to \mathbb{R}$ is said to be uniformly continuous if for all $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x_1)−f(x_2)| < \varepsilon$ whenever$|x_1 −x_2| < \delta$.

Let $U \subseteq \mathbb{R}^\mathbb{R}$ denote the set of uniformly continuous functions $\mathbb{R}\to\mathbb{R}$. Equip $\mathbb{R}^\mathbb{R}$ with the product topology and give $U$ the subspace topology.

Let $\tau : U \times \mathbb{R} \to U$ be defined to be the function $(f, x)\mapsto f_x$ where $f_x$ is the function $f_x(y) = f(y − x)$. Show that $\tau$ is not continuous, but that it is continuous in each variable separately, i.e. for a fixed $x$, the map $f \mapsto f_x$ is continuous, and for a fixed $f$ the map $x \mapsto f_x$ is continuous.


I have the outline of the proof sketched out, but need some help filling in the details. I'm going to prove that $\tau$ is continuous by contradiction.

Assume that it is continuous as a map from $U \times\mathbb{R} \to U$. Then if the map is indeed continuous then the set $\{ (f,x) | 0 < f(-x) <1 \}$ is open. So in particular it must contain a basic open set. Then there must be finitely many real numbers $a_1, a_2, \ldots, a_n$ and finitely many open subsets of the real line $V_1, V_2, \ldots, V_n$ and an interval $(a,b)$ such that for any uniformly continuous function $f$ with $f(a_i) \in V_i$ for all $i =1,2,\ldots,n$ the image of $(a,b)$ under $f$ i.e. $f(a,b)$ is contained in $(0,1)$.

This is false since you can always find a uniformly continuous function that is as big as you like away from a finite set of numbers.

Does this make sense? does anyone see a counterexample? Part two, where you prove things the variables separately, could someone walk me through the case where $x$ is fixed?

Thanks in advance!

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For the fixed $x$ case, isn't it enough to note that the inverse image of $\{f \mid f(y) \in V \}$ is $\{f \mid f(y+x) \in V \}$ for arbitrary $y \in \mathbb{R}$ and open $V \subset \mathbb{R}$? – Niels Diepeveen Oct 3 '11 at 1:58

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