Extension of a function which is uniformly continuous on compact subsets

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be open
• $-\infty<a<b<c<\infty$

If $f:\Lambda\times\left((a,c)\setminus\left\{b\right\}\right)\to\mathbb R$ is uniformly continuous on any compact subset of $\Lambda\times\left((a,c)\setminus\left\{b\right\}\right)$, are we able to deduce that $f$ has a unique extension $\tilde f$ to $\Lambda\times(a,c)$ and that $\tilde f$ is uniformly continuous?

I know that a uniformly continuous function into a Banach space has a unique extension to the closure of its domain.

Answer 1

This is false. The important thing to note here is that "uniformly continuous on compact subsets" is not the same as uniform continuity. Every continuous function on a compact metric space is uniformly continuous, so the requirement that $f$ be uniformly continuous on compact subsets of $\Lambda \times ((a,c) - \{b\})$ is satisfied by any continuous function.

There are plenty of continuous functions on such a domain that cannot be continously extended as you wish to be able to say. Take, for instance $f : (0,1) \times ((-1, 1) - \{0\}) \rightarrow \mathbb{R}$ defined by the map $$ (x, y) \mapsto \bigg(\frac{1}{y}\bigg).$$ Here $f$ is clearly continuous (hence uniformly continuous on compact sets), but cannot be continuously extended to any point of the form $(a, 0)$, where $0 < a < 1$.