Extension by continuity on metric spaces

Let $X,Y$ metric spaces and let $X_0 \subset X$ be dense. Consider $T \colon X_0 \to Y$ be continuous. Then can $T$ be extended uniquely by continuity on all $X$ by $$Tx=\lim T x_n = T \lim x_n$$ where $x_n \to x$ and the limit doesn't depend on the approximating sequence?

Here you find the result proved where $X,Y$ are Frechet spaces(so metric spaces with translation invariance) https://www-users.cse.umn.edu/~garrett/m/real/notes_2019-20/08c_extension_by_continuity.pdf

and here

https://en.wikipedia.org/wiki/Continuous_linear_extension

for bounded linear operators. Is there something more general?

Solution 1:

All the sources you mention suppose that $T$ is linear. Since you are asking for a generalization, I will avoid this condition.

If $T$ is only supposed to be continuous, the result is false. As an example, take $T=\sin(1/x)$, $X=[0,1], X_0=(0,1], Y=\mathbb{R}$.

We need an assumption like uniform continuity. This by itself is still not enough; we need an assumption on the metric space itself, since otherwise the result is false: take $T(x)=x$, $X=\mathbb{R}$, $X_0=\mathbb{R}-\{0\}=Y$.

Keeping in mind the two counterexamples above, the most general theorem we can state is: Let $(X,d_1)$ be a metric space with dense subet $X_0$, $(Y,d_2)$ a complete metric space and $T:X_0\to Y$ a uniformly continuous function. Then there is a unique continuous extension $\tilde T:X\to Y$ of $T$.

This result is not hard to prove and I encourage you to try and do it (its proof is not too different from the one for linear operators that Garrett gives in its notes)