Continuous function on colimit

Question

Let $X$ be a Banach space and $f:X\rightarrow \mathbb{R}$ be continuous. Suppose that $\{X_n\}_{n \in \mathbb{N}}$ is a strictly nested sequence of sub-Banach spaces, for which $\cup_{n \in \mathbb{N}} X_n$ dense in $X$. Then the colimit is an LF-space which is not metrizable.

Since $f|_{X_n}$ is continuous to $\mathbb{R}$, then by the universal property of the colimit (in the category LCS of locally convex spaces and continuous linear maps) it should extend to $f':X\rightarrow \mathbb{R}$ (where not $X$ is considered with the colimit topology and not its original Banach space topology).

What is $f'$ (explicitly in tems of $f|_{X_n}$)? My intuition is that it is either $f$ or an infinite-sum of the $f|_{X_n}$..

Note: In particular it shouldn't be $f$ because the colimit topology is strictly finer.

Answer 1

I did not understand fully your setting (for example, are the $X_n$ given the topology of a subspace or simply the embedding map $X_n \to X$ is required to be continuous... Or is $f$ linear... etc.). But the colimit topology on $X$ will be finer than the original topology on $X$, and therefore the functional $f$, being continuous w.r.t. the original topology will also be continuous w.r.t. the colimit topology, and hence it satisfies the demand of the universal property - the unique continuous map from the colimit whose restrictions to the $X_n$'s are the given... So yes, I think $f^{\prime}$ will be equal to the original $f$.