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. 



Continuous function on colimit
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 $f$.


