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Let $X$ be a normed linear space. The strong topology on $X$ is the topology induced by the norm, i.e. it has the open balls $B_r(x)$ as its basis.

Is there an alternative characterization of strong topology? If not, are there additional assumptions on the space (such as $\dim X < \infty $) that yield an alternative characterization for the given space?

I was thinking about this in the context of other topologies, such as the weak topology, which can be defined as the coarsest topology such that all continuous linear functionals on $X$ (i.e. in $X^*$, the topological dual) are continuous in that weak topology.

For example, could we claim that the coarsest topology such that all continuous functions (w.r.t. the norm) are still continuous in that topology is the strong topology?

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As to your final question: for any Tychonoff (aka completely regular) space $(X,\mathcal{T})$ and suppose we consider the set $\mathcal{F}$ of all real-valued continuous functions on $(X,\mathcal{T})$; we then have that $\mathcal{T}$ is the coarsest topology that makes all functions in $\mathcal{F}$ continuous.

So that fact is completely general for all Tychonoff spaces and all topologies on topological vector spaces that are $T_1$ and make the linear operations continuous are Tychonoff. So it holds for both the weak and the strong topology.

Note that the weak topology is defined differently: it is the coarsest topology that makes all functionals (!) continuous that were already continuous in the strong topology.

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