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Pseudocompactness in the $m$-topology in $C(X)$
The sets $\{f \in C(X) : |g-f| \leq u , g\in C(X) , $u$ \text{ a positive unit of } C(X)\}$ form a base for some topology on $C(X)$ which is called the $m$-topology on $C(X)$.
A norm on $C^*(X)$ is given by $||f|| = \sup |f(x)|$ and the resulting metric topology is called the uniform norm topology on $C^*(X)$.
How to show the above two topologies on $C^*(X)$ (consider relative top. for the m-topology) coincide iff $X$ is pseudocompact.
(Hints: When $X$ is not pseudo-compact, the set of constant functions in $C^*(X)$ is discrete, in the m-topology, so that $C^*(X)$ is not even a topological vector space whereas $C^*(X)$ forms a Banach algebra w.r.t. uniform norm topology).