If $X$ is a completely regular space, then the strict topology on the algebra of its bounded continuous functions is usually taken to be given by the seminorms $f \mapsto \| f \|_\varphi = \sup \varphi |f|$, where $\varphi$ ranges over the positive bounded functions that vanish at infinity.
To my surprise, I have found a related, equivalent description (first page), that claims that $\varphi$ may range only over the positive bounded upper semi-continuous functions that vanish at infinity. Why is it so?
The reduction to usc case is made by the upper semicontinuous regularization of $\varphi$, defined as $$ \varphi^* (x) = \inf_{V\ni x} \sup_V \varphi $$ where the infimum is taken over all neighborhoods $V$ of $x$. By construction, the set $\{x : \varphi^* (x) < t\}$ is open for every $t$, which means $\varphi^*$ is upper semicontinuous.
Note that $\varphi\le \varphi^*$ (also, $\varphi^*$ is the smallest usc majorant of $\varphi$).
For every bounded continuous $f$ we have $\sup \varphi |f| = \sup \varphi ^*|f|$. Indeed, $\le$ is clear. To prove $\ge$, take $t$ such that $t < \sup \varphi^* |f|$. Then there exists a point $x$ where $\varphi^*(x) |f(x)| > t$. By definition of $\varphi^*$, for every neighborhood $V\ni x$ we have $|f(x)| \sup_V \varphi > t$. Taking a sufficiently small neighborhood $V$ and using the continuity of $f$ we arrive at $\sup_V (\varphi|f|) > t$.
Aside: In a number of places it is written that "$\varphi^* (x) = \limsup_{y\to x} \varphi(y)$" which I think is a byproduct of terminological confusion around $\limsup$. The way $\limsup$ is defined, it is possible that $\limsup_{y\to x} \varphi(y)$ can be less than $\varphi(x)$. It would be correct to write $\varphi^* (x) = \max(\varphi(x), \limsup_{y\to x} \varphi(y))$.