Let $X$ be a Tychonoff topological space and let $x\in X$. Let $B\subset C(X)$ be convex and compact in the topology of pointwise convergence, and such that $f(x)=1$, for every $f\in B$.



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> Is there an open neighborhood $U$ of $x$ such that $f(y)\ne 0$, for every $f\in B$ and $y\in U$?
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Equicontinuity-like property of a convex compact set
Since $C(X)$ is not complete one cannot take the closed convex hull of the example in the comment. But what about this:


Let $g\_n=1-f\_n$ with $f\_n$ as in my comment.
Since the $g\_n$ are bounded by one, the linear map $T:\ell^1\to C(X)$, $\lambda\mapsto \sum\limits\_{n=1}^\infty \lambda\_ng\_n$ is well defined. We would like to have $T$ continuous as a map $(\ell^1,\sigma(\ell^1,c\_0)) \to (C(X),pw)$ (where $c\_0$ is the space of all null sequences so that $\ell^1$ is its dual). This follows from the pointwise convergence to $0$ of all sequences $(g\_n(x))\_{n\in\mathbb N}$. The unit ball $K$ of $\ell^1$ is weak$^\*$ compact à la Alaoglu and hence $T(K)$ is compact and convex in $(C(X),pw)$. Then $B=\{1-g: g\in T(K)\}$ is convex and compact in the pointwise topology, it satisfies $f(0)=1$ for all $f\in B$ (because $g\_n(0)=0$ for all $n$), and it contains all $f\_n=1-g\_n=1-T(e\_n)$ with the standard unit vectors $e\_n$ of $\ell^1$.


