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Let $X$ be a Tychonoff space, let $A,B\subset X$ be closed. Let $J_A$ be the set of all continuous on $X$ real-valued functions which vanish on $A$.

For which $X$'s is it true that $J_A+J_B=J_{A\cap B}$?

I can prove this if $X$ is hereditary normal, as well as in the case when $C(X)$ is complete and sequential in the compact open topology (in particular if $X$ is hemi-compact compactly generated). However, in the second case the proof is somewhat indirect, so as a side here is another question:

Is there a constructive-ish proof of the property in question for compact $X$'s?

It would be particularly nice if there was a pair of monotone maps $\varphi,\psi$ from $J_{A\cap B}$ into $J_A$ and $J_B$ such that $\varphi(f)+\psi(f)=f$, for every $f\in J_{A\cap B}$.

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    $\begingroup$ The condition implies $X$ is normal, right? For if $A, B \subseteq X$ are disjoint closed subsets, then $J_{A \cap B} = C(X)$ so $1 = f+g$ for $f \in J_A$ and $g \in J_B$. Then $U = \{f < \tfrac{1}{2}\}$ and $V = \{g < \tfrac{1}{2}\}$ are disjoint open sets containing $A$ and $B$ respectively. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 10, 2022 at 13:42

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This characterizes normality. That it implies normality was observed above by Remy.

Conversely, assume $f$ vanishes on $A\cap B$. Define $h:A\cup B\to\mathbb{R}$ by $h(x)=f(x)$ if $x\in A$ and $h(x)=0$ if $x\in B$. By the Tieze-Urysohn theorem $h$ has a continuous extension $H:X\to\mathbb{R}$. That extension belongs to $J_B$. But then $G=f-H$ belongs to $J_A$ because $f(x)=H(x)$ on $A$. We have $f=G+H$.

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