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 $A$.



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> For which $X$'s is it true that $J\_A+J\_B=J\_{A\cap B}$?
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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:



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> Is there a constructive-ish proof of the property in question for compact $X$'s?
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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 $f\in J\_{A\cap B}$.



On the equality $\{f\in C(X), f|_A=0\}+\{f\in C(X), f|_B=0\}=\{f\in C(X), f|_{A\cap B}=0\}$
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 $f=G+H$.


