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Let $(X,\mathcal T)$ be a topological space and $A\subset X$ be a closed subset. Assume $g\in C(A,\mathbb C)$ is such that $g=0$ on $\partial A$. Define the extension $\tilde g$ by $\tilde g = g$ on $A$ and $\tilde g = 0$ on $A^c$. Prove that $\tilde g$ is continuous.

This is Exercise 4.15 in Folland's Real Analysis. I've worked out a proof (see below). I prove that $\tilde g$ is continuous at each $x\in X$ by considering cases : $x\in \text{int} A$, $x\in \partial A$ and $x\in A^c$. I want to know if this is the best possible proof (I find it quite tedious).

Let $x\in \text{int} A$ and $\epsilon >0$. Since $g\in C(A,\mathbb C)$, there is some $U\in \mathcal T$ such that $x\in A\cap U$ and $g(A\cap U) \subset B(g(x),\epsilon)$. Then $x\in \text{int}(A)\cap U \in \mathcal T$ and $\tilde g(\text{int}(A)\cap U) = g(\text{int}(A)\cap U) \subset B(g(x),\epsilon)= B(\tilde g(x),\epsilon)$.

Let $x\in \partial A$ and $\epsilon >0$. Since $g\in C(A,\mathbb C)$, there is some $U\in \mathcal T$ such that $x\in A\cap U$ and $g(A\cap U) \subset B(0,\epsilon)$. Note that $\tilde g(A^c\cap U)\subset \{0\}$, thus $\tilde g(U) \subset B(0,\epsilon)=B(\tilde g(x),\epsilon)$.

Let $x\in A^c$ and $\epsilon >0$. Note that $A^c\in \mathcal T$ and $\tilde g(A^c)\subset \{0\} \subset B(\tilde g(x),\epsilon)$.

Gabriel Romon
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1 Answers1

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You have that $$\tilde{g}(x)= \begin{cases}g(x) & x \in A\\ 0 & x \in \overline{A^\complement} \end{cases}$$

where both cases overlap on $A \cap \overline{A^\complement} = \partial A$, where both are $0$ by assumption.

So the pasting lemma for the case of two closed subsets (a proof of which in general topology terms can be found on this site) implies that $\tilde{g}$ is continuous on $X$ iff $g$ is continuous on $A$ (the constantly $0$ map being trivially continuous anyway).

Henno Brandsma
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