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I got stuck at showing a function is continuous, here is the setting: Suppose $X$ is locally compact hausdorff topological space. $\omega \in C_{c}(X,\mathbb{R}^n)$ is a compactly supported continuous function from $X$ to $\mathbb{R}^n$. While $f,g \in C_{c}(X,[0,\infty))$ such that $|\omega| \leq f+g $, which $| \cdot |$ is the Euclidean norm

We define $\omega'$ as following: \begin{equation*} \omega' = \left\{ \begin{array}{ll} \frac{f}{f+g}\omega & \quad \textit{if} \quad f+g>0 \\ 0 & \quad \textit{if} \quad f+g=0 \end{array} \right. \end{equation*}

Is $\omega'$ continuous? I have tried and failed to show $\omega'$ is uniform limit of some continuous function. And also I think the way proving preimage set is open here is kind of tedious.\

Any hints, thoughts, answers are appreciated.

mikeqwertyuiop
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Clearly $\omega'$ is continuous on $\{f+g > 0\}$. If $x\in \overline{\{f+g > 0\}}$, take a net $x_\alpha \in \{f+g > 0\}$ such that $x_\alpha\to x$. Then from the inequality $|\omega'(x_\alpha)| = \frac{f(x_\alpha)}{f(x_\alpha)+g(x_\alpha)}|\omega(x_\alpha)| \leq f(x_\alpha)$ we see that if $f(x) = 0$ then $\omega'(x_\alpha)\to 0 = \omega'(x)$. If $f(x)\neq 0$ then we already know that $\omega'(x_\alpha)\to \omega'(x)$ from continuity of $\omega$ on $\{f+g > 0\}$.

Thus from the following post it follows that $\omega'$ is continuous on $\overline{\{f+g > 0\}}$. Since $\omega'$ is continuous on $\{f+g = 0\}$ as well and both sets are closed, $\overline{\{f+ g > 0\}}\cup \{f+g = 0\} = X$, it follows from gluing lemma that $\omega'$ is continuous.

Remark: We didn't use anything about the space $X$ or that $f, g, \omega$ are compactly supported.

Here's an argument using open sets instead. Let $x_0$ be such that $f(x_0) = 0$ and $\varepsilon > 0$. Take a neighourhood $U$ of $x_0$ such that $f(x)\leq \varepsilon$ for $x\in U$. Then $$|\omega'(x_0)-\omega'(x)| = |\omega'(x)| \leq f(x)\leq \varepsilon$$ for $x\in U\cap \{f+g > 0\}$ and $|\omega'(x_0)-\omega'(x)| = 0$ for $x\in U\cap \{f+g = 0\}$. Thus $$\omega'[U]\subseteq \{a\in\mathbb{R}^n : |\omega'(x_0)-a|\leq \varepsilon\}.$$ So $\omega'$ is continuous at $x_0$.

If $f(x_0)\neq 0$ then $x_0\in \{f+g > 0\}$, but we know that $\{f+g > 0\}$ is open and the restriction of $\omega'$ to this set is continuous, so $\omega'$ is continuous at $x_0$.

The above argument shows that $\omega'$ is continuous at each $x_0\in X$, thus continuous.

Jakobian
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  • Thanks for your answer, I am not familiar with nets, is it similar to sequence limit? Like if X is metric space we have [$\lim_{n \to \infty} f(x_n) = f(a) $ whenever $x_n \to a$ ] $\iff$ $f$ is continuous. While how net is possible to use to prove continuity that even $X$ is not metric space? ( Again, I familiar with nets, it would be helpful if you can suggest me some reference or briefy tell me how it works. Thank you – mikeqwertyuiop Nov 26 '23 at 22:19
  • @mikeqwertyuiop If $X$ is a metric space, you can use sequences. In general, for topological spaces sequences aren't enough (this is the case for locally compact Hausdorff spaces as well). Nets are completely analogous to sequences, but instead of on $\mathbb{N}$ they can be defined on any directed set (like $(0, 1)$), which is a special type of a pre-ordered set. For definition see Engelking's General topology. – Jakobian Nov 26 '23 at 22:20
  • So why using net like the post you suggested can show continuity on $\overline{{f+g >0}}$ ? – mikeqwertyuiop Nov 26 '23 at 22:22
  • @mikeqwertyuiop Because of the post I linked. If $X$ is any topological space, $Y$ is regular and $f:X\to Y$ is such that $f\restriction_D$ is continuous, where $D\subseteq X$ is dense, and $f(x_\alpha)\to f(x)$ for all nets $x_\alpha\in D$ converging to $x$ then $f$ is continuous. – Jakobian Nov 26 '23 at 22:24
  • Thank you, I will try to go through the proof and definitions – mikeqwertyuiop Nov 26 '23 at 22:25
  • Here regularity of $Y$ is important, in general you only have something like $f:X\to Y$ is continuous at $x$ iff for any net $x_\alpha \to x$ we have $f(x_\alpha)\to f(x)$. – Jakobian Nov 26 '23 at 22:26
  • It might be difficult for you to go through all definitions, I'd just accept the argument at face value for now. – Jakobian Nov 26 '23 at 22:28
  • Probably , actually I am justifying steps of Riesz Representation Theorem and this is like the last hurdle of the proof :( – mikeqwertyuiop Nov 26 '23 at 22:30
  • @mikeqwertyuiop I gave you another proof which is more elementary – Jakobian Nov 26 '23 at 22:48