An Application of Urysohn's Lemma?

Question

Let $X$ be a compact Hausdorff (hence, normal) space. Let $\epsilon>0$, $x_{0}\in X$, $U\ni x_{0}$ be open, and $f\colon X\to \mathbb{R}$ be continuous.

I am trying to find a continuous function $g\colon X\to\mathbb{R}$ such that:

i) $\|g-f\|_{\infty}<\epsilon$,

ii) $g|_{U^{c}}=f|_{U^{c}}$,

iii) $g|_{V}\equiv f(x_{0})$ in some neighbourhood $V$ of $x_{0}$?

Using Urysohn's lemma, I was able to come up with a $g$ that satisfied all but condition i): I wasn't able to guarantee that $\|g_{U\setminus V}-f_{U\setminus V}\|_{\infty}<\epsilon$.

Is there a way to show the existence of such a $g$, either by Urysohn or some similar variant?

Answer 1

Yes.

1) Choose an open neighborhood $W$ of $x_0$ such that $W \subset U$ and $\lvert f(x) - f(x_0) \rvert < \epsilon/2$ for $x \in W$.

2) Choose an open neighborhood $V$ of $x_0$ such that $\overline{V} \subset W$.

3) Choose a continuous $h : X \to [0,1]$ such that $h(x) = 0$ for $x \in \overline{V}$ and $h(x) = 1$ for $x \in X \setminus W$.

Define $$g(x) = f(x_0) + h(x)(f(x) - f(x_0)).$$ Then $g(x) = f(x)$ for $x \in X \setminus W \supset X \setminus U$ and $g(x) = f(x_0)$ for $x \in \overline{V} \supset V$. It remains to show that $\|g-f\|_{\infty}<\epsilon$, that is, $\sup_{x \in X} \lvert g(x) - f(x) \rvert = \sup_{x \in W} \lvert g(x) - f(x) \rvert < \epsilon$. But for $x \in W$ we have

$$\lvert g(x) - f(x) \rvert = \lvert (h(x) - 1)(f(x) - f(x_0)) \rvert = \lvert h(x) - 1 \rvert \cdot \lvert f(x) - f(x_0) \rvert$$ $$\le \lvert f(x) - f(x_0) \rvert < \epsilon/2 .$$

Answer 2

Let $h(x) = = mid(-\delta, g(x)-f(x), \delta)$ (i.e. the middle value) for some $0 < \delta < \epsilon$, and then replace $g$ by $f + h$ and $V$ by $f^{-1}((f(x_0)-\delta,f(x_o)+\delta)) \cap V$.