A uniform continuity problem

Question

Let $A$ be a set of real numbers and $f:A \to \mathbb R$ be a function such that for every $\epsilon >0$ , there exist a uniformly continuous function $g_\epsilon :A \to \mathbb R$ such that $|f(x)-g_\epsilon(x)|<\epsilon , \forall x \in A$ , then how to prove that $f$ is uniformly continuous on $A$ ?

Answer 1

Let $\varepsilon>0$. Then, as $g_{\varepsilon/3}$ is uniformly continuous, there exists a $\delta>0$, such that $\lvert x-y\rvert<\delta$, implies that $\lvert g_{\varepsilon/3}(x)-g_{\varepsilon/3}(y)\rvert<\varepsilon/3$. Then, whenever $\lvert x-y\rvert<\delta$, we have $$ \lvert f(x)-f(y)\rvert\le \lvert f(x)-g_{\varepsilon/3}(x)\rvert+\lvert g_{\varepsilon/3}(x)-g_{\varepsilon/3}(y)\rvert+\lvert g_{\varepsilon/3}(y)-f(y)\rvert<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon, $$ which proves that $f$ is uniformly continuous.

Answer 2

Let $\tilde \epsilon > 0$, then there is some $g_{\epsilon}$ (take $\epsilon = \tilde \epsilon/3$) such that $$ | f(z)-g_{\epsilon}(z)| < \frac{\tilde \epsilon}{3}$$ for every $z \in A$. Now $g_\epsilon$ is unformly continuous and so there are is some $\delta > 0$ such that $x,y \in A, |x-y|< \delta $ implies $$ |g_\epsilon(x)- g_\epsilon (y)| < \frac{\tilde \epsilon}{3},$$ it follows that $x,y \in A, |x-y|< \delta $ implies $$\begin{array}{rcl}|f(x)-f(y)| &=& |f(x)-g_\epsilon(x)+ g_\epsilon (x)+g_\epsilon(y)- g_\epsilon (y)-f(y)| \\ &\leq & \underbrace{|f(x)-g_\epsilon(x)|}_{< \tilde\epsilon/3} + \underbrace{|g_\epsilon (x)-g_\epsilon(y)|}_{< \tilde\epsilon/3} + \underbrace{| g_\epsilon (y)-f(y)|}_{< \tilde\epsilon/3} < \tilde \epsilon \end{array}$$ and thus $f$ is uniformly continuous.

Answer 3

Let $\epsilon>0$, and take $G:=g_{\epsilon/3}$. Since $G$ is uniformly continuous, there is $\delta>0$ such that for every $a,b\in A$ with $|a-b|<\delta$, $|G(a)-G(b)|<\frac{\epsilon}{3}$. thus for every such $a,b$:

$$|f(a)-f(b)|\leq|f(a)-g(a)|+|g(a)-g(b)|+|g(b)-f(b)|\leq3\cdot\frac{\epsilon}{3}=\epsilon.$$