Prove that $f$ is uniformly continuous on $(-1,3)$ using the definition of UC.

Question

$$f(x)= \left\{ \begin{array}{ll} \frac{2-\sqrt{4-x}}{x} & x\neq 0 \\ \frac{1}{4} & x=0 \end{array} \right.$$ Either $x\neq 0$ and $y\neq0$, $x=0$ and $y\neq 0$, $x\neq 0$ and $y=0$, or $x=0$ and $y=0$ (I have already proved cases 2, 3, and 4). If $x\neq 0$ and $y\neq 0$, then $\forall \ \varepsilon>0$ pick $\displaystyle\delta=\min\{1,\text{Something else with } \varepsilon\}$ so that if $x,y\in(-1,3)\backslash\{0\}$ and $|x-y|<\delta$ then $$|f(x)-f(y)|=\left| \frac{2-\sqrt{4-x}}{x}-\frac{2-\sqrt{4-y}}{y}\right|=\left| \frac{2y-y\sqrt{4-x}-2x+x\sqrt{4-y}}{xy}\right|$$ since $x,y\in(-1,3)$, we have $-x\leq 1$ and $-y\leq 1$ $$\leq \left| \frac{2(y-x)+\sqrt{4+1}(x-y)}{xy}\right|$$ since $\delta\leq 1$ we have $x-1<y$ and $y-1<x$...

It was at this point when I had set my delta as $\displaystyle\delta=\min\left\{1,\frac{9\varepsilon}{2+\sqrt{5}}\right\}$, that I realized I was maximizing my denominator making the function smaller rather than larger. I don't see a way to continue down this route as I can pick $x$ and $y$ to be arbitrarially small. What did I miss here? I have since been playing around with rationalizing my denominator and adding possibly adding in $+1/4-1/4$ to try and simplify, but I can't quite get it.

Edit 1: I don't think this can be proven in the direction I took because I basically have $$\left|\frac{x-y}{xy}\right|=\left|\frac{1}{x}-\frac{1}{y}\right|$$ and this is the same proof for $f=1/x$ which is not uniformly continuous on $(0,1)$.

Answer 1

We have that $$\frac{2-\sqrt{4-x}}{x}=\frac{\left(2-\sqrt{4-x}\right)\left(2+\sqrt{4-x}\right)}{x\left(2+\sqrt{4-x}\right)}=\frac{1}{2+\sqrt{4-x}} $$ and $$\begin{align} \left|f(x)-f(y)\right|&=\left|\frac{1}{2+\sqrt{4-x}}-\frac{1}{2+\sqrt{4-y}}\right| \\ &= \left|\frac{\sqrt{4-y}-\sqrt{4-x}}{\left(2+\sqrt{4-x}\right)\left(2+\sqrt{4-y}\right)}\right| \\ &= \left|\frac{x-y}{\left(2+\sqrt{4-x}\right)\left(2+\sqrt{4-y}\right)\left(\sqrt{4-x}+\sqrt{4-y}\right)}\right| \\ &= \frac{|x-y|}{\left(2+\sqrt{4-x}\right)\left(2+\sqrt{4-y}\right)\left(\sqrt{4-x}+\sqrt{4-y}\right)} \end{align} $$ Notice that each of the square roots are $>1$. Hence, $$\left(2+\sqrt{4-x}\right)\left(2+\sqrt{4-y}\right)\left(\sqrt{4-x}+\sqrt{4-y}\right)>3\cdot 3\cdot 2=18 \\ \Rightarrow \frac{|x-y|}{\left(2+\sqrt{4-x}\right)\left(2+\sqrt{4-y}\right)\left(\sqrt{4-x}+\sqrt{4-y}\right)}\leq \frac{|x-y|}{18} $$ So, choosing $\delta_1=\varepsilon$ works. The final choice of $\delta$ should be $\min(\delta_1,\delta_2,\delta_3)$ where the different deltas are for each of the three cases (for the 4th one, any choice works).

Answer 2

There's a quicker way using the fact that a function that is continuous on a compact set $M$ is also uniformly continuous on $M$ (this is called Heine-Cantor theorem).

Check that $f$ is continuous on $[-1,3]$ by checking that $\lim_{x\to 0}f(x)=1/4$ (continuity at the other points is obvious). Thus, it's also uniformly continuous on $[-1,3]$. A function that is uniformly continuous on a set $A$ is uniformly continuous on any subset of $A$. The claim then follows.