Proving $f(x)=\sqrt{x}$ is uniformly continuous.

Question

Where did I go wrong? The text I'm using gave a different proof. Help.

Attempt:

Given any $\epsilon>0$. Let $\delta=\epsilon$ so that if $\vert x-y\vert<\delta$, then \begin{align} \vert x-y\vert<\epsilon &\Leftrightarrow \frac{\vert x-y\vert}{\vert \sqrt{x}+\sqrt{y}\vert}<\vert x-y\vert<\epsilon\\ &\Leftrightarrow \frac{\vert x-y\vert}{\vert \sqrt{x}+\sqrt{y}\vert}<\epsilon\\ &\Leftrightarrow \vert \sqrt{x}-\sqrt{y}\vert<\epsilon,\quad\text{since}\quad \vert\sqrt{x}-\sqrt{y}\vert=\frac{\vert x-y\vert}{\vert \sqrt{x}+\sqrt{y}\vert}\end{align}

Answer 1

The main issue here, from my perspective, is that $|x - y| < \varepsilon$ need not imply $\frac{|x - y|}{|\sqrt{x} + \sqrt{y}|} < \varepsilon.$ For example, take $x = 0.01$, $y = 0$, and $\varepsilon = 0.02$.

Answer 2

Your first step should definitely not be to assume that $|\sqrt{x} - \sqrt{y}| < \epsilon$. This is what you are trying to prove!! The definition of uniform continuity:

A function $f:U\to \mathbb{R}$, $U\subseteq \mathbb{R}$, is uniformly continuous if for any $\epsilon > 0$ and any $x,y\in U$, there exists $\delta > 0$ such that whenever $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon$.

You're trying to show something about every $\epsilon > 0$. So you should start by taking an arbitrary $\epsilon > 0$ (and a pair $x,y\in U$) and showing that it has the desired property, in this case $|\sqrt{x} - \sqrt{y}| < \epsilon$. This is a fundamental proof method in real analysis and it will come up over and over, so you should start getting used to it now.

It's also not true that $\frac{|x-y|}{\sqrt{x} + \sqrt{y}} < |x-y|$ (consider $x = y = \frac{1}{4}$ -- in fact, whenever $\sqrt{x} + \sqrt{y} < 1$).

Answer 3

As mentioned below, your proof is wrong. But here's a easy solution that doesn't use many inequalities and just Weiestrass and MVT:

$\textbf{Solution}:$ Fix $\epsilon > 0$. We shall split up $\mathbb{R}^+$ to $[0,1]$, $[1,\infty)$ to make use of the powerful tool MVT. Indeed, by MVT for any $x,y \in [1,\infty)$, we have \begin{align}|f(x)-f(y)| = |x-y||f'(\xi)| \leq |x-y|/2 \end{align} for some $\xi \in [1,\infty)$, and for the second inequality we used the derivative of $f$ is bounded by 1/2 on $[1,\infty)$. Now let $\delta_{1} = 2 \epsilon$

Now for $x \in [0,1]$, we have by Heine Borel that $[0,1]$ is compact, and $f$ is continous over $[0,1]$, so by Weiestrass (a continuous function over a compact interval is uniformly continuous) that $f([0,1])$ is uniformly continous. Hence, there's some $\delta_{2}$ s.t. for any $|x-y| < \delta_{2}, |f(x)-f(y)| < \epsilon$ for $x,y \in [0,1]$.

Now let $\delta = \min(\delta_{1}, \delta_{2})$, and we have uniformly continuity on $\mathbb{R}^+$