Is $\sqrt{x}$ uniformly continuous in $\mathbb{R}^+$?

Question

We are given this function: $f:R^+\rightarrow R,x\rightarrow \sqrt{x}$.

We need to prove that this function is uniformly continuous. My proof is this one but i'm not sure is it complete and right.

Let pick up $\epsilon>0$ and $\gamma>\epsilon^2$. We need to find $|f(x)-f(y)|<\epsilon$ for all $x,y\in R^+$ with $|x-y|<\gamma$

Without loss of generality we can assume that $x<y$.

We see that $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}-\sqrt{y}|\rightarrow|x-y|<\gamma$

$|\sqrt{x}-\sqrt{y}|^2\leq|\sqrt{x}-\sqrt{y}||\sqrt{x}-\sqrt{y}|=|x-y|<\epsilon^2\rightarrow||\sqrt{x}-\sqrt{y}|<\epsilon$

Since $|\sqrt{x}-\sqrt{y}|<\epsilon^2$ This show that $f(x)=\sqrt{x}$ is uniformly continuous on $f:\mathbb R^+\rightarrow \mathbb R$

Answer 1

It's very unclear what you are doing:

1. What is $\gamma$ and why did you set $\gamma > \epsilon^2$?
2. I assume that $\gamma$ is the $\delta$ from standard definitions. Even then, you have the line: $$|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}-\sqrt{y}|\to |x-y|<\gamma$$ which makes no sense. What you SHOULD do is start with the fact that $|x-y|<\gamma$ and finish with $|\sqrt x-\sqrt y|<\epsilon$, but that's not what I see here.
3. You say that $|\sqrt{x} - \sqrt{y}|^2 = |x-y|$, which is just plain wrong.

Answer 2

Use the inequality: $|x-y|\geq|x+y - 2\sqrt{xy}|$

Fix $\delta > |x-y|≥0$

$\Rightarrow$ $\delta > |x-y| = |x+y - 2y| = |x+y - 2\sqrt{yy}| ≥ |x+y - 2\sqrt{xy}| = |\sqrt{x} - \sqrt{y}|^2$

Hence, $|\sqrt{x} - \sqrt{y}| < \sqrt{\delta} = \epsilon$

Since, $\epsilon$ only depends on $\delta$, we have a bound independent of the values of x and y and only dependent on their distance. Hence, we have uniform continuity.