<s>[INST] Prove $f(x) = \sqrt x$ is uniformly continuous on $[0,\infty)$ I have seen a proof in $\sqrt x$ is uniformly continuous
Below shows an alternative proof. Please correct me if im wrong.
Proof:
For any given $\varepsilon >0$,
Let $\delta_1 = \frac{\varepsilon}{2}$, $\forall x,y \in [1,\infty)$ with $|x-y|<\delta_1$
Since $|\sqrt x + \sqrt y|\geq2$
$$|\sqrt x - \sqrt y|  = \frac{|x-y|}{|\sqrt x+\sqrt y|} < |x-y| < \delta_1 = \frac{\varepsilon}{2}$$
Hence, $\sqrt x$ is uniformly continuous on $[1,\infty)$.
$\sqrt x$ is continuous on [0,1] , so $\sqrt x$ is uniformly continuous on [0,1].
So, there exist $\delta_2 > 0$ such that $\forall x,y \in [0,1], |x-y|<\delta_2$, $|\sqrt x -\sqrt y| <\frac{\varepsilon}{2}$
Let $\delta = \min{(\delta_1,\delta_2)}$
$\forall x,y \in [0,\infty)$ with $|x-y|<\delta$,
Case 1: $x,y \in [0,1]$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 2: $x,y \in [1,\infty)$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 3: $x \in [0,1] , y \in [1,\infty]$
$$|\sqrt x-\sqrt y| = |\sqrt x -1+1-\sqrt y| \leq |\sqrt x-1| + |\sqrt y -1| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}= \varepsilon$$
by applying case 1 and case 2.
 [/INST] Let $0 < \epsilon < 1$ and $x,y \in [0,\infty)$ with $|x-y| < \epsilon^2$.
Consider $|\sqrt{x} - \sqrt{y}| = ||\sqrt{x}| - |\sqrt{y}|| \leq |\sqrt{x} + \sqrt{y}|$.
So, $|\sqrt{x} - \sqrt{y}|^2 = |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} + \sqrt{y}| = |x-y|$, that is, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}}$ which implies that $\sqrt{\centerdot}$ is Hölder-continuous.
Finally, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}} < \sqrt{\epsilon^2} = \epsilon$.
Q.E.D.
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