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Prove that the function $\sqrt x$ is uniformly continuous on $\{x\in \mathbb{R} | x \ge 0\}$.

To show uniformly continuity I must show for a given $\epsilon > 0$ there exists a $\delta>0$ such that for all $x_1, x_2 \in \mathbb{R}$ we have $|x_1 - x_2| < \delta$ implies that $|f(x_1) - f(x_2)|< \epsilon.$

What I did was $\left|\sqrt x - \sqrt x_0\right| = \left|\frac{(\sqrt x - \sqrt x_0)(\sqrt x + \sqrt x_0)}{(\sqrt x + \sqrt x_0)}\right| = \left|\frac{x - x_0}{\sqrt x + \sqrt x_0}\right| < \frac{\delta}{\sqrt x + \sqrt x_0}$

but I found some proof online that made $\delta = \epsilon^2$ where I don't understand how they got? So, in order for $\delta =\epsilon^2$ then $\sqrt x + \sqrt x_0$ must $\le$ $\epsilon$ then $\frac{\delta}{\sqrt x + \sqrt x_0} \le \frac{\delta}{\epsilon} = \epsilon$. But then why would $\epsilon \le \sqrt x + \sqrt x_0? $ Ah, I think I understand it now just by typing this out and from an earlier hint by Michael Hardy here.

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    $\begingroup$ It's not defined on $\mathbb R$, which is one of the first conditions for continuity. $\endgroup$ – Thomas Andrews Nov 17 '13 at 3:07
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    $\begingroup$ What you get for delta really depends on how you approach the problem. It can be different and not really necessary to reconcile. $\endgroup$ – Chris C Nov 17 '13 at 3:10
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    $\begingroup$ Why do you think it is uniformly continuous on $\mathbb R^{\geq 0}$? $\endgroup$ – Thomas Andrews Nov 17 '13 at 3:10
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Let $\epsilon > 0.$ Pick $\delta = \epsilon^2.$ Then for $|x-y| < \delta$ we have

$$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y| = |x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon. $$

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    $\begingroup$ From where the motivation of $\delta = \epsilon^2$ came from? $\endgroup$ – ramanujan May 2 '15 at 17:16
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    $\begingroup$ @ramanujan Intuitively, my guess is, since the function involves a square root, maybe that's why the idea of $\varepsilon^2$ came.... $\endgroup$ – Diya May 10 '15 at 17:36
  • $\begingroup$ Why does $\left|\sqrt{x}-\sqrt{y}\right|^2<\varepsilon^2$ imply $\left|\sqrt{x}-\sqrt{y}\right|<\varepsilon$? $\endgroup$ – Al Jebr Apr 17 '16 at 20:44
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    $\begingroup$ @AlJebr The square root function is strictly increasing and the quantities involved are non-negative. Therefore $\lvert \sqrt x - \sqrt y \rvert = \sqrt{\lvert \sqrt x - \sqrt y \rvert^2} < \sqrt{\epsilon^2} = \epsilon$. $\endgroup$ – Alex Provost Apr 18 '16 at 1:59
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    $\begingroup$ @Anakhand By the triangle inequality, $|a-b| \leq |a| + |b| = a+b = |a+b|$ whenever $a,b$ are non-negative. Apply this to one of the factors $|\sqrt x - \sqrt y|$ in $|\sqrt x - \sqrt y|^2$. $\endgroup$ – Alex Provost Jan 16 at 12:12
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We'll prove that $f(x) = \sqrt{x}$ is uniformly continuous on $\mathbb{R}_+$. Indeed, $[0,1]$ being a compact set, $f$ is uniformly continuous on this interval. On the other hand, on $[1,\infty), f$ is Lipschitz, and hence is uniformly continuous. Hence we are now done.

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  • $\begingroup$ Thanks but I wanted to know how to do it the way I mentioned it above. $\endgroup$ – user104235 Nov 17 '13 at 3:28
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    $\begingroup$ Sure, you can detail the above answer using $\epsilon-\delta.$ Indeed, by uniform continuity on $[0,1],$ you'll get a $\delta_1.$ On $[1,\infty),$ you'll get a $\delta_2 = \epsilon/M$ say, where $M$ is the Lipschitz constant. Then the minimum of the two deltas works. $\endgroup$ – Raghav Nov 17 '13 at 3:33
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Let's try a more general approach using the mean value theorem. Let $f=x^\alpha$ and suppose $x<y$. Since $y^{\alpha}-x^\alpha=(y-x) \alpha c^{\alpha-1}$ for $x<c<y$. So for $0< \alpha<1$ we thus have

$$y^\alpha-x^{\alpha}\le (y-x)\alpha y^{\alpha-1}\le \alpha(y-x)$$

for $y\ge 1$, which shows that $f$ is uniformly continuous on $[1, \infty)$ and clearly is uniformly continuous on $[0,1]$. Thus, if $0<\alpha<1$, $f$ is uniformly continuous on $[0,\infty)$.

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The explanation is from Jonathan Kane's textbook "Writing Proofs in Analysis". Which asks the reader to observe the behavior of the function $f(x,y)=\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}=\vert\sqrt{x}-\sqrt{y}\vert$. The natural step is to restrict the "size" of $\vert x-y\vert$ so that as $x,y\to\infty$ then so does $\sqrt{x}+\sqrt{y}$, which leads to $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}\to 0$. But a seeming roadblock arises as $x,y\to 0$ since that would make the denominator approach $0$ as well. However, the problem disappears when we realize that if $\sqrt{x}+\sqrt{y}\to 0$ then $\vert \sqrt{x}-\sqrt{y}\vert\to 0$ or if $\sqrt{x}+\sqrt{y}\to \infty$ then $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}=\vert\sqrt{x}-\sqrt{y}\vert\to 0$. Hence, if the given $\epsilon>0$ is such that $\sqrt{x}+\sqrt{y}<\epsilon$, then $\vert\sqrt{x}-\sqrt{y}\vert<\epsilon$ and we are done. On the other hand, if $\sqrt{x}+\sqrt{y}\geq\epsilon$, then $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}<\frac{\vert x-y\vert}{\epsilon}$ and we only need to compute for $ \frac{\vert x-y\vert}{\epsilon}<\epsilon$ to get $\vert x-y\vert<\epsilon^2$.

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In $[0,1]$ we want $$ \begin{split} |f(x)-f(y)|=&\frac {|x-y|}{\sqrt{x}+\sqrt{y}} <\varepsilon\\ \Updownarrow\\ |x-y|<&\varepsilon(\sqrt{x}+\sqrt{y})<2\varepsilon\:\:\text{ so }\:\:\delta=2\varepsilon \end{split} $$ In $[1, \infty]$, $$ \frac {|x-y|}{\sqrt{x}+\sqrt{y}} < |x-y|\:\:\text{ so }\:\:\delta=2\varepsilon $$

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  • $\begingroup$ Sorry to comment on an older post but in the first case, instead of $|x-y|<\varepsilon(\sqrt{x}+\sqrt{y})<2\varepsilon$ wouldn't it make more sense to say $|x-y|<\varepsilon(\sqrt{x}+\sqrt{y})\leq\varepsilon^2$ and then have $\delta = \varepsilon^2$? $\endgroup$ – user639631 Apr 10 at 9:41

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