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 a proof online that made $\delta = \epsilon^2$ which I don't understand how they got it? 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. $\endgroup$ Commented Nov 17, 2013 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
    Commented Nov 17, 2013 at 3:10
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    $\begingroup$ Why do you think it is uniformly continuous on $\mathbb R^{\geq 0}$? $\endgroup$ Commented Nov 17, 2013 at 3:10

8 Answers 8


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
    Commented May 2, 2015 at 17:16
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    $\begingroup$ Why does $\left|\sqrt{x}-\sqrt{y}\right|^2<\varepsilon^2$ imply $\left|\sqrt{x}-\sqrt{y}\right|<\varepsilon$? $\endgroup$
    – user5826
    Commented Apr 17, 2016 at 20:44
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    $\begingroup$ @AlJebr The square root function is strictly increasing and the quantities involved are non-negative. Therefore $\lvert \sqrt x - \sqrt y \rvert = \sqrt{\lvert \sqrt x - \sqrt y \rvert^2} < \sqrt{\epsilon^2} = \epsilon$. $\endgroup$ Commented Apr 18, 2016 at 1:59
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    $\begingroup$ $\newcommand{\abs}[1]{\left\vert #1 \right\vert}$I'm struggling to justify the step $\abs{\sqrt{x} - \sqrt{y}}^2 \le \abs{\sqrt{x} - \sqrt{y}}\abs{\sqrt{x} + \sqrt{y}}$. Is it a general truth of expressions of the form $\abs{a - b}$? $\endgroup$
    – Anakhand
    Commented Jan 16, 2019 at 11:31
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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 $|\sqrt x - \sqrt y|^2$. $\endgroup$ Commented Jan 16, 2019 at 12:12

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.

  • $\begingroup$ Thanks but I wanted to know how to do it the way I mentioned it above. $\endgroup$
    – user104235
    Commented Nov 17, 2013 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
    Commented Nov 17, 2013 at 3:33
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    $\begingroup$ @Raghav what, if $x\in [0,1]$ and $y\in [1,∞)$ ? Is your value of delta works for this case too? $\endgroup$ Commented Nov 20, 2020 at 12:29
  • $\begingroup$ @Akash Patalwanshi Good point, I've always wondered I wonder the same $\endgroup$ Commented Aug 15, 2021 at 14:50
  • $\begingroup$ @AkashPatalwanshi Fix $\varepsilon>0$. There are $\delta_1, \delta_2>0$ such that 1. for all $x,y\in [0,1]$, $|x-y|<\delta_1$ implies $|f(x)-f(y)| <\varepsilon/2$ and 2. for all $x,y\in [1, +\infty)$, $|x-y|<\delta_2$ implies $|f(x)-f(y)| <\varepsilon/2$. Let $\delta := \min \{\delta_1, \delta_2\}$ and $x,y\in [0, +\infty)$ such that $|x-y| < \delta$. [...] $\endgroup$
    – Analyst
    Commented Apr 27, 2022 at 9:47

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 $\vert x-y\vert<\epsilon^2$.


We let $ \epsilon > 0 $

We have that $ | x - y | < \delta $ .

We choose that $ \delta = \epsilon^2 $

We then note the following result.

$ | f(x) - f(y) | = | \sqrt{x} - \sqrt{y} | = \sqrt{ | \sqrt{x} - \sqrt{y}|^2} \leq \sqrt{ | \sqrt{x} - \sqrt{y}| | \sqrt{x} + \sqrt{y} |} = \sqrt{ | x - y | } < \sqrt{\delta} = \epsilon$

$ \square $


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 $[0,\infty)$.

  • $\begingroup$ Hi, can you please explain why f is uniformly continuous on [0, 1]? $\endgroup$ Commented Dec 20, 2020 at 7:08
  • $\begingroup$ @user1742188 It follows from Heine-Cantor Theorem, that a continuous function over a compact set (In the case of $\mathbb{R}$, compact sets are closed and bounded) is uniformly continuous. $\endgroup$
    – Newton
    Commented Mar 22, 2023 at 19:35

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 $\delta = \varepsilon^2$? $\endgroup$
    – user619755
    Commented Apr 10, 2019 at 9:41

Although many good answers are already here, let me write one approach.

Let $\epsilon>0$ and $\delta:=\epsilon^2.$

Suppose $\mathbb R\ni x,y\geqq 0$ and $|x-y|<\delta.$ We can assume $x\geqq y.$

Let $t:=x-y\geqq 0.$ Then, we have $x=y+t$, and from $|x-y|<\delta$, we get $0\leqq t<\delta.$ Thus, \begin{align*} |\sqrt x-\sqrt y| &=\sqrt x-\sqrt y\\ &=\sqrt{y+t}-\sqrt y\\ &<\sqrt{y+\delta}-\sqrt y\\ &\leqq \sqrt y+\sqrt\delta -\sqrt y\\ &=\sqrt\delta\\ &=\sqrt\epsilon^2\\ &=\epsilon. \end{align*}


Here is justification of $\sqrt{y+\delta}\leqq\sqrt y+\sqrt\delta.$

\begin{align} \sqrt{y+\delta}^2 &=y+\delta\\ &\leqq y+2\sqrt{y}\sqrt \delta+\delta\\ &=(\sqrt y+\sqrt\delta)^2. \end{align} Thus $\sqrt{y+\delta}^2\leqq(\sqrt y+\sqrt\delta)^2$, and since $\sqrt{y+\delta},\sqrt y+\sqrt\delta\geqq 0$, we get $\sqrt{y+\delta}\leqq\sqrt y+\sqrt\delta.$

  • $\begingroup$ Can you justify $ \sqrt{y+\delta} \leqq \sqrt y +\sqrt\delta \;$? $\endgroup$
    – 311411
    Commented Apr 10, 2023 at 14:14
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    $\begingroup$ Yes, I have added the explanation. @311411 $\endgroup$
    – So_M.
    Commented Apr 10, 2023 at 22:35

Here is my attempt of proving this, as a first year student of CompSci.

Since $[0,1]$ is a compact set, accoring to the Heine-Cantor theorem $f(x)={\sqrt x}$ is uniformly continuous on $[0,1]$. Moving along to $[1,+{\infty})$, we need to prove that $\forall \epsilon>0$ $\exists \delta = \delta(\epsilon)$ such that $|f(x_1)-f(x_2)|<\epsilon$ once $|x_1-x_2|<\delta$.

Now, \begin{align*} |\sqrt x_1-\sqrt x_2|<\epsilon &\implies\left|\frac{x_1+x_2}{\sqrt x_1+\sqrt x_2}\right|\lt\epsilon\\ \end{align*} Since we are observing the interval $[1,+\infty)$, it holds that $\sqrt x_1+\sqrt x_2\ge2$. Furthermore, we have that

\begin{align*} \left|\frac{x_1+x_2}{\sqrt x_1+\sqrt x_2}\right| &\le\left|\frac{x_1+x_2}2\right|\\ &\le\frac{|x_1+x_2|}2<\epsilon \end{align*}

Finally, we have that $|x_1+x_2|<2\epsilon$ which means that the $\delta$ we are looking for is $\delta=2\epsilon.$

To wrap up, we have proved that $\forall \epsilon>0$ $\exists \delta = 2\epsilon$ such that $|\sqrt{x_1}-\sqrt{x_2}|<\epsilon$ once $|x_1-x_2|<2\epsilon$. Since $f(x)$ is uniformly continous both on the compact set $[0,1]$ and the interval $[1,+\infty]$, $f(x)$ is uniformly continous on $\mathbb{R}_+$. $\blacksquare$


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