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

Question

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.

Answer 1

Let $\epsilon > 0.$ Pick $\delta = \epsilon^2.$ Then for $|x-y| < \delta$ we have

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

Answer 2

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.

Answer 3

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

Answer 4

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 $

Answer 5

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 $[0,\infty)$.

Answer 6

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 $[1, \infty]$, $$ \frac {|x-y|}{\sqrt{x}+\sqrt{y}} < |x-y|\:\:\text{ so }\:\:\delta=2\varepsilon $$

Answer 7

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*}

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Edit

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 $\sqrt{y+\delta}\leqq\sqrt y+\sqrt\delta.$

Answer 8

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 $\mathbb{R}_+$. $\blacksquare$