Show that the function $f(x) = \sqrt{x(x-1)}$ is uniformly continuous on $[1, \infty)$.

Attempt: \begin{align} \left|\sqrt{x(x-1)} - \sqrt{y(y-1)}\right| &= \left|\frac{x^2 - x - y^2 - y}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right| \\ &= \left(\frac{x + y + 1}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right)|x-y| \\ &< \left(\frac{1}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right)|x-y| \end{align}

From here, I have no idea how to get rid of the denominator so that I will only be left with $|x-y|$. At this point, I don't even know if my approach is correct. Please give me some hint on how I should proceed. Thanks in advance!

  • 1
    $\begingroup$ The denominator is sufficiently large when either of $x,y$ is large enough, so the term in the parenthesis is bounded. The rest can be contained in a closed interval so continuity implies uniform continuity $\endgroup$ Sep 5, 2021 at 6:34
  • 3
    $\begingroup$ It seems that you try to prove that the function is Lipschitz continuous on $[1, \infty)$ (which is isn't). $\endgroup$
    – Martin R
    Sep 5, 2021 at 6:45
  • $\begingroup$ It is Lipschitz on $[1+\epsilon, \infty)$ though! Arguably not useful here though. $\endgroup$
    – parsiad
    Sep 5, 2021 at 6:49
  • $\begingroup$ Why is the $<$ sign there? It should be $>$ following from $x+y+1>1$. Instead, split $[1,\infty)$ into $[1,M+1]$ and $[M,\infty)$ and work separately. $\endgroup$ Sep 5, 2021 at 7:10

3 Answers 3


First notice that $\lim_{x\to1^{+}}f(x)=0$ clearly. So fix $\varepsilon>0$; $\exists\delta'>0$ such that $0<x-1<\delta', x\in[1,\infty)$ implies $|f(x)|<\varepsilon$.

Then consider the intervals $[1, 1+\delta'+1]$ and $[1+\delta',\infty)$.

On $[1+\delta',\infty)$, by notice that the coefficient in the bracket of your attempt is bounded above (though your coefficient seems to be incorrect, after correcting it still you can check it's bounded, for example by fixing an arbitrary $y$ to check it's always bounded in $x$ by 1), we know easily that $f(x)$ is Lipschitz on this interval, thus uniformly continuous on it; hence, $\exists\delta_{1}>0$ such that $|x-y|<\delta_{1}$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in[1+\delta',\infty)$.

On compact $[1, 1+\delta'+1]$, since $f(x)$ is continuous clearly, we know it's uniformly continuous on this interval. So $\exists\delta_{2}>0$ such that $|x-y|<\delta_{2}$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in[1, 1+\delta'+1]$.

Finally consider $\delta=min\{\delta_{1}, \delta_{2}, 1\}>0$; then for any $|x-y|<\delta$, we know $x$ and $y$ must be either both in $[1+\delta',\infty)$ or both in $[1, 1+\delta'+1]$; so $|f(x)-f(y)|<\varepsilon$, which shows the uniform continuity as desired.



You can use or prove the following general result

if $f: D \to \mathbb R$ is a continuous map defined on $D \subseteq \mathbb R$ which is uniformly continuous on $D \setminus [a,b]$ where $[a,b]$ is a closed bounded interval, then $f$ is uniformly continuous on $D$.

The prove is fairly easy using Heine–Cantor theorem that you probably know.

You can apply that result to $f(x) = \sqrt{x(x-1)}$ with for example $a=1, b=2$ and follow on with the inequalities you mention in the question.

  • $\begingroup$ You need e.g. $D \cap [a, b]$ to be closed for the fact to work. $\endgroup$
    – Adayah
    Sep 5, 2021 at 9:00

You can use the following

Lemma: Let $f, g: [1,\infty) \to \mathbb R$ two continuous functions such that $\lim_{x \to \infty} \lvert f(x) - g(x) \rvert = 0$. If $g$ is uniformly continuous, then also $f$ is uniformly continuous.

Proof: Let $\epsilon > 0$. Choose $\delta' > 0$ such that $\lvert x - y \vert < \delta'$ implies $\lvert g(x) - g(y) \rvert < \epsilon/3$. W.l.o.g. we may assume that $\delta' \le 1$. Choose $R \ge 1$ such that $\lvert f(x) - g(x) \rvert < \epsilon/3$ for $x \ge R$. Since $f$ is uniformly continuous on the compact interval $[1,R+1]$, we find $\delta'' > 0$ such that $\lvert x - y \vert < \delta''$ implies $\lvert f(x) - f(y) \rvert < \epsilon$ whenever $x, y \in [1,R+1]$. Now let $\delta = \min (\delta',\delta'') \le 1$. Consider $x,y \in [1,\infty)$ such that $\lvert x - y \rvert < \delta$. If one of $x,y$ is contained in $[1,R]$, then both $x,y$ are contained in $[1,R+1]$ and therefore $\lvert f(x) - f(y) \rvert < \epsilon$. If both $x,y$ are contained in $[R,\infty)$, then $$\lvert f(x) - f(y) \rvert = \lvert f(x) - g(x) + g(x) - g(y) + g(y) - f(y) \rvert \\ \le \lvert f(x) - g(x) \rvert + \lvert g(x) - g(y) \rvert + \lvert f(y) - g(y) \rvert < \epsilon .$$

The uniform continuity of $f(x) = \sqrt{x(x-1)} = \sqrt{(x -\frac 1 2)^2 - \frac 1 4}$ follows from the Lemma by taking $g(x) = x - \frac 1 2$ which is clearly uniformly continuous. Note that with $z = x - \frac 1 2$ we have $\sqrt{z^2 - \frac 1 4} < z$ for all $z$ and $$z - \epsilon \le \sqrt{z^2 - \frac 1 4}$$ for $z \ge \frac{\epsilon^2 + \frac 1 4 }{2\epsilon}$.


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