# Show that sqrt(x) is uniformly continuous on [0,infinity)

• Mar 24th 2010, 05:48 PM
tn11631
Show that sqrt(x) is uniformly continuous on [0,infinity)
Yikes! I feel like were all having trouble with uniform continuity...

Use the definition of uniform continuity to show that $\displaystyle \sqrt{x}$ is uniformly continuous on [0,$\displaystyle \infty$) (relative to [0,$\displaystyle \infty$))

For this, do I have to show that $\displaystyle \sqrt{x}$ is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1,$\displaystyle \infty$) before I can show its uniformly continuous on [0,$\displaystyle \infty$)?
• Mar 25th 2010, 01:38 AM
Opalg
Quote:

Originally Posted by tn11631
Yikes! I feel like were all having trouble with uniform continuity...

Use the definition of uniform continuity to show that $\displaystyle \sqrt{x}$ is uniformly continuous on [0,$\displaystyle \infty$) (relative to [0,$\displaystyle \infty$))

For this, do I have to show that $\displaystyle \sqrt{x}$ is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1,$\displaystyle \infty$) before I can show its uniformly continuous on [0,$\displaystyle \infty$)?

That is certainly the most convenient way to do it. The usual conditions ensuring that a function is uniformly continuous are either (a) it is continuous on a bounded closed interval, or (b) it is differentiable with a bounded derivative.

The function $\displaystyle \sqrt{x}$ on the interval $\displaystyle [0,\infty)$ does not satisfy either of those conditions. But it satisfies (a) and (b) on the overlapping subintervals $\displaystyle [0,2]$ and $\displaystyle [1,\infty)$ respectively. From that, you can deduce that it is uniformly continuous on the whole interval.
• Mar 25th 2010, 05:46 AM
Pinkk
Lemma: $\displaystyle |\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$ for any $\displaystyle x,y \ge 0$.

Proof:

Let $\displaystyle \epsilon > 0$ and choose $\displaystyle \delta = \epsilon^{2}$. Then, for $\displaystyle x,y \in [0,\infty )$, $\displaystyle |x-y|<\delta$ implies that $\displaystyle |\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|} < \sqrt{\epsilon^{2}} = \epsilon$. Thus, $\displaystyle f(x) = \sqrt{x}$ is uniformly continuous on $\displaystyle [0, \infty )$. Q.E.D.