A. $f$ is bounded

B. $f$ may not be uniformly continuous

C. $f$ is uniformly continuous

D. $f$ is unbounded.

Let $f(x)=\sqrt x$. Then $g(x)=x$ is uniformly continuous and unbounded. Hence option A can not be true.

Let $f(x)=1$. Then $g(x)=1$ is uniformly continuous and bounded. Hence the option D can not be true.

How should I check uniform continuity?


Since $g$ is uniformly continuous, hence for a given $\epsilon \gt 0$, $\exists \delta \gt 0$ depending only upon $\epsilon$ such that $|g(x)-g(y)| \lt \epsilon$ whenever $|x-y| \lt \delta$ $\forall x,y$.

$\Rightarrow |\sqrt g(x) -\sqrt g(y)||\sqrt g(x)+\sqrt g(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.

$\Rightarrow |f(x)-f(y)||f(x)+f(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.

What should be my next step now? How can I use the continuity of $f$?

  • $\begingroup$ Which one according to you is correct? $\endgroup$ – Learnmore Dec 16 '15 at 14:47
  • $\begingroup$ @learnmore I am proving that $f$ is uniformly continuous. If $f$ is not uniformly continuous,then I must get stuck somewhere or have an unusual situation. $\endgroup$ – Error 404 Dec 16 '15 at 15:18

Hint. You want to prove that $f(x) = \sqrt{g(x)}$ is uniformly continuous. The statement of this looks like "$|f(x) - f(y)| \leq \epsilon$ whenever...(condition on $x$ and $y$ being close enough)." Since $g$ is known to be uniformly continuous, you can always write "$|g(x) - g(y)| \leq \alpha$ whenever..."

So it will be enough for you to find some function $h \colon (0,+\infty) \to (0,+\infty)$ with the property that whenever $|g(x) - g(y)| \leq \alpha = h(\epsilon)$, you necessarily have $|f(x) - f(y)|\leq \epsilon$. Since the only thing you really know is that $f(x)$ is the square root of $g(x)$, this clarifies what conditions you would want the function $h$ to satisfy. Now your job is to find a function $h$ that works.

Essentially, the proof amounts to showing that the square root function is uniformly continuous.

Edit. Do you have a theorem that says a composite of uniformly continuous functions is uniformly continuous? If so, all you need to do is prove that $j(t) = \sqrt{t}$ is uniformly continuous, since $f(x) = j(g(x))$. (If not, you will need to include a small additional step that amounts to proving this is the special case you're dealing with.)

To prove the uniform continuity of $j$, it will be enough to prove that, for $a, b \geq 0$, $$|a - b| \leq \epsilon^2 \Longrightarrow |\sqrt{a} - \sqrt{b}| \leq \epsilon.$$

An alternative is to show that $j$ has a Lipschitz constant on $[1,+\infty)$ and to note that $[0,1]$ is a closed bounded interval, hence $j$ must be uniformly continuous there.

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  • $\begingroup$ I have edited my work above. But still getting nowhere. :( $\endgroup$ – Error 404 Dec 16 '15 at 7:48
  • $\begingroup$ Ah! Great trick using square root function!! $\endgroup$ – Error 404 Dec 17 '15 at 11:51

in your second example (where $f(x)=1$), $f$ is a continuous function and $g(x)= 1$ is uniform continuous. But note that, $f(x)$ is also uniform continuous ! Hence option b) is also discarded! Therefore, option c) MUST be right :p

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