$f:\mathbb R\rightarrow [0,\infty )$ is continuous such that $g(x)={(f(x))}^2$ is uniformly continuous . Then which of the following is always true $?$

$A.$ $f$ is bounded.

$B.$ $f$ may not be uniformly continuous.

$C.$ $f$ is uniformly continuous.

$D.$ $f$ is unbounded.

I ticked option $C.$ Following is my logic :

$$f:\mathbb R\rightarrow [0,\infty)$$ $$h:[0,\infty)\rightarrow [0,\infty) ; h(x)=x^2$$ Then $$g=h\circ f$$ Now the function $h$ is bijective(no $?$) and $$h^{-1}(x)=\sqrt x$$ So composing both sides with $h^{-1}$ we get $$f=h^{-1}\circ g$$

Also both $g$ and $h^{-1}$ being uniformly continuous , $f$ is uniformly continuous.

Was I correct $?$

Also , if my proof was correct I could really use some help to find counterexamples for the rest or say , why they cannot be true .

Thank you .


marked as duplicate by José Carlos Santos, Rolf Hoyer, Peter Taylor, loup blanc, Lee Mosher Nov 11 '17 at 14:54

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  • $\begingroup$ $h$ is not bijective, but that's not the big issue here. The function $x\mapsto\sqrt{x}$ is actually not uniformly continuous on any closed interval containing $0$. So, if $f(x) = \sqrt{|x|}$, then $g(x) = |x|$ is uniformly continuous, but $f$ is not. $\endgroup$ – Joey Zou Dec 18 '15 at 5:32
  • $\begingroup$ @JoeyZou : Sorry I might be confused here . The function $x\mapsto \sqrt x$ is continuous and any closed interval on $\mathbb R$ , whether contains $0$ or not , is compact. And continuous functions on compact sets are uniformly continuous right $?$ Also , this link math.stackexchange.com/questions/569928/… . And is not positive square root of a positive number unique $?$ $\endgroup$ – user118494 Dec 18 '15 at 5:48
  • $\begingroup$ I was thinking of Lipschitz continuity and not uniform continuity. My bad. Now that I've reread your question, I think everything is ok. $\endgroup$ – Joey Zou Dec 18 '15 at 6:06
  • $\begingroup$ @joey Zou why $h$ is not bijective ??? it's domain and range is $\mathbb{R}^+$ $\endgroup$ – RAM_3R Jul 27 '18 at 20:49

$f$ is bounded, as if there were a sequence $\{x_n\}$ with $$\lim_{n\to\infty}x_n= \lim_{n\to\infty}f(x_n)=\infty,$$ then we would have $\lim_{n\to\infty}f^2(x_n)=\infty$, which contradicts the assumption that $f^2$ converges uniformly on $\mathbb R$. Hence there exists $M$ so that $\sup_{x\in\mathbb R}|f(x)|<M$. Let $\varepsilon>0$. Then we may choose $\delta>0$ so that $|x-y|<\delta$ implies $$|f(x)-f(y)|<\frac\varepsilon{2M}\left|f^2(x)-f^2(y)\right|, $$ which yields $$|f(x)-f(y)| = \frac{\left|f^2(x)-f^2(y)\right|}{|f(x)+f(y)|}<\varepsilon. $$ It follows that $f$ is uniformly continuous.

  • $\begingroup$ does unboundedness of any continuous function on an unbounded interval means that the function is not uniformly continuous?.... you can consider $g(x)= log(x)$ defined on $[1,\infty)$... $\endgroup$ – DEEP Feb 12 '16 at 6:50

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