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Let $f$ be a continuous function from $\mathbb{R}$ to $[0,\infty)$ and $g(x)= f(x)^2$ is uniformly continuous. Then I want to prove that $f$ is uniformly continuous. I would like to mention one thing that is $f(x)^2=f(x)f(x)$, not the composition $f\circ f$.

My approach: Since $f(x)\geq 0$ we have $f(x)=\sqrt{g(x)}$. Now for any arbitrary $x,y\in \mathbb{R}$ we have $f(x)+f(y)=m$ and therefore $f(x)-f(y)= \frac{g(x)-g(y)}{m)}$ this gives that $f$ is uniformly continuous.

MANI
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    Where do we appeal to uniform continuity of $f^2$? – AlvinL Oct 08 '22 at 15:34
  • Composition of uniformly continuous functions is uniformly continuous, how this is giving answer of my question? – MANI Oct 08 '22 at 15:43
  • @AnneBauval if f And g are uniformly continuous then their composition is also. But we can say anything conversly and in my problem – MANI Oct 08 '22 at 15:51
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    $f=\sqrt~\circ f^2$. – Anne Bauval Oct 08 '22 at 15:53
  • @AnneBauval Is $\sqrt$ uniformly continuous? How do we claim this? Please let me know whether my approach is correct or not? – MANI Oct 08 '22 at 15:56
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    What happens when you try to prove that $\sqrt{;}$ is uniformly continuous? This is a special case of the question. And, according to Anne's comments, will imply the whole thing. – GEdgar Oct 08 '22 at 16:03
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    Here is a complete duplicate https://math.stackexchange.com/questions/1576885 Your approach was not correct (see first comment by Alvin + there is no reason why $f(x)-f(y)= \frac{g(x)-g(y)}{f(x)+f(y)}$ would "give that $f$ is uniformly continuous" + $m$ could even be $0$ for some $(x,y)$'s). – Anne Bauval Oct 08 '22 at 16:03
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    @AnneBauval thanks, fixed (it was within 5 minutes). – GEdgar Oct 08 '22 at 16:06
  • @AnneBauval thanks I got the answer. This is a duplicate question. Please let me know what should I do in such scenario. – MANI Oct 08 '22 at 16:10
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    You can delete your question if you want (by clicking on "Delete" on the line "Share Cite Edit Following Close 1 Flag" below your post). – Anne Bauval Oct 08 '22 at 16:13

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