If $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous












1












$begingroup$


I'm looking for help with a proof that I still cannot figure it out. Here is the statement:



"If $f:mathbb{R} rightarrow mathbb{R}$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous."



I know that if $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$, $f$ cannot be Lipschitz. So I tried to prove that if $f$ is uniformly continuous $Rightarrow$ $f$ is Lipschitz, because if it were true, if $f$ is NOT Lipschitz would mean $f$ is NOT uniformly continuous, and the proof would be resolved.



However, uniform continuity does not imply Lipschitz.






analysis uniform-continuity lipschitz-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
    $endgroup$
    – D. Thomine
    Dec 23 '18 at 19:33






  • 1




    $begingroup$
    Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:39












  • $begingroup$
    Thank you you both for your asnwers!
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:49
















1












$begingroup$


I'm looking for help with a proof that I still cannot figure it out. Here is the statement:



"If $f:mathbb{R} rightarrow mathbb{R}$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous."



I know that if $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$, $f$ cannot be Lipschitz. So I tried to prove that if $f$ is uniformly continuous $Rightarrow$ $f$ is Lipschitz, because if it were true, if $f$ is NOT Lipschitz would mean $f$ is NOT uniformly continuous, and the proof would be resolved.



However, uniform continuity does not imply Lipschitz.






analysis uniform-continuity lipschitz-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
    $endgroup$
    – D. Thomine
    Dec 23 '18 at 19:33






  • 1




    $begingroup$
    Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:39












  • $begingroup$
    Thank you you both for your asnwers!
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:49














1












1








1





$begingroup$


I'm looking for help with a proof that I still cannot figure it out. Here is the statement:



"If $f:mathbb{R} rightarrow mathbb{R}$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous."



I know that if $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$, $f$ cannot be Lipschitz. So I tried to prove that if $f$ is uniformly continuous $Rightarrow$ $f$ is Lipschitz, because if it were true, if $f$ is NOT Lipschitz would mean $f$ is NOT uniformly continuous, and the proof would be resolved.



However, uniform continuity does not imply Lipschitz.






analysis uniform-continuity lipschitz-functions







share|cite|improve this question











$endgroup$




I'm looking for help with a proof that I still cannot figure it out. Here is the statement:



"If $f:mathbb{R} rightarrow mathbb{R}$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous."



I know that if $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$, $f$ cannot be Lipschitz. So I tried to prove that if $f$ is uniformly continuous $Rightarrow$ $f$ is Lipschitz, because if it were true, if $f$ is NOT Lipschitz would mean $f$ is NOT uniformly continuous, and the proof would be resolved.



However, uniform continuity does not imply Lipschitz.






analysis uniform-continuity lipschitz-functions




analysis uniform-continuity lipschitz-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 19:25









Bernard

124k742117




124k742117










asked Dec 23 '18 at 19:23









HallowHallow

84




84








  • 1




    $begingroup$
    Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
    $endgroup$
    – D. Thomine
    Dec 23 '18 at 19:33






  • 1




    $begingroup$
    Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:39












  • $begingroup$
    Thank you you both for your asnwers!
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:49














  • 1




    $begingroup$
    Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
    $endgroup$
    – D. Thomine
    Dec 23 '18 at 19:33






  • 1




    $begingroup$
    Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:39












  • $begingroup$
    Thank you you both for your asnwers!
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:49








1




1




$begingroup$
Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
$endgroup$
– D. Thomine
Dec 23 '18 at 19:33




$begingroup$
Wait, that statement sounds very false (take a function with oscillations which get steeper and steeper, but smaller and smaller).
$endgroup$
– D. Thomine
Dec 23 '18 at 19:33




1




1




$begingroup$
Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:39






$begingroup$
Perhaps the author intended to assume $lim_{xtoinfty} |f'(x)|=infty$ instead? If the derivative is unbounded on a compact set instead, obviously the result will fail, as @D.Thomine suggested.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:39














$begingroup$
Thank you you both for your asnwers!
$endgroup$
– Hallow
Dec 23 '18 at 20:49




$begingroup$
Thank you you both for your asnwers!
$endgroup$
– Hallow
Dec 23 '18 at 20:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

The statement is not true. Consider the function
$$f(x)=frac{sin(x^5)}{x} ,.$$
with $f(0)=0$. Then, it is easy to see that $f in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.



But
$$f'(x)=frac{5x^5cos(x^5)-sin(x^5)}{x^2}$$is clearly unbounded as $$f'(sqrt[5]{2npi})=5cdot (2npi)^frac{3}{5}$$



Note If $lim_{x to infty} |f'(x)|=infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $delta$ such that $|x-y|< delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+frac{delta}{2})$ and take the limit as $z to infty$ to get a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think $f(x)=x^2$ is the simplest example..
    $endgroup$
    – Empty
    Dec 23 '18 at 20:34










  • $begingroup$
    Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:45










  • $begingroup$
    I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:47








  • 1




    $begingroup$
    @Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 20:56










  • $begingroup$
    @Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:03












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

The statement is not true. Consider the function
$$f(x)=frac{sin(x^5)}{x} ,.$$
with $f(0)=0$. Then, it is easy to see that $f in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.



But
$$f'(x)=frac{5x^5cos(x^5)-sin(x^5)}{x^2}$$is clearly unbounded as $$f'(sqrt[5]{2npi})=5cdot (2npi)^frac{3}{5}$$



Note If $lim_{x to infty} |f'(x)|=infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $delta$ such that $|x-y|< delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+frac{delta}{2})$ and take the limit as $z to infty$ to get a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think $f(x)=x^2$ is the simplest example..
    $endgroup$
    – Empty
    Dec 23 '18 at 20:34










  • $begingroup$
    Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:45










  • $begingroup$
    I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:47








  • 1




    $begingroup$
    @Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 20:56










  • $begingroup$
    @Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:03
















1












$begingroup$

The statement is not true. Consider the function
$$f(x)=frac{sin(x^5)}{x} ,.$$
with $f(0)=0$. Then, it is easy to see that $f in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.



But
$$f'(x)=frac{5x^5cos(x^5)-sin(x^5)}{x^2}$$is clearly unbounded as $$f'(sqrt[5]{2npi})=5cdot (2npi)^frac{3}{5}$$



Note If $lim_{x to infty} |f'(x)|=infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $delta$ such that $|x-y|< delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+frac{delta}{2})$ and take the limit as $z to infty$ to get a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think $f(x)=x^2$ is the simplest example..
    $endgroup$
    – Empty
    Dec 23 '18 at 20:34










  • $begingroup$
    Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:45










  • $begingroup$
    I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:47








  • 1




    $begingroup$
    @Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 20:56










  • $begingroup$
    @Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:03














1












1








1





$begingroup$

The statement is not true. Consider the function
$$f(x)=frac{sin(x^5)}{x} ,.$$
with $f(0)=0$. Then, it is easy to see that $f in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.



But
$$f'(x)=frac{5x^5cos(x^5)-sin(x^5)}{x^2}$$is clearly unbounded as $$f'(sqrt[5]{2npi})=5cdot (2npi)^frac{3}{5}$$



Note If $lim_{x to infty} |f'(x)|=infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $delta$ such that $|x-y|< delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+frac{delta}{2})$ and take the limit as $z to infty$ to get a contradiction.






share|cite|improve this answer









$endgroup$



The statement is not true. Consider the function
$$f(x)=frac{sin(x^5)}{x} ,.$$
with $f(0)=0$. Then, it is easy to see that $f in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.



But
$$f'(x)=frac{5x^5cos(x^5)-sin(x^5)}{x^2}$$is clearly unbounded as $$f'(sqrt[5]{2npi})=5cdot (2npi)^frac{3}{5}$$



Note If $lim_{x to infty} |f'(x)|=infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $delta$ such that $|x-y|< delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+frac{delta}{2})$ and take the limit as $z to infty$ to get a contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 19:44









N. S.N. S.

105k7116210




105k7116210












  • $begingroup$
    I think $f(x)=x^2$ is the simplest example..
    $endgroup$
    – Empty
    Dec 23 '18 at 20:34










  • $begingroup$
    Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:45










  • $begingroup$
    I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:47








  • 1




    $begingroup$
    @Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 20:56










  • $begingroup$
    @Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:03


















  • $begingroup$
    I think $f(x)=x^2$ is the simplest example..
    $endgroup$
    – Empty
    Dec 23 '18 at 20:34










  • $begingroup$
    Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:45










  • $begingroup$
    I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
    $endgroup$
    – Hallow
    Dec 23 '18 at 20:47








  • 1




    $begingroup$
    @Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 20:56










  • $begingroup$
    @Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:03
















$begingroup$
I think $f(x)=x^2$ is the simplest example..
$endgroup$
– Empty
Dec 23 '18 at 20:34




$begingroup$
I think $f(x)=x^2$ is the simplest example..
$endgroup$
– Empty
Dec 23 '18 at 20:34












$begingroup$
Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
$endgroup$
– Hallow
Dec 23 '18 at 20:45




$begingroup$
Thank you @N. S. for your answer. You all might be right, the author (my teacher) has probable made a mistake, it wouldn't be the first time.
$endgroup$
– Hallow
Dec 23 '18 at 20:45












$begingroup$
I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
$endgroup$
– Hallow
Dec 23 '18 at 20:47






$begingroup$
I just would like to get clarified why $sup_{x in mathbb{R}}frac{5x^{5}cos(x^{5}) - sin(x^{5})}{x^{2}} = infty$. @N. S.
$endgroup$
– Hallow
Dec 23 '18 at 20:47






1




1




$begingroup$
@Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 20:56




$begingroup$
@Empty: I don't understand. $f(x)=x^2$ is not uniformly continuous on $Bbb R$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 20:56












$begingroup$
@Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 21:03




$begingroup$
@Hallow: To answer your last question: Take the sequence $x_n = root5of{2pi n}$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 21:03


















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