If $f$ is $C^{1}$ and $sup_{x in mathbb{R}}|f'(x)| = infty$ , prove that $f$ cannot be uniformly continuous
$begingroup$
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
edited Dec 23 '18 at 19:25
Bernard
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
$endgroup$
add a comment |
$begingroup$
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
edited Dec 23 '18 at 19:25
Bernard
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 23 '18 at 19:25
Bernard
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
$endgroup$
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
edited Dec 23 '18 at 19:25
Bernard
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
edited Dec 23 '18 at 19:25
Bernard
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
edited Dec 23 '18 at 19:25
Bernard
124k742117
edited Dec 23 '18 at 19:25
Bernard
124k742117
edited Dec 23 '18 at 19:25
Bernard
124k742117
124k742117
asked Dec 23 '18 at 19:23
HallowHallow
84
asked Dec 23 '18 at 19:23
HallowHallow
84
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
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
|
show 5 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
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oldest
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oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
$endgroup$
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.
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
answered Dec 23 '18 at 19:44
N. S.N. S.
105k7116210
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
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$begingroup$
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