Polynomial growth implies locally Lipschitz?

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Let $f:mathbbR^mtomathbbR^m$ satisfy $|f(x)|le c|x|^n$ for some (re-edit:) $ninmathbbN$ with some constant $c>0$. Is $f$ locally Lipschitz?



I see that it is around $x=0$: $|f(x)-f(0)|le frac^n+c0|x|=c|x|^n-1|x|le delta |x|$ for a suitable $delta$.



Apparently, a polarization argument is a way to go but I do not know how to apply it for $|f(x)-f(x')|$.






real-analysis lipschitz-functions nonlinear-analysis







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  • What definition of "locally Lipschitz" are you using?
    – zhw.
    27 mins ago














up vote
1
down vote

favorite












Let $f:mathbbR^mtomathbbR^m$ satisfy $|f(x)|le c|x|^n$ for some (re-edit:) $ninmathbbN$ with some constant $c>0$. Is $f$ locally Lipschitz?



I see that it is around $x=0$: $|f(x)-f(0)|le frac^n+c0|x|=c|x|^n-1|x|le delta |x|$ for a suitable $delta$.



Apparently, a polarization argument is a way to go but I do not know how to apply it for $|f(x)-f(x')|$.






real-analysis lipschitz-functions nonlinear-analysis







share|cite|improve this question























  • What definition of "locally Lipschitz" are you using?
    – zhw.
    27 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $f:mathbbR^mtomathbbR^m$ satisfy $|f(x)|le c|x|^n$ for some (re-edit:) $ninmathbbN$ with some constant $c>0$. Is $f$ locally Lipschitz?



I see that it is around $x=0$: $|f(x)-f(0)|le frac^n+c0|x|=c|x|^n-1|x|le delta |x|$ for a suitable $delta$.



Apparently, a polarization argument is a way to go but I do not know how to apply it for $|f(x)-f(x')|$.






real-analysis lipschitz-functions nonlinear-analysis







share|cite|improve this question















Let $f:mathbbR^mtomathbbR^m$ satisfy $|f(x)|le c|x|^n$ for some (re-edit:) $ninmathbbN$ with some constant $c>0$. Is $f$ locally Lipschitz?



I see that it is around $x=0$: $|f(x)-f(0)|le frac^n+c0|x|=c|x|^n-1|x|le delta |x|$ for a suitable $delta$.



Apparently, a polarization argument is a way to go but I do not know how to apply it for $|f(x)-f(x')|$.






real-analysis lipschitz-functions nonlinear-analysis




real-analysis lipschitz-functions nonlinear-analysis






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edited 1 hour ago

























asked 1 hour ago









shall.i.am

366217




366217











  • What definition of "locally Lipschitz" are you using?
    – zhw.
    27 mins ago
















  • What definition of "locally Lipschitz" are you using?
    – zhw.
    27 mins ago















What definition of "locally Lipschitz" are you using?
– zhw.
27 mins ago




What definition of "locally Lipschitz" are you using?
– zhw.
27 mins ago










3 Answers
3






active

oldest

votes

















up vote
2
down vote













I think the function $f$ from $mathbbR$ to $mathbbR$ define by $f(x)=x sin (1/x)$ is a counter example to your problem.






share|cite|improve this answer








New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • 2




    This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
    – Hans
    1 hour ago






  • 2




    There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
    – Hans
    1 hour ago











  • Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
    – shall.i.am
    1 hour ago






  • 1




    @Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
    – zhw.
    36 mins ago







  • 1




    @Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
    – zhw.
    34 mins ago

















up vote
1
down vote













A counterexample is
$$f(x)=begincases |x|^n-1x, |x|inmathbb Q\
-|x|^n-1x, |x|notinmathbb Q
endcases.
$$






share|cite|improve this answer






















  • Is it obvious that this is a counter example ?
    – Surb
    39 mins ago







  • 1




    @Surb: Yes, since it is discontinuous.
    – Hans
    38 mins ago






  • 1




    @Surb: I have modified my counterexample for it to work for arbitrary $m$.
    – Hans
    26 mins ago










  • I have already upvoted your answer :)
    – Surb
    26 mins ago










  • @Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
    – Hans
    21 mins ago

















up vote
1
down vote













A continuous nowhere differentiable function such as
$$sum_n=1^infty sin(2^n x)over n^2$$
is bounded but it most certainly is not locally lipschitz.






share|cite|improve this answer




















  • really? continuous implies uniform continuity on any compact isn't it?
    – Surb
    26 mins ago











  • This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
    – ncmathsadist
    24 mins ago










  • right makes sense, I mixed up two definitions$
    – Surb
    23 mins ago










  • I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
    – Hans
    11 mins ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













I think the function $f$ from $mathbbR$ to $mathbbR$ define by $f(x)=x sin (1/x)$ is a counter example to your problem.






share|cite|improve this answer








New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • 2




    This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
    – Hans
    1 hour ago






  • 2




    There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
    – Hans
    1 hour ago











  • Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
    – shall.i.am
    1 hour ago






  • 1




    @Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
    – zhw.
    36 mins ago







  • 1




    @Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
    – zhw.
    34 mins ago














up vote
2
down vote













I think the function $f$ from $mathbbR$ to $mathbbR$ define by $f(x)=x sin (1/x)$ is a counter example to your problem.






share|cite|improve this answer








New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • 2




    This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
    – Hans
    1 hour ago






  • 2




    There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
    – Hans
    1 hour ago











  • Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
    – shall.i.am
    1 hour ago






  • 1




    @Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
    – zhw.
    36 mins ago







  • 1




    @Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
    – zhw.
    34 mins ago












up vote
2
down vote










up vote
2
down vote









I think the function $f$ from $mathbbR$ to $mathbbR$ define by $f(x)=x sin (1/x)$ is a counter example to your problem.






share|cite|improve this answer








New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









I think the function $f$ from $mathbbR$ to $mathbbR$ define by $f(x)=x sin (1/x)$ is a counter example to your problem.







share|cite|improve this answer








New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 1 hour ago









AlexL

591




591




New contributor




AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 2




    This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
    – Hans
    1 hour ago






  • 2




    There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
    – Hans
    1 hour ago











  • Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
    – shall.i.am
    1 hour ago






  • 1




    @Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
    – zhw.
    36 mins ago







  • 1




    @Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
    – zhw.
    34 mins ago












  • 2




    This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
    – Hans
    1 hour ago






  • 2




    There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
    – Hans
    1 hour ago











  • Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
    – shall.i.am
    1 hour ago






  • 1




    @Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
    – zhw.
    36 mins ago







  • 1




    @Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
    – zhw.
    34 mins ago







2




2




This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
– Hans
1 hour ago




This is not a counterexample, because local Lipschitz allows constants be dependent on $x$.
– Hans
1 hour ago




2




2




There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
– Hans
1 hour ago





There is nothing to reconcile. The definition of local Lipschitz en.wikipedia.org/wiki/Lipschitz_continuity#Definitions is satisfied for this example in the answer.
– Hans
1 hour ago













Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
– shall.i.am
1 hour ago




Ah, that's what I thought. Great, thanks. The example "Differentiable functions that are not (locally) Lipschitz continuous" in the wikipedia page is very confusing.
– shall.i.am
1 hour ago




1




1




@Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
– zhw.
36 mins ago





@Hans The definition of "locally Lipschitz" you cite requires that for each $x,$ there exists a full neighborhood $U_x$ of $x$ in which $f$ is Lipschitz. There is no such neighborhood of $0$ for the example of AlexL, so it is a valid counterexample.
– zhw.
36 mins ago





1




1




@Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
– zhw.
34 mins ago




@Surb No, the example offered by AlexL is not locally Lipschitz. Perhaps you are thinking of "pointwise Lipschitz"?
– zhw.
34 mins ago










up vote
1
down vote













A counterexample is
$$f(x)=begincases |x|^n-1x, |x|inmathbb Q\
-|x|^n-1x, |x|notinmathbb Q
endcases.
$$






share|cite|improve this answer






















  • Is it obvious that this is a counter example ?
    – Surb
    39 mins ago







  • 1




    @Surb: Yes, since it is discontinuous.
    – Hans
    38 mins ago






  • 1




    @Surb: I have modified my counterexample for it to work for arbitrary $m$.
    – Hans
    26 mins ago










  • I have already upvoted your answer :)
    – Surb
    26 mins ago










  • @Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
    – Hans
    21 mins ago














up vote
1
down vote













A counterexample is
$$f(x)=begincases |x|^n-1x, |x|inmathbb Q\
-|x|^n-1x, |x|notinmathbb Q
endcases.
$$






share|cite|improve this answer






















  • Is it obvious that this is a counter example ?
    – Surb
    39 mins ago







  • 1




    @Surb: Yes, since it is discontinuous.
    – Hans
    38 mins ago






  • 1




    @Surb: I have modified my counterexample for it to work for arbitrary $m$.
    – Hans
    26 mins ago










  • I have already upvoted your answer :)
    – Surb
    26 mins ago










  • @Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
    – Hans
    21 mins ago












up vote
1
down vote










up vote
1
down vote









A counterexample is
$$f(x)=begincases |x|^n-1x, |x|inmathbb Q\
-|x|^n-1x, |x|notinmathbb Q
endcases.
$$






share|cite|improve this answer














A counterexample is
$$f(x)=begincases |x|^n-1x, |x|inmathbb Q\
-|x|^n-1x, |x|notinmathbb Q
endcases.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 28 mins ago

























answered 50 mins ago









Hans

4,42621029




4,42621029











  • Is it obvious that this is a counter example ?
    – Surb
    39 mins ago







  • 1




    @Surb: Yes, since it is discontinuous.
    – Hans
    38 mins ago






  • 1




    @Surb: I have modified my counterexample for it to work for arbitrary $m$.
    – Hans
    26 mins ago










  • I have already upvoted your answer :)
    – Surb
    26 mins ago










  • @Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
    – Hans
    21 mins ago
















  • Is it obvious that this is a counter example ?
    – Surb
    39 mins ago







  • 1




    @Surb: Yes, since it is discontinuous.
    – Hans
    38 mins ago






  • 1




    @Surb: I have modified my counterexample for it to work for arbitrary $m$.
    – Hans
    26 mins ago










  • I have already upvoted your answer :)
    – Surb
    26 mins ago










  • @Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
    – Hans
    21 mins ago















Is it obvious that this is a counter example ?
– Surb
39 mins ago





Is it obvious that this is a counter example ?
– Surb
39 mins ago





1




1




@Surb: Yes, since it is discontinuous.
– Hans
38 mins ago




@Surb: Yes, since it is discontinuous.
– Hans
38 mins ago




1




1




@Surb: I have modified my counterexample for it to work for arbitrary $m$.
– Hans
26 mins ago




@Surb: I have modified my counterexample for it to work for arbitrary $m$.
– Hans
26 mins ago












I have already upvoted your answer :)
– Surb
26 mins ago




I have already upvoted your answer :)
– Surb
26 mins ago












@Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
– Hans
21 mins ago




@Surb: I understand. Thank you. :-) I just want to address your previous concern for $m>1$.
– Hans
21 mins ago










up vote
1
down vote













A continuous nowhere differentiable function such as
$$sum_n=1^infty sin(2^n x)over n^2$$
is bounded but it most certainly is not locally lipschitz.






share|cite|improve this answer




















  • really? continuous implies uniform continuity on any compact isn't it?
    – Surb
    26 mins ago











  • This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
    – ncmathsadist
    24 mins ago










  • right makes sense, I mixed up two definitions$
    – Surb
    23 mins ago










  • I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
    – Hans
    11 mins ago















up vote
1
down vote













A continuous nowhere differentiable function such as
$$sum_n=1^infty sin(2^n x)over n^2$$
is bounded but it most certainly is not locally lipschitz.






share|cite|improve this answer




















  • really? continuous implies uniform continuity on any compact isn't it?
    – Surb
    26 mins ago











  • This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
    – ncmathsadist
    24 mins ago










  • right makes sense, I mixed up two definitions$
    – Surb
    23 mins ago










  • I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
    – Hans
    11 mins ago













up vote
1
down vote










up vote
1
down vote









A continuous nowhere differentiable function such as
$$sum_n=1^infty sin(2^n x)over n^2$$
is bounded but it most certainly is not locally lipschitz.






share|cite|improve this answer












A continuous nowhere differentiable function such as
$$sum_n=1^infty sin(2^n x)over n^2$$
is bounded but it most certainly is not locally lipschitz.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 27 mins ago









ncmathsadist

41.5k258100




41.5k258100











  • really? continuous implies uniform continuity on any compact isn't it?
    – Surb
    26 mins ago











  • This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
    – ncmathsadist
    24 mins ago










  • right makes sense, I mixed up two definitions$
    – Surb
    23 mins ago










  • I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
    – Hans
    11 mins ago

















  • really? continuous implies uniform continuity on any compact isn't it?
    – Surb
    26 mins ago











  • This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
    – ncmathsadist
    24 mins ago










  • right makes sense, I mixed up two definitions$
    – Surb
    23 mins ago










  • I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
    – Hans
    11 mins ago
















really? continuous implies uniform continuity on any compact isn't it?
– Surb
26 mins ago





really? continuous implies uniform continuity on any compact isn't it?
– Surb
26 mins ago













This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
– ncmathsadist
24 mins ago




This is uniformly continuous, but it is not locally lipschitz. That is a far stronger condition than continuity.
– ncmathsadist
24 mins ago












right makes sense, I mixed up two definitions$
– Surb
23 mins ago




right makes sense, I mixed up two definitions$
– Surb
23 mins ago












I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
– Hans
11 mins ago





I like the example but can you prove it is not locally Lipschitz or provide a reference to the proof of this claim?
– Hans
11 mins ago


















 

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One-line joke

A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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