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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share|cite|improve this question













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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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