Prove that $f(x)=xsin(1/x)$ for $xne0$, $f(0)=0$, is not Lipschitz on $[0,1]$












0














Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$



MY TRIAL



My idea is to show that $f$ does not have a bounded derivative.



So, suppose for contradiction that there exists $K>0$ such that
$$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
As $xto 0$,
begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
Please, I'm I right? Any other way of showing this is also accepted.






real-analysis analysis uniform-continuity lipschitz-functions







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    Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$



    MY TRIAL



    My idea is to show that $f$ does not have a bounded derivative.



    So, suppose for contradiction that there exists $K>0$ such that
    $$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
    As $xto 0$,
    begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
    which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
    Please, I'm I right? Any other way of showing this is also accepted.






    real-analysis analysis uniform-continuity lipschitz-functions







    share|cite|improve this question



























      0












      0








      0


      1





      Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$



      MY TRIAL



      My idea is to show that $f$ does not have a bounded derivative.



      So, suppose for contradiction that there exists $K>0$ such that
      $$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
      As $xto 0$,
      begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
      which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
      Please, I'm I right? Any other way of showing this is also accepted.






      real-analysis analysis uniform-continuity lipschitz-functions







      share|cite|improve this question















      Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$



      MY TRIAL



      My idea is to show that $f$ does not have a bounded derivative.



      So, suppose for contradiction that there exists $K>0$ such that
      $$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
      As $xto 0$,
      begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
      which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
      Please, I'm I right? Any other way of showing this is also accepted.






      real-analysis analysis uniform-continuity lipschitz-functions




      real-analysis analysis uniform-continuity lipschitz-functions






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      edited 2 days ago







      Mike

















      asked 2 days ago









      MikeMike

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          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$



          for $n>C.$ So, $f$ can't be Lipschitz.






          share|cite|improve this answer























          • (+1) for that. I'm I right too?
            – Mike
            2 days ago










          • Your approach was right, but the reasoning was a little bit shaky.
            – John_Wick
            2 days ago










          • Oh, at what point?
            – Mike
            2 days ago










          • You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            – John_Wick
            2 days ago










          • Alright, let me edit it.
            – Mike
            2 days ago



















          1














          $f'(0)$ does not exist so your approach won't work. But the idea is correct:



          You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$



          Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.






          share|cite|improve this answer

















          • 1




            Thanks for that, Matematleta!
            – Mike
            2 days ago










          • You are welcome!
            – Matematleta
            2 days ago











          Your Answer





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          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$



          for $n>C.$ So, $f$ can't be Lipschitz.






          share|cite|improve this answer























          • (+1) for that. I'm I right too?
            – Mike
            2 days ago










          • Your approach was right, but the reasoning was a little bit shaky.
            – John_Wick
            2 days ago










          • Oh, at what point?
            – Mike
            2 days ago










          • You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            – John_Wick
            2 days ago










          • Alright, let me edit it.
            – Mike
            2 days ago
















          1














          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$



          for $n>C.$ So, $f$ can't be Lipschitz.






          share|cite|improve this answer























          • (+1) for that. I'm I right too?
            – Mike
            2 days ago










          • Your approach was right, but the reasoning was a little bit shaky.
            – John_Wick
            2 days ago










          • Oh, at what point?
            – Mike
            2 days ago










          • You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            – John_Wick
            2 days ago










          • Alright, let me edit it.
            – Mike
            2 days ago














          1












          1








          1






          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$



          for $n>C.$ So, $f$ can't be Lipschitz.






          share|cite|improve this answer














          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$



          for $n>C.$ So, $f$ can't be Lipschitz.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered 2 days ago









          John_WickJohn_Wick

          1,464111




          1,464111












          • (+1) for that. I'm I right too?
            – Mike
            2 days ago










          • Your approach was right, but the reasoning was a little bit shaky.
            – John_Wick
            2 days ago










          • Oh, at what point?
            – Mike
            2 days ago










          • You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            – John_Wick
            2 days ago










          • Alright, let me edit it.
            – Mike
            2 days ago


















          • (+1) for that. I'm I right too?
            – Mike
            2 days ago










          • Your approach was right, but the reasoning was a little bit shaky.
            – John_Wick
            2 days ago










          • Oh, at what point?
            – Mike
            2 days ago










          • You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            – John_Wick
            2 days ago










          • Alright, let me edit it.
            – Mike
            2 days ago
















          (+1) for that. I'm I right too?
          – Mike
          2 days ago




          (+1) for that. I'm I right too?
          – Mike
          2 days ago












          Your approach was right, but the reasoning was a little bit shaky.
          – John_Wick
          2 days ago




          Your approach was right, but the reasoning was a little bit shaky.
          – John_Wick
          2 days ago












          Oh, at what point?
          – Mike
          2 days ago




          Oh, at what point?
          – Mike
          2 days ago












          You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
          – John_Wick
          2 days ago




          You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
          – John_Wick
          2 days ago












          Alright, let me edit it.
          – Mike
          2 days ago




          Alright, let me edit it.
          – Mike
          2 days ago











          1














          $f'(0)$ does not exist so your approach won't work. But the idea is correct:



          You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$



          Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.






          share|cite|improve this answer

















          • 1




            Thanks for that, Matematleta!
            – Mike
            2 days ago










          • You are welcome!
            – Matematleta
            2 days ago
















          1














          $f'(0)$ does not exist so your approach won't work. But the idea is correct:



          You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$



          Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.






          share|cite|improve this answer

















          • 1




            Thanks for that, Matematleta!
            – Mike
            2 days ago










          • You are welcome!
            – Matematleta
            2 days ago














          1












          1








          1






          $f'(0)$ does not exist so your approach won't work. But the idea is correct:



          You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$



          Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.






          share|cite|improve this answer












          $f'(0)$ does not exist so your approach won't work. But the idea is correct:



          You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$



          Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          MatematletaMatematleta

          10.1k2918




          10.1k2918








          • 1




            Thanks for that, Matematleta!
            – Mike
            2 days ago










          • You are welcome!
            – Matematleta
            2 days ago














          • 1




            Thanks for that, Matematleta!
            – Mike
            2 days ago










          • You are welcome!
            – Matematleta
            2 days ago








          1




          1




          Thanks for that, Matematleta!
          – Mike
          2 days ago




          Thanks for that, Matematleta!
          – Mike
          2 days ago












          You are welcome!
          – Matematleta
          2 days ago




          You are welcome!
          – Matematleta
          2 days ago


















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