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












0












$begingroup$


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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$endgroup$

















    0












    $begingroup$


    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











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 5 at 14:58







      Omojola Micheal

















      asked Jan 5 at 14:29









      Omojola MichealOmojola Micheal

      2,049424




      2,049424






















          2 Answers
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          1












          $begingroup$

          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ 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











          $endgroup$













          • $begingroup$
            (+1) for that. I'm I right too?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:51










          • $begingroup$
            Your approach was right, but the reasoning was a little bit shaky.
            $endgroup$
            – John_Wick
            Jan 5 at 14:53










          • $begingroup$
            Oh, at what point?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:54










          • $begingroup$
            You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            $endgroup$
            – John_Wick
            Jan 5 at 14:56










          • $begingroup$
            Alright, let me edit it.
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:57



















          1












          $begingroup$

          $f'(0)$ 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









          $endgroup$









          • 1




            $begingroup$
            Thanks for that, Matematleta!
            $endgroup$
            – Omojola Micheal
            Jan 5 at 16:23










          • $begingroup$
            You are welcome!
            $endgroup$
            – Matematleta
            Jan 5 at 16:25












          Your Answer





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          1












          $begingroup$

          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ 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











          $endgroup$













          • $begingroup$
            (+1) for that. I'm I right too?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:51










          • $begingroup$
            Your approach was right, but the reasoning was a little bit shaky.
            $endgroup$
            – John_Wick
            Jan 5 at 14:53










          • $begingroup$
            Oh, at what point?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:54










          • $begingroup$
            You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            $endgroup$
            – John_Wick
            Jan 5 at 14:56










          • $begingroup$
            Alright, let me edit it.
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:57
















          1












          $begingroup$

          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ 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











          $endgroup$













          • $begingroup$
            (+1) for that. I'm I right too?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:51










          • $begingroup$
            Your approach was right, but the reasoning was a little bit shaky.
            $endgroup$
            – John_Wick
            Jan 5 at 14:53










          • $begingroup$
            Oh, at what point?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:54










          • $begingroup$
            You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            $endgroup$
            – John_Wick
            Jan 5 at 14:56










          • $begingroup$
            Alright, let me edit it.
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:57














          1












          1








          1





          $begingroup$

          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ 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











          $endgroup$



          $fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ 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 Jan 6 at 4:24

























          answered Jan 5 at 14:42









          John_WickJohn_Wick

          1,616111




          1,616111












          • $begingroup$
            (+1) for that. I'm I right too?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:51










          • $begingroup$
            Your approach was right, but the reasoning was a little bit shaky.
            $endgroup$
            – John_Wick
            Jan 5 at 14:53










          • $begingroup$
            Oh, at what point?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:54










          • $begingroup$
            You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            $endgroup$
            – John_Wick
            Jan 5 at 14:56










          • $begingroup$
            Alright, let me edit it.
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:57


















          • $begingroup$
            (+1) for that. I'm I right too?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:51










          • $begingroup$
            Your approach was right, but the reasoning was a little bit shaky.
            $endgroup$
            – John_Wick
            Jan 5 at 14:53










          • $begingroup$
            Oh, at what point?
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:54










          • $begingroup$
            You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
            $endgroup$
            – John_Wick
            Jan 5 at 14:56










          • $begingroup$
            Alright, let me edit it.
            $endgroup$
            – Omojola Micheal
            Jan 5 at 14:57
















          $begingroup$
          (+1) for that. I'm I right too?
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:51




          $begingroup$
          (+1) for that. I'm I right too?
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:51












          $begingroup$
          Your approach was right, but the reasoning was a little bit shaky.
          $endgroup$
          – John_Wick
          Jan 5 at 14:53




          $begingroup$
          Your approach was right, but the reasoning was a little bit shaky.
          $endgroup$
          – John_Wick
          Jan 5 at 14:53












          $begingroup$
          Oh, at what point?
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:54




          $begingroup$
          Oh, at what point?
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:54












          $begingroup$
          You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
          $endgroup$
          – John_Wick
          Jan 5 at 14:56




          $begingroup$
          You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
          $endgroup$
          – John_Wick
          Jan 5 at 14:56












          $begingroup$
          Alright, let me edit it.
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:57




          $begingroup$
          Alright, let me edit it.
          $endgroup$
          – Omojola Micheal
          Jan 5 at 14:57











          1












          $begingroup$

          $f'(0)$ 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









          $endgroup$









          • 1




            $begingroup$
            Thanks for that, Matematleta!
            $endgroup$
            – Omojola Micheal
            Jan 5 at 16:23










          • $begingroup$
            You are welcome!
            $endgroup$
            – Matematleta
            Jan 5 at 16:25
















          1












          $begingroup$

          $f'(0)$ 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









          $endgroup$









          • 1




            $begingroup$
            Thanks for that, Matematleta!
            $endgroup$
            – Omojola Micheal
            Jan 5 at 16:23










          • $begingroup$
            You are welcome!
            $endgroup$
            – Matematleta
            Jan 5 at 16:25














          1












          1








          1





          $begingroup$

          $f'(0)$ 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









          $endgroup$



          $f'(0)$ 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 Jan 5 at 16:17









          MatematletaMatematleta

          12.1k21020




          12.1k21020








          • 1




            $begingroup$
            Thanks for that, Matematleta!
            $endgroup$
            – Omojola Micheal
            Jan 5 at 16:23










          • $begingroup$
            You are welcome!
            $endgroup$
            – Matematleta
            Jan 5 at 16:25














          • 1




            $begingroup$
            Thanks for that, Matematleta!
            $endgroup$
            – Omojola Micheal
            Jan 5 at 16:23










          • $begingroup$
            You are welcome!
            $endgroup$
            – Matematleta
            Jan 5 at 16:25








          1




          1




          $begingroup$
          Thanks for that, Matematleta!
          $endgroup$
          – Omojola Micheal
          Jan 5 at 16:23




          $begingroup$
          Thanks for that, Matematleta!
          $endgroup$
          – Omojola Micheal
          Jan 5 at 16:23












          $begingroup$
          You are welcome!
          $endgroup$
          – Matematleta
          Jan 5 at 16:25




          $begingroup$
          You are welcome!
          $endgroup$
          – Matematleta
          Jan 5 at 16:25


















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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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