Exercise on Sobolev Spaces: Show that this function belong to $W^{1,infty}$












2












$begingroup$



Let $u in C^0(I)$ be a bounded function on the open interval $I=(a,b)
subset mathbb R$. Suppose that there exists a partition $a=t_0 lt
t_1 lt dots lt t_n=b$ such that:




  • $f in C^1((t_i,t_{i+1}))$

  • $f' in L^{infty}((t_i,t_{i+1}))$


for every $i=0,dots,n-1$. Show that $fin W^{1,infty}$




MY ATTEMPT:



It suffices to show that there exists a constant $Cgt 0$ such that $vert u(x)-u(y)vert le C vert x-yvert$ for a.e $x,y in I$. However I have only managed to prove that $u$ satisfies a Lipschitz condition in every subinterval $(t_i,t_{i+1})$ since:



if $t_i lt x lt y lt t_{i+1}$, then $vert u(x)-u(y) vert le int_{t_i}^{t_{i+1}} vert u'(x) vert le {vert vert u vert vert}_{L^{infty}}((t_i,t_{i+1})) vert x-y vert$



I believe that somehow I should take the sum over $i$ in order to pass to the whole $I$ but I don't see how. Moreover I haven't used that $u$ is also bounded and I 'm wondering what I miss here.



Any help or hint is much appreciated.



Thanks in advance!






functional-analysis sobolev-spaces lipschitz-functions







share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Let $u in C^0(I)$ be a bounded function on the open interval $I=(a,b)
    subset mathbb R$    . Suppose that there exists a partition $a=t_0 lt
    t_1 lt dots lt t_n=b$     such that:




    • $f in C^1((t_i,t_{i+1}))$

    • $f' in L^{infty}((t_i,t_{i+1}))$


    for every $i=0,dots,n-1$. Show that $fin W^{1,infty}$




    MY ATTEMPT:



    It suffices to show that there exists a constant $Cgt 0$ such that $vert u(x)-u(y)vert le C vert x-yvert$ for a.e $x,y in I$. However I have only managed to prove that $u$ satisfies a Lipschitz condition in every subinterval $(t_i,t_{i+1})$ since:



    if $t_i lt x lt y lt t_{i+1}$, then $vert u(x)-u(y) vert le int_{t_i}^{t_{i+1}} vert u'(x) vert le {vert vert u vert vert}_{L^{infty}}((t_i,t_{i+1})) vert x-y vert$



    I believe that somehow I should take the sum over $i$ in order to pass to the whole $I$ but I don't see how. Moreover I haven't used that $u$ is also bounded and I 'm wondering what I miss here.



    Any help or hint is much appreciated.



    Thanks in advance!






    functional-analysis sobolev-spaces lipschitz-functions







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Let $u in C^0(I)$ be a bounded function on the open interval $I=(a,b)
      subset mathbb R$      . Suppose that there exists a partition $a=t_0 lt
      t_1 lt dots lt t_n=b$       such that:




      • $f in C^1((t_i,t_{i+1}))$

      • $f' in L^{infty}((t_i,t_{i+1}))$


      for every $i=0,dots,n-1$. Show that $fin W^{1,infty}$




      MY ATTEMPT:



      It suffices to show that there exists a constant $Cgt 0$ such that $vert u(x)-u(y)vert le C vert x-yvert$ for a.e $x,y in I$. However I have only managed to prove that $u$ satisfies a Lipschitz condition in every subinterval $(t_i,t_{i+1})$ since:



      if $t_i lt x lt y lt t_{i+1}$, then $vert u(x)-u(y) vert le int_{t_i}^{t_{i+1}} vert u'(x) vert le {vert vert u vert vert}_{L^{infty}}((t_i,t_{i+1})) vert x-y vert$



      I believe that somehow I should take the sum over $i$ in order to pass to the whole $I$ but I don't see how. Moreover I haven't used that $u$ is also bounded and I 'm wondering what I miss here.



      Any help or hint is much appreciated.



      Thanks in advance!






      functional-analysis sobolev-spaces lipschitz-functions







      share|cite|improve this question









      $endgroup$





      Let $u in C^0(I)$ be a bounded function on the open interval $I=(a,b)
      subset mathbb R$      . Suppose that there exists a partition $a=t_0 lt
      t_1 lt dots lt t_n=b$       such that:




      • $f in C^1((t_i,t_{i+1}))$

      • $f' in L^{infty}((t_i,t_{i+1}))$


      for every $i=0,dots,n-1$. Show that $fin W^{1,infty}$




      MY ATTEMPT:



      It suffices to show that there exists a constant $Cgt 0$ such that $vert u(x)-u(y)vert le C vert x-yvert$ for a.e $x,y in I$. However I have only managed to prove that $u$ satisfies a Lipschitz condition in every subinterval $(t_i,t_{i+1})$ since:



      if $t_i lt x lt y lt t_{i+1}$, then $vert u(x)-u(y) vert le int_{t_i}^{t_{i+1}} vert u'(x) vert le {vert vert u vert vert}_{L^{infty}}((t_i,t_{i+1})) vert x-y vert$



      I believe that somehow I should take the sum over $i$ in order to pass to the whole $I$ but I don't see how. Moreover I haven't used that $u$ is also bounded and I 'm wondering what I miss here.



      Any help or hint is much appreciated.



      Thanks in advance!






      functional-analysis sobolev-spaces lipschitz-functions




      functional-analysis sobolev-spaces lipschitz-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Feb 1 at 13:46









      kaithkolesidoukaithkolesidou

      1,262512




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          $begingroup$

          You have already proved the Lipschitz estimate inside a single interval $(t_i, t_{i+1})$. To complete the proof of Lipschitz condition on $I$ it is enough to show that
          $$
          tag{1} |u(x) - u(t)| leq C |x - t|, text{ for all } x in [t_i, t_{i+1}] text{ and } t in {t_i, t_{i+1}}.
          $$
          Indeed, once we have $(1)$, then the Lipschitz estimate propagates across intervals, as for any $xin (t_i, t_{i+1})$ and any $yin (t_{i+1}, t_{i+2})$ we get
          $$
          |u(x) - u(y)| leq |u(x) - u(t_{i+1})| + |u(t_{i+1}) - u(y)| leq C (|x - t_{i+1}| + |t_{i+1} - y|) = C |x - y|,
          $$
          and similarly, if $x$ and $y$ are not necessarily in neighboring intervals.



          To prove $(1)$ for the endpoint $t = t_i$, take $z in (t_i, t_{i+1}) $ and write
          $$
          tag{2} |u(x) - u(t_i)| leq |u(x) - u(z)| + |u(z) - u(t_i)| leq C |x - z| + |u(z) - u(t_i)|,
          $$
          where we used the already proved Lipschitz estimate in $(t_i, t_{i+1})$. Notice that
          the constant $C$ in $(2)$ does not depend on $z$ (it can be taken to be $||u'||_{L^infty(t_i, t_{i+1})}$ as you showed above), hence we may pass to the limit in $(2)$ with respect to $z$ when $zto t_i$. Taking into account the continuity of $u$ at $t_i$ we get
          $$
          |u(x) - u(t_i)| leq C |x - t_i|,
          $$
          which is what was required to prove.






          share|cite|improve this answer









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            $begingroup$

            You have already proved the Lipschitz estimate inside a single interval $(t_i, t_{i+1})$. To complete the proof of Lipschitz condition on $I$ it is enough to show that
            $$
            tag{1} |u(x) - u(t)| leq C |x - t|, text{ for all } x in [t_i, t_{i+1}] text{ and } t in {t_i, t_{i+1}}.
            $$
            Indeed, once we have $(1)$, then the Lipschitz estimate propagates across intervals, as for any $xin (t_i, t_{i+1})$ and any $yin (t_{i+1}, t_{i+2})$ we get
            $$
            |u(x) - u(y)| leq |u(x) - u(t_{i+1})| + |u(t_{i+1}) - u(y)| leq C (|x - t_{i+1}| + |t_{i+1} - y|) = C |x - y|,
            $$
            and similarly, if $x$ and $y$ are not necessarily in neighboring intervals.



            To prove $(1)$ for the endpoint $t = t_i$, take $z in (t_i, t_{i+1}) $ and write
            $$
            tag{2} |u(x) - u(t_i)| leq |u(x) - u(z)| + |u(z) - u(t_i)| leq C |x - z| + |u(z) - u(t_i)|,
            $$
            where we used the already proved Lipschitz estimate in $(t_i, t_{i+1})$. Notice that
            the constant $C$ in $(2)$ does not depend on $z$ (it can be taken to be $||u'||_{L^infty(t_i, t_{i+1})}$ as you showed above), hence we may pass to the limit in $(2)$ with respect to $z$ when $zto t_i$. Taking into account the continuity of $u$ at $t_i$ we get
            $$
            |u(x) - u(t_i)| leq C |x - t_i|,
            $$
            which is what was required to prove.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You have already proved the Lipschitz estimate inside a single interval $(t_i, t_{i+1})$. To complete the proof of Lipschitz condition on $I$ it is enough to show that
              $$
              tag{1} |u(x) - u(t)| leq C |x - t|, text{ for all } x in [t_i, t_{i+1}] text{ and } t in {t_i, t_{i+1}}.
              $$
              Indeed, once we have $(1)$, then the Lipschitz estimate propagates across intervals, as for any $xin (t_i, t_{i+1})$ and any $yin (t_{i+1}, t_{i+2})$ we get
              $$
              |u(x) - u(y)| leq |u(x) - u(t_{i+1})| + |u(t_{i+1}) - u(y)| leq C (|x - t_{i+1}| + |t_{i+1} - y|) = C |x - y|,
              $$
              and similarly, if $x$ and $y$ are not necessarily in neighboring intervals.



              To prove $(1)$ for the endpoint $t = t_i$, take $z in (t_i, t_{i+1}) $ and write
              $$
              tag{2} |u(x) - u(t_i)| leq |u(x) - u(z)| + |u(z) - u(t_i)| leq C |x - z| + |u(z) - u(t_i)|,
              $$
              where we used the already proved Lipschitz estimate in $(t_i, t_{i+1})$. Notice that
              the constant $C$ in $(2)$ does not depend on $z$ (it can be taken to be $||u'||_{L^infty(t_i, t_{i+1})}$ as you showed above), hence we may pass to the limit in $(2)$ with respect to $z$ when $zto t_i$. Taking into account the continuity of $u$ at $t_i$ we get
              $$
              |u(x) - u(t_i)| leq C |x - t_i|,
              $$
              which is what was required to prove.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You have already proved the Lipschitz estimate inside a single interval $(t_i, t_{i+1})$. To complete the proof of Lipschitz condition on $I$ it is enough to show that
                $$
                tag{1} |u(x) - u(t)| leq C |x - t|, text{ for all } x in [t_i, t_{i+1}] text{ and } t in {t_i, t_{i+1}}.
                $$
                Indeed, once we have $(1)$, then the Lipschitz estimate propagates across intervals, as for any $xin (t_i, t_{i+1})$ and any $yin (t_{i+1}, t_{i+2})$ we get
                $$
                |u(x) - u(y)| leq |u(x) - u(t_{i+1})| + |u(t_{i+1}) - u(y)| leq C (|x - t_{i+1}| + |t_{i+1} - y|) = C |x - y|,
                $$
                and similarly, if $x$ and $y$ are not necessarily in neighboring intervals.



                To prove $(1)$ for the endpoint $t = t_i$, take $z in (t_i, t_{i+1}) $ and write
                $$
                tag{2} |u(x) - u(t_i)| leq |u(x) - u(z)| + |u(z) - u(t_i)| leq C |x - z| + |u(z) - u(t_i)|,
                $$
                where we used the already proved Lipschitz estimate in $(t_i, t_{i+1})$. Notice that
                the constant $C$ in $(2)$ does not depend on $z$ (it can be taken to be $||u'||_{L^infty(t_i, t_{i+1})}$ as you showed above), hence we may pass to the limit in $(2)$ with respect to $z$ when $zto t_i$. Taking into account the continuity of $u$ at $t_i$ we get
                $$
                |u(x) - u(t_i)| leq C |x - t_i|,
                $$
                which is what was required to prove.






                share|cite|improve this answer









                $endgroup$



                You have already proved the Lipschitz estimate inside a single interval $(t_i, t_{i+1})$. To complete the proof of Lipschitz condition on $I$ it is enough to show that
                $$
                tag{1} |u(x) - u(t)| leq C |x - t|, text{ for all } x in [t_i, t_{i+1}] text{ and } t in {t_i, t_{i+1}}.
                $$
                Indeed, once we have $(1)$, then the Lipschitz estimate propagates across intervals, as for any $xin (t_i, t_{i+1})$ and any $yin (t_{i+1}, t_{i+2})$ we get
                $$
                |u(x) - u(y)| leq |u(x) - u(t_{i+1})| + |u(t_{i+1}) - u(y)| leq C (|x - t_{i+1}| + |t_{i+1} - y|) = C |x - y|,
                $$
                and similarly, if $x$ and $y$ are not necessarily in neighboring intervals.



                To prove $(1)$ for the endpoint $t = t_i$, take $z in (t_i, t_{i+1}) $ and write
                $$
                tag{2} |u(x) - u(t_i)| leq |u(x) - u(z)| + |u(z) - u(t_i)| leq C |x - z| + |u(z) - u(t_i)|,
                $$
                where we used the already proved Lipschitz estimate in $(t_i, t_{i+1})$. Notice that
                the constant $C$ in $(2)$ does not depend on $z$ (it can be taken to be $||u'||_{L^infty(t_i, t_{i+1})}$ as you showed above), hence we may pass to the limit in $(2)$ with respect to $z$ when $zto t_i$. Taking into account the continuity of $u$ at $t_i$ we get
                $$
                |u(x) - u(t_i)| leq C |x - t_i|,
                $$
                which is what was required to prove.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 1 at 21:17









                HaykHayk

                2,7121215




                2,7121215






























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