Bounded Gradient implies Lipschitz proof with the mean value theorem












1












$begingroup$


Let $f:mathbb{R}^n to mathbb{R}$ with $|| nabla f(x)|| leq M$ (say it is the Euclidean norm), then f is Lipschitz.



I have seen proofs that do this for the case where $f:mathbb{R} to mathbb{R}$ by applying the mean value theorem. I am wondering if there is a proof available that shows how the mean value theorem is applied to the problem for a function from $f:mathbb{R}^n to mathbb{R}$?






multivariable-calculus lipschitz-functions







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$endgroup$

















    1












    $begingroup$


    Let $f:mathbb{R}^n to mathbb{R}$ with $|| nabla f(x)|| leq M$ (say it is the Euclidean norm), then f is Lipschitz.



    I have seen proofs that do this for the case where $f:mathbb{R} to mathbb{R}$ by applying the mean value theorem. I am wondering if there is a proof available that shows how the mean value theorem is applied to the problem for a function from $f:mathbb{R}^n to mathbb{R}$?






    multivariable-calculus lipschitz-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $f:mathbb{R}^n to mathbb{R}$ with $|| nabla f(x)|| leq M$ (say it is the Euclidean norm), then f is Lipschitz.



      I have seen proofs that do this for the case where $f:mathbb{R} to mathbb{R}$ by applying the mean value theorem. I am wondering if there is a proof available that shows how the mean value theorem is applied to the problem for a function from $f:mathbb{R}^n to mathbb{R}$?






      multivariable-calculus lipschitz-functions







      share|cite|improve this question









      $endgroup$




      Let $f:mathbb{R}^n to mathbb{R}$ with $|| nabla f(x)|| leq M$ (say it is the Euclidean norm), then f is Lipschitz.



      I have seen proofs that do this for the case where $f:mathbb{R} to mathbb{R}$ by applying the mean value theorem. I am wondering if there is a proof available that shows how the mean value theorem is applied to the problem for a function from $f:mathbb{R}^n to mathbb{R}$?






      multivariable-calculus lipschitz-functions




      multivariable-calculus lipschitz-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 22:40









      geo17geo17

      1038




      1038






















          2 Answers
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          active

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          1












          $begingroup$

          I don't understand where is your problem. This is exactcly the same proof in higher-dimension.



          By the mean value theorem we have :



          $$| f(x) - f(y) | leq sup_{x in mathbb{R}^n} | nabla f(x) | |x -y | leq M | x - y|$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
            $endgroup$
            – geo17
            Jan 24 at 23:03



















          0












          $begingroup$

          Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:



          Let



          $x, y in Bbb R^n; tag 1$



          let



          $gamma:[0, 1] to Bbb R^n tag 2$



          be given by



          $gamma(t) = x + t(y - x); tag 3$



          then $gamma(t)$ is a line segment 'twixt



          $gamma(0) = x ; text{and} ; gamma(1) = y; tag 4$



          then



          $f(y) - f(x) = f(gamma(1)) - f(gamma(0))$
          $= displaystyle int_0^1 dfrac{df(gamma(t))}{dt} ; dt = int_0^1 nabla f(gamma(t)) cdot dot gamma(t) ; dt = int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt tag 5$



          therefore,
          22nd
          $vert f(y) - f(x) vert = left vert displaystyle int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt right vert le displaystyle int_0^1 vert nabla f(gamma(t)) vert vert y - x vert ; dt le vert y - x vert int_0^1 M ; dt = Mvert y - x vert, tag 6$



          which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OEDelta$.



          It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $vert nabla f(x) vert le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write



          $f(y) - f(x) = f(gamma(1)) - f(gamma(0)) = (f(gamma(r))'(1 - 0) = (f(gamma(r))', 0 < r < 1; tag 7$



          and note that



          $f(gamma(r))' = nabla f(gamma(r)) cdot dot gamma(r) = nabla f(gamma(r)) cdot (y - x); tag 8$



          if we combine (7) and (8) and take norms, the desired result is obtained.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

            oldest

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            active

            oldest

            votes









            1












            $begingroup$

            I don't understand where is your problem. This is exactcly the same proof in higher-dimension.



            By the mean value theorem we have :



            $$| f(x) - f(y) | leq sup_{x in mathbb{R}^n} | nabla f(x) | |x -y | leq M | x - y|$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
              $endgroup$
              – geo17
              Jan 24 at 23:03
















            1












            $begingroup$

            I don't understand where is your problem. This is exactcly the same proof in higher-dimension.



            By the mean value theorem we have :



            $$| f(x) - f(y) | leq sup_{x in mathbb{R}^n} | nabla f(x) | |x -y | leq M | x - y|$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
              $endgroup$
              – geo17
              Jan 24 at 23:03














            1












            1








            1





            $begingroup$

            I don't understand where is your problem. This is exactcly the same proof in higher-dimension.



            By the mean value theorem we have :



            $$| f(x) - f(y) | leq sup_{x in mathbb{R}^n} | nabla f(x) | |x -y | leq M | x - y|$$






            share|cite|improve this answer









            $endgroup$



            I don't understand where is your problem. This is exactcly the same proof in higher-dimension.



            By the mean value theorem we have :



            $$| f(x) - f(y) | leq sup_{x in mathbb{R}^n} | nabla f(x) | |x -y | leq M | x - y|$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 22:56









            ThinkingThinking

            1,22716




            1,22716








            • 1




              $begingroup$
              I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
              $endgroup$
              – geo17
              Jan 24 at 23:03














            • 1




              $begingroup$
              I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
              $endgroup$
              – geo17
              Jan 24 at 23:03








            1




            1




            $begingroup$
            I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
            $endgroup$
            – geo17
            Jan 24 at 23:03




            $begingroup$
            I wasn't sure where the first inequality came from, now I see it is from considering the line $(1-t)x + ty$ and then applying the Cauchy-Schwarz inequality
            $endgroup$
            – geo17
            Jan 24 at 23:03











            0












            $begingroup$

            Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:



            Let



            $x, y in Bbb R^n; tag 1$



            let



            $gamma:[0, 1] to Bbb R^n tag 2$



            be given by



            $gamma(t) = x + t(y - x); tag 3$



            then $gamma(t)$ is a line segment 'twixt



            $gamma(0) = x ; text{and} ; gamma(1) = y; tag 4$



            then



            $f(y) - f(x) = f(gamma(1)) - f(gamma(0))$
            $= displaystyle int_0^1 dfrac{df(gamma(t))}{dt} ; dt = int_0^1 nabla f(gamma(t)) cdot dot gamma(t) ; dt = int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt tag 5$



            therefore,
            22nd
            $vert f(y) - f(x) vert = left vert displaystyle int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt right vert le displaystyle int_0^1 vert nabla f(gamma(t)) vert vert y - x vert ; dt le vert y - x vert int_0^1 M ; dt = Mvert y - x vert, tag 6$



            which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OEDelta$.



            It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $vert nabla f(x) vert le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write



            $f(y) - f(x) = f(gamma(1)) - f(gamma(0)) = (f(gamma(r))'(1 - 0) = (f(gamma(r))', 0 < r < 1; tag 7$



            and note that



            $f(gamma(r))' = nabla f(gamma(r)) cdot dot gamma(r) = nabla f(gamma(r)) cdot (y - x); tag 8$



            if we combine (7) and (8) and take norms, the desired result is obtained.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:



              Let



              $x, y in Bbb R^n; tag 1$



              let



              $gamma:[0, 1] to Bbb R^n tag 2$



              be given by



              $gamma(t) = x + t(y - x); tag 3$



              then $gamma(t)$ is a line segment 'twixt



              $gamma(0) = x ; text{and} ; gamma(1) = y; tag 4$



              then



              $f(y) - f(x) = f(gamma(1)) - f(gamma(0))$
              $= displaystyle int_0^1 dfrac{df(gamma(t))}{dt} ; dt = int_0^1 nabla f(gamma(t)) cdot dot gamma(t) ; dt = int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt tag 5$



              therefore,
              22nd
              $vert f(y) - f(x) vert = left vert displaystyle int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt right vert le displaystyle int_0^1 vert nabla f(gamma(t)) vert vert y - x vert ; dt le vert y - x vert int_0^1 M ; dt = Mvert y - x vert, tag 6$



              which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OEDelta$.



              It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $vert nabla f(x) vert le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write



              $f(y) - f(x) = f(gamma(1)) - f(gamma(0)) = (f(gamma(r))'(1 - 0) = (f(gamma(r))', 0 < r < 1; tag 7$



              and note that



              $f(gamma(r))' = nabla f(gamma(r)) cdot dot gamma(r) = nabla f(gamma(r)) cdot (y - x); tag 8$



              if we combine (7) and (8) and take norms, the desired result is obtained.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:



                Let



                $x, y in Bbb R^n; tag 1$



                let



                $gamma:[0, 1] to Bbb R^n tag 2$



                be given by



                $gamma(t) = x + t(y - x); tag 3$



                then $gamma(t)$ is a line segment 'twixt



                $gamma(0) = x ; text{and} ; gamma(1) = y; tag 4$



                then



                $f(y) - f(x) = f(gamma(1)) - f(gamma(0))$
                $= displaystyle int_0^1 dfrac{df(gamma(t))}{dt} ; dt = int_0^1 nabla f(gamma(t)) cdot dot gamma(t) ; dt = int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt tag 5$



                therefore,
                22nd
                $vert f(y) - f(x) vert = left vert displaystyle int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt right vert le displaystyle int_0^1 vert nabla f(gamma(t)) vert vert y - x vert ; dt le vert y - x vert int_0^1 M ; dt = Mvert y - x vert, tag 6$



                which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OEDelta$.



                It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $vert nabla f(x) vert le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write



                $f(y) - f(x) = f(gamma(1)) - f(gamma(0)) = (f(gamma(r))'(1 - 0) = (f(gamma(r))', 0 < r < 1; tag 7$



                and note that



                $f(gamma(r))' = nabla f(gamma(r)) cdot dot gamma(r) = nabla f(gamma(r)) cdot (y - x); tag 8$



                if we combine (7) and (8) and take norms, the desired result is obtained.






                share|cite|improve this answer









                $endgroup$



                Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:



                Let



                $x, y in Bbb R^n; tag 1$



                let



                $gamma:[0, 1] to Bbb R^n tag 2$



                be given by



                $gamma(t) = x + t(y - x); tag 3$



                then $gamma(t)$ is a line segment 'twixt



                $gamma(0) = x ; text{and} ; gamma(1) = y; tag 4$



                then



                $f(y) - f(x) = f(gamma(1)) - f(gamma(0))$
                $= displaystyle int_0^1 dfrac{df(gamma(t))}{dt} ; dt = int_0^1 nabla f(gamma(t)) cdot dot gamma(t) ; dt = int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt tag 5$



                therefore,
                22nd
                $vert f(y) - f(x) vert = left vert displaystyle int_0^1 nabla f(gamma(t)) cdot (y - x) ; dt right vert le displaystyle int_0^1 vert nabla f(gamma(t)) vert vert y - x vert ; dt le vert y - x vert int_0^1 M ; dt = Mvert y - x vert, tag 6$



                which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OEDelta$.



                It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $vert nabla f(x) vert le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write



                $f(y) - f(x) = f(gamma(1)) - f(gamma(0)) = (f(gamma(r))'(1 - 0) = (f(gamma(r))', 0 < r < 1; tag 7$



                and note that



                $f(gamma(r))' = nabla f(gamma(r)) cdot dot gamma(r) = nabla f(gamma(r)) cdot (y - x); tag 8$



                if we combine (7) and (8) and take norms, the desired result is obtained.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 7:20









                Robert LewisRobert Lewis

                48.1k23167




                48.1k23167






























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