From Hessian to Lipschitz continuity












0












$begingroup$


Let $f$ be convex and twice differentiable. I can't prove that
$$
nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
$$
I only established the converse.



edit:



There exists a $zintext{conv}{x,y}$ such that
$$
nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
$$
By multiplying on the left by $(y-x)^T$ we get
$$
begin{align}
(y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
&leq (y-x)^T nabla f(x) + L||y-x||_2^2
end{align}
$$
Now how do we prove that
$$
(nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
$$
I think I'm missing something very easy.






convex-analysis







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $f$ be convex and twice differentiable. I can't prove that
    $$
    nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
    $$
    I only established the converse.



    edit:



    There exists a $zintext{conv}{x,y}$ such that
    $$
    nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
    $$
    By multiplying on the left by $(y-x)^T$ we get
    $$
    begin{align}
    (y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
    &leq (y-x)^T nabla f(x) + L||y-x||_2^2
    end{align}
    $$
    Now how do we prove that
    $$
    (nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
    $$
    I think I'm missing something very easy.






    convex-analysis







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $f$ be convex and twice differentiable. I can't prove that
      $$
      nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
      $$
      I only established the converse.



      edit:



      There exists a $zintext{conv}{x,y}$ such that
      $$
      nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
      $$
      By multiplying on the left by $(y-x)^T$ we get
      $$
      begin{align}
      (y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
      &leq (y-x)^T nabla f(x) + L||y-x||_2^2
      end{align}
      $$
      Now how do we prove that
      $$
      (nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
      $$
      I think I'm missing something very easy.






      convex-analysis







      share|cite|improve this question











      $endgroup$




      Let $f$ be convex and twice differentiable. I can't prove that
      $$
      nabla^2f(x)preceq LI implies ||nabla f(x)-nabla f(y)||_2leq L||x-y||_2
      $$
      I only established the converse.



      edit:



      There exists a $zintext{conv}{x,y}$ such that
      $$
      nabla f(y) = nabla f(x) + nabla^2 f(z)(y-x)
      $$
      By multiplying on the left by $(y-x)^T$ we get
      $$
      begin{align}
      (y-x)^T nabla f(y) &= (y-x)^T nabla f(x) + (y-x)^T nabla^2 f(z)(y-x)\
      &leq (y-x)^T nabla f(x) + L||y-x||_2^2
      end{align}
      $$
      Now how do we prove that
      $$
      (nabla f(y) - nabla f(x))^T (y-x) leq L||y-x||_2^2 implies nabla ftext{ is }Ltext{-lipschitz continuous}?
      $$
      I think I'm missing something very easy.






      convex-analysis




      convex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 23 '17 at 12:26







      Kiuhnm

















      asked May 23 '17 at 1:24









      KiuhnmKiuhnm

      361210




      361210






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
          begin{align}
          |nabla f(y) - nabla f(x)|
          &=
          |g(1) - g(0)|
          \
          &= left| int_0^1 g'(t) mathop{dt}right|
          \
          &le int_0^1 |g'(t)| mathop{dt}
          \
          &= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
          \
          &le int_0^1 L |x-y| mathop{dt}
          \
          &= L|x-y|.
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What property do you use for the last $leq$? Thanks.
            $endgroup$
            – Kiuhnm
            May 23 '17 at 23:06








          • 1




            $begingroup$
            @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
            $endgroup$
            – angryavian
            May 23 '17 at 23:21










          • $begingroup$
            Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
            $endgroup$
            – Kiuhnm
            May 24 '17 at 0:12








          • 1




            $begingroup$
            Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
            $endgroup$
            – angryavian
            May 24 '17 at 1:33











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2292769%2ffrom-hessian-to-lipschitz-continuity%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
          begin{align}
          |nabla f(y) - nabla f(x)|
          &=
          |g(1) - g(0)|
          \
          &= left| int_0^1 g'(t) mathop{dt}right|
          \
          &le int_0^1 |g'(t)| mathop{dt}
          \
          &= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
          \
          &le int_0^1 L |x-y| mathop{dt}
          \
          &= L|x-y|.
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What property do you use for the last $leq$? Thanks.
            $endgroup$
            – Kiuhnm
            May 23 '17 at 23:06








          • 1




            $begingroup$
            @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
            $endgroup$
            – angryavian
            May 23 '17 at 23:21










          • $begingroup$
            Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
            $endgroup$
            – Kiuhnm
            May 24 '17 at 0:12








          • 1




            $begingroup$
            Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
            $endgroup$
            – angryavian
            May 24 '17 at 1:33
















          1












          $begingroup$

          Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
          begin{align}
          |nabla f(y) - nabla f(x)|
          &=
          |g(1) - g(0)|
          \
          &= left| int_0^1 g'(t) mathop{dt}right|
          \
          &le int_0^1 |g'(t)| mathop{dt}
          \
          &= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
          \
          &le int_0^1 L |x-y| mathop{dt}
          \
          &= L|x-y|.
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What property do you use for the last $leq$? Thanks.
            $endgroup$
            – Kiuhnm
            May 23 '17 at 23:06








          • 1




            $begingroup$
            @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
            $endgroup$
            – angryavian
            May 23 '17 at 23:21










          • $begingroup$
            Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
            $endgroup$
            – Kiuhnm
            May 24 '17 at 0:12








          • 1




            $begingroup$
            Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
            $endgroup$
            – angryavian
            May 24 '17 at 1:33














          1












          1








          1





          $begingroup$

          Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
          begin{align}
          |nabla f(y) - nabla f(x)|
          &=
          |g(1) - g(0)|
          \
          &= left| int_0^1 g'(t) mathop{dt}right|
          \
          &le int_0^1 |g'(t)| mathop{dt}
          \
          &= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
          \
          &le int_0^1 L |x-y| mathop{dt}
          \
          &= L|x-y|.
          end{align}






          share|cite|improve this answer









          $endgroup$



          Let $g(t) = nabla f(y + t(x-y))$ and note $g'(t) = [nabla^2 f(y + t(x-y))] (x-y)$. By the fundamental theorem of calculus,
          begin{align}
          |nabla f(y) - nabla f(x)|
          &=
          |g(1) - g(0)|
          \
          &= left| int_0^1 g'(t) mathop{dt}right|
          \
          &le int_0^1 |g'(t)| mathop{dt}
          \
          &= int_0^1 |[nabla^2 f(y + t(x-y))] (x-y)| mathop{dt}
          \
          &le int_0^1 L |x-y| mathop{dt}
          \
          &= L|x-y|.
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 23 '17 at 21:36









          angryavianangryavian

          41.4k23380




          41.4k23380












          • $begingroup$
            What property do you use for the last $leq$? Thanks.
            $endgroup$
            – Kiuhnm
            May 23 '17 at 23:06








          • 1




            $begingroup$
            @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
            $endgroup$
            – angryavian
            May 23 '17 at 23:21










          • $begingroup$
            Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
            $endgroup$
            – Kiuhnm
            May 24 '17 at 0:12








          • 1




            $begingroup$
            Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
            $endgroup$
            – angryavian
            May 24 '17 at 1:33


















          • $begingroup$
            What property do you use for the last $leq$? Thanks.
            $endgroup$
            – Kiuhnm
            May 23 '17 at 23:06








          • 1




            $begingroup$
            @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
            $endgroup$
            – angryavian
            May 23 '17 at 23:21










          • $begingroup$
            Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
            $endgroup$
            – Kiuhnm
            May 24 '17 at 0:12








          • 1




            $begingroup$
            Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
            $endgroup$
            – angryavian
            May 24 '17 at 1:33
















          $begingroup$
          What property do you use for the last $leq$? Thanks.
          $endgroup$
          – Kiuhnm
          May 23 '17 at 23:06






          $begingroup$
          What property do you use for the last $leq$? Thanks.
          $endgroup$
          – Kiuhnm
          May 23 '17 at 23:06






          1




          1




          $begingroup$
          @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
          $endgroup$
          – angryavian
          May 23 '17 at 23:21




          $begingroup$
          @Kiuhnm Use $nabla^2 f(y+t(x-y)) preceq LI$.
          $endgroup$
          – angryavian
          May 23 '17 at 23:21












          $begingroup$
          Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
          $endgroup$
          – Kiuhnm
          May 24 '17 at 0:12






          $begingroup$
          Yes, but are you using $|Ax|leq |A||x|$? Is this always true?
          $endgroup$
          – Kiuhnm
          May 24 '17 at 0:12






          1




          1




          $begingroup$
          Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
          $endgroup$
          – angryavian
          May 24 '17 at 1:33




          $begingroup$
          Since $f$ is convex, the eigenvalues of $nabla^2 f(u)$ are nonnegative. The condition $preceq LI$ means the eigenvalues are upper bounded by $L$. Thus the operator norm of the Hessian is $le L$.
          $endgroup$
          – angryavian
          May 24 '17 at 1:33


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2292769%2ffrom-hessian-to-lipschitz-continuity%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more