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




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      edited May 23 '17 at 12:26







      Kiuhnm

















      asked May 23 '17 at 1:24









      KiuhnmKiuhnm

      361211




      361211






















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          $begingroup$

          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











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          $begingroup$

          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

          41.7k23381




          41.7k23381












          • $begingroup$
            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


















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