Does $f(x) = frac12x^TAx$ have an L-Lipschitz continuous gradient

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Does $f(x) = frac12x^TAx$ have an L-Lipschitz continuous gradient i.e there is a constant L>0 such that
$$||nabla f(x) - nabla f(y)||_2 le L||x-y||_2$$ for any $x,y$?
I tried to derive it but could not go further than this: $$||(x^T-y^T)A||_2 le L||x-y||_2$$
I would appreciate any help.




lipschitz-functions gradient-descent non-convex-optimization




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    Does $f(x) = frac12x^TAx$ have an L-Lipschitz continuous gradient i.e there is a constant L>0 such that
    $$||nabla f(x) - nabla f(y)||_2 le L||x-y||_2$$ for any $x,y$?
    I tried to derive it but could not go further than this: $$||(x^T-y^T)A||_2 le L||x-y||_2$$
    I would appreciate any help.




    lipschitz-functions gradient-descent non-convex-optimization




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      Does $f(x) = frac12x^TAx$ have an L-Lipschitz continuous gradient i.e there is a constant L>0 such that
      $$||nabla f(x) - nabla f(y)||_2 le L||x-y||_2$$ for any $x,y$?
      I tried to derive it but could not go further than this: $$||(x^T-y^T)A||_2 le L||x-y||_2$$
      I would appreciate any help.




      lipschitz-functions gradient-descent non-convex-optimization




      share|cite|improve this question












      Does $f(x) = frac12x^TAx$ have an L-Lipschitz continuous gradient i.e there is a constant L>0 such that
      $$||nabla f(x) - nabla f(y)||_2 le L||x-y||_2$$ for any $x,y$?
      I tried to derive it but could not go further than this: $$||(x^T-y^T)A||_2 le L||x-y||_2$$
      I would appreciate any help.





      lipschitz-functions gradient-descent non-convex-optimization





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      asked Aug 8 at 18:01









      Pumpkin

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          The function is $L$-Lipschitz, where $L$ is the operator norm of $A$.
          The operator norm is defined as
          $$|A|_op := sup_x ne 0 fracA x.$$
          Indeed, from your work we have
          $$|nabla f(x) - nabla f(y)|_2 = |A(x-y)|_2 le |A|_op |x-y|_2$$
          simply by the definition of operator norm.



          One can show that the operator norm is equal to the largest singular value of $A$. Further, if $A$ is symmetric (and thus orthogonally diagonalizable), it is equal to the largest absolute eigenvalue, $max_i |lambda_i|$.






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            up vote
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            down vote



            accepted










            The function is $L$-Lipschitz, where $L$ is the operator norm of $A$.
            The operator norm is defined as
            $$|A|_op := sup_x ne 0 fracA x.$$
            Indeed, from your work we have
            $$|nabla f(x) - nabla f(y)|_2 = |A(x-y)|_2 le |A|_op |x-y|_2$$
            simply by the definition of operator norm.



            One can show that the operator norm is equal to the largest singular value of $A$. Further, if $A$ is symmetric (and thus orthogonally diagonalizable), it is equal to the largest absolute eigenvalue, $max_i |lambda_i|$.






            share|cite|improve this answer
























              up vote
              4
              down vote



              accepted










              The function is $L$-Lipschitz, where $L$ is the operator norm of $A$.
              The operator norm is defined as
              $$|A|_op := sup_x ne 0 fracA x.$$
              Indeed, from your work we have
              $$|nabla f(x) - nabla f(y)|_2 = |A(x-y)|_2 le |A|_op |x-y|_2$$
              simply by the definition of operator norm.



              One can show that the operator norm is equal to the largest singular value of $A$. Further, if $A$ is symmetric (and thus orthogonally diagonalizable), it is equal to the largest absolute eigenvalue, $max_i |lambda_i|$.






              share|cite|improve this answer






















                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                The function is $L$-Lipschitz, where $L$ is the operator norm of $A$.
                The operator norm is defined as
                $$|A|_op := sup_x ne 0 fracA x.$$
                Indeed, from your work we have
                $$|nabla f(x) - nabla f(y)|_2 = |A(x-y)|_2 le |A|_op |x-y|_2$$
                simply by the definition of operator norm.



                One can show that the operator norm is equal to the largest singular value of $A$. Further, if $A$ is symmetric (and thus orthogonally diagonalizable), it is equal to the largest absolute eigenvalue, $max_i |lambda_i|$.






                share|cite|improve this answer












                The function is $L$-Lipschitz, where $L$ is the operator norm of $A$.
                The operator norm is defined as
                $$|A|_op := sup_x ne 0 fracA x.$$
                Indeed, from your work we have
                $$|nabla f(x) - nabla f(y)|_2 = |A(x-y)|_2 le |A|_op |x-y|_2$$
                simply by the definition of operator norm.



                One can show that the operator norm is equal to the largest singular value of $A$. Further, if $A$ is symmetric (and thus orthogonally diagonalizable), it is equal to the largest absolute eigenvalue, $max_i |lambda_i|$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 8 at 18:14









                angryavian

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