Addition of two L-smooth function is also L-smooth?

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Assume $f(x)$ has an L-Lipschitz continuous gradient say $L_1$ i.e there is a constant L>0 such that
$$|nabla f(x) - nabla f(y)|_2 le L|x-y|_2$$ for any $x,y$.



Also $g(x)$ has an L-Lipschitz continuous gradient, say $L_2$. Is $f(x)+g(x)$ has an L-Lipschitz continuous gradient?



I tried to use property of L2 norm but couldn't be sure if this is correct. Since
$| x + y|_2 leq | x|_2 + | y|_2$, the $L$ of $f(x)+g(x)$ would be $L_1+L_2$



I would appreciate any help.




lipschitz-functions gradient-descent non-convex-optimization




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    Assume $f(x)$ has an L-Lipschitz continuous gradient say $L_1$ i.e there is a constant L>0 such that
    $$|nabla f(x) - nabla f(y)|_2 le L|x-y|_2$$ for any $x,y$.



    Also $g(x)$ has an L-Lipschitz continuous gradient, say $L_2$. Is $f(x)+g(x)$ has an L-Lipschitz continuous gradient?



    I tried to use property of L2 norm but couldn't be sure if this is correct. Since
    $| x + y|_2 leq | x|_2 + | y|_2$, the $L$ of $f(x)+g(x)$ would be $L_1+L_2$



    I would appreciate any help.




    lipschitz-functions gradient-descent non-convex-optimization




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Assume $f(x)$ has an L-Lipschitz continuous gradient say $L_1$ i.e there is a constant L>0 such that
      $$|nabla f(x) - nabla f(y)|_2 le L|x-y|_2$$ for any $x,y$.



      Also $g(x)$ has an L-Lipschitz continuous gradient, say $L_2$. Is $f(x)+g(x)$ has an L-Lipschitz continuous gradient?



      I tried to use property of L2 norm but couldn't be sure if this is correct. Since
      $| x + y|_2 leq | x|_2 + | y|_2$, the $L$ of $f(x)+g(x)$ would be $L_1+L_2$



      I would appreciate any help.




      lipschitz-functions gradient-descent non-convex-optimization




      share|cite|improve this question














      Assume $f(x)$ has an L-Lipschitz continuous gradient say $L_1$ i.e there is a constant L>0 such that
      $$|nabla f(x) - nabla f(y)|_2 le L|x-y|_2$$ for any $x,y$.



      Also $g(x)$ has an L-Lipschitz continuous gradient, say $L_2$. Is $f(x)+g(x)$ has an L-Lipschitz continuous gradient?



      I tried to use property of L2 norm but couldn't be sure if this is correct. Since
      $| x + y|_2 leq | x|_2 + | y|_2$, the $L$ of $f(x)+g(x)$ would be $L_1+L_2$



      I would appreciate any help.





      lipschitz-functions gradient-descent non-convex-optimization





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      share|cite|improve this question








      edited Aug 20 at 14:07









      Martin Sleziak

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      asked Aug 20 at 12:56









      Pumpkin

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          I may be wrong but this seems like a simple use of sub-additivity of the norm and additivity of the gradient:



          $||nabla big( f+g big)(x) - nabla big(f+g)(y)||_2=BigVert nabla f(x)+nabla g(x)- big( nabla f(y)+nabla g(y) big) Big Vert_2=$



          $=Big Vert big( nabla f(x)-nabla f(y) big) + big( nabla g(x)-nabla g(y) big) Big Vert_2leq big Vert nabla f(x)-nabla f(y) big Vert_2+ big Vert nabla g(x)-nabla g(y) big Vert_2leq$



          $leq L_1cdot Vert x-y Vert_2+ L_2 cdot Vert x-y Vert_2= big( L_1+L_2 big) cdot Vert x-y Vert_2 $






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            up vote
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            I may be wrong but this seems like a simple use of sub-additivity of the norm and additivity of the gradient:



            $||nabla big( f+g big)(x) - nabla big(f+g)(y)||_2=BigVert nabla f(x)+nabla g(x)- big( nabla f(y)+nabla g(y) big) Big Vert_2=$



            $=Big Vert big( nabla f(x)-nabla f(y) big) + big( nabla g(x)-nabla g(y) big) Big Vert_2leq big Vert nabla f(x)-nabla f(y) big Vert_2+ big Vert nabla g(x)-nabla g(y) big Vert_2leq$



            $leq L_1cdot Vert x-y Vert_2+ L_2 cdot Vert x-y Vert_2= big( L_1+L_2 big) cdot Vert x-y Vert_2 $






            share|cite|improve this answer
























              up vote
              0
              down vote













              I may be wrong but this seems like a simple use of sub-additivity of the norm and additivity of the gradient:



              $||nabla big( f+g big)(x) - nabla big(f+g)(y)||_2=BigVert nabla f(x)+nabla g(x)- big( nabla f(y)+nabla g(y) big) Big Vert_2=$



              $=Big Vert big( nabla f(x)-nabla f(y) big) + big( nabla g(x)-nabla g(y) big) Big Vert_2leq big Vert nabla f(x)-nabla f(y) big Vert_2+ big Vert nabla g(x)-nabla g(y) big Vert_2leq$



              $leq L_1cdot Vert x-y Vert_2+ L_2 cdot Vert x-y Vert_2= big( L_1+L_2 big) cdot Vert x-y Vert_2 $






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                I may be wrong but this seems like a simple use of sub-additivity of the norm and additivity of the gradient:



                $||nabla big( f+g big)(x) - nabla big(f+g)(y)||_2=BigVert nabla f(x)+nabla g(x)- big( nabla f(y)+nabla g(y) big) Big Vert_2=$



                $=Big Vert big( nabla f(x)-nabla f(y) big) + big( nabla g(x)-nabla g(y) big) Big Vert_2leq big Vert nabla f(x)-nabla f(y) big Vert_2+ big Vert nabla g(x)-nabla g(y) big Vert_2leq$



                $leq L_1cdot Vert x-y Vert_2+ L_2 cdot Vert x-y Vert_2= big( L_1+L_2 big) cdot Vert x-y Vert_2 $






                share|cite|improve this answer












                I may be wrong but this seems like a simple use of sub-additivity of the norm and additivity of the gradient:



                $||nabla big( f+g big)(x) - nabla big(f+g)(y)||_2=BigVert nabla f(x)+nabla g(x)- big( nabla f(y)+nabla g(y) big) Big Vert_2=$



                $=Big Vert big( nabla f(x)-nabla f(y) big) + big( nabla g(x)-nabla g(y) big) Big Vert_2leq big Vert nabla f(x)-nabla f(y) big Vert_2+ big Vert nabla g(x)-nabla g(y) big Vert_2leq$



                $leq L_1cdot Vert x-y Vert_2+ L_2 cdot Vert x-y Vert_2= big( L_1+L_2 big) cdot Vert x-y Vert_2 $







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 20 at 13:19









                Keen-ameteur

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