Proving inequality involving scalar product on $mathbb{R}^n$












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Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.



I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.



For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.



Any hint?



Thanks a lot in advance






normed-spaces inner-product-space supremum-and-infimum







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    0












    $begingroup$


    Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.



    I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.



    For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.



    Any hint?



    Thanks a lot in advance






    normed-spaces inner-product-space supremum-and-infimum







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.



      I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.



      For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.



      Any hint?



      Thanks a lot in advance






      normed-spaces inner-product-space supremum-and-infimum







      share|cite|improve this question











      $endgroup$




      Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.



      I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.



      For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.



      Any hint?



      Thanks a lot in advance






      normed-spaces inner-product-space supremum-and-infimum




      normed-spaces inner-product-space supremum-and-infimum






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      edited Jan 27 at 22:17









      max_zorn

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      3,43061429










      asked Jan 27 at 21:35









      eleguitareleguitar

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          $begingroup$

          When $leftlVert yrightrVertge 1$, we find that
          $$
          sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
          $$           When $leftlVert yrightrVertle 1$, we can observe
          $$
          sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
          $$           Thus we have
          $$
          sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
          $$           as desired.






          share|cite|improve this answer











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            $begingroup$

            When $leftlVert yrightrVertge 1$, we find that
            $$
            sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
            $$             When $leftlVert yrightrVertle 1$, we can observe
            $$
            sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
            $$             Thus we have
            $$
            sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
            $$             as desired.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              When $leftlVert yrightrVertge 1$, we find that
              $$
              sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
              $$               When $leftlVert yrightrVertle 1$, we can observe
              $$
              sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
              $$               Thus we have
              $$
              sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
              $$               as desired.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                When $leftlVert yrightrVertge 1$, we find that
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
                $$                 When $leftlVert yrightrVertle 1$, we can observe
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
                $$                 Thus we have
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
                $$                 as desired.






                share|cite|improve this answer











                $endgroup$



                When $leftlVert yrightrVertge 1$, we find that
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
                $$                 When $leftlVert yrightrVertle 1$, we can observe
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
                $$                 Thus we have
                $$
                sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
                $$                 as desired.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 27 at 21:55

























                answered Jan 27 at 21:47









                SongSong

                18.5k21651




                18.5k21651






























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