Show an inequality












1












$begingroup$


Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$






inequality hilbert-spaces







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$endgroup$








  • 1




    $begingroup$
    Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 17:02
















1












$begingroup$


Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$






inequality hilbert-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 17:02














1












1








1





$begingroup$


Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$






inequality hilbert-spaces







share|cite|improve this question









$endgroup$




Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$






inequality hilbert-spaces




inequality hilbert-spaces






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asked Jan 7 at 16:59









PhilipPhilip

826




826








  • 1




    $begingroup$
    Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 17:02














  • 1




    $begingroup$
    Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 17:02








1




1




$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02




$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02










2 Answers
2






active

oldest

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$begingroup$

Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)



Take the square root, then you get
$$|x+Cy|geq |x|$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$



    $||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$



    This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$






    share|cite|improve this answer









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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

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      3












      $begingroup$

      Observe that, for given $Cin mathbb{R},$
      $$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
      due to the orthogonality ($xperp y implies left<x,y right>=0.$)



      Take the square root, then you get
      $$|x+Cy|geq |x|$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Observe that, for given $Cin mathbb{R},$
        $$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
        due to the orthogonality ($xperp y implies left<x,y right>=0.$)



        Take the square root, then you get
        $$|x+Cy|geq |x|$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Observe that, for given $Cin mathbb{R},$
          $$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
          due to the orthogonality ($xperp y implies left<x,y right>=0.$)



          Take the square root, then you get
          $$|x+Cy|geq |x|$$






          share|cite|improve this answer









          $endgroup$



          Observe that, for given $Cin mathbb{R},$
          $$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
          due to the orthogonality ($xperp y implies left<x,y right>=0.$)



          Take the square root, then you get
          $$|x+Cy|geq |x|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 17:10









          LeBLeB

          1,080317




          1,080317























              1












              $begingroup$

              $||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$



              $||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$



              This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$



                $||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$



                This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$



                  $||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$



                  This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$






                  share|cite|improve this answer









                  $endgroup$



                  $||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$



                  $||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$



                  This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 17:37









                  ecrinecrin

                  3547




                  3547






























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