Conservation of simple dot product inequalities

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Let us suppose you have a mapping $~f colon mathbbR^d to mathbbR^d$ such that
$$forall (a,b) in (mathbbR^d)^2,~langle a-b, f(a)-f(b)rangle geq 0$$



Is it true that
$$forall (a,b, c) in (mathbbR^d)^2,~langle a-c, b-crangle geq 0 Rightarrow langle f(a)-f(c), f(b)-(c)rangle geq 0$$



If no, is there an intuitive counter-example to this property ?



Discussion



In general for $(a,b,c) in (mathbbR^d)^3$



  • $langle a-c, f(a)-f(c)rangle > 0$

  • $langle b-c, f(b)-f(c)rangle > 0$

  • $langle a-c, b-crangle > 0$

does not implies anything about $langle f(a) - f(c), f(b) - f(c) rangle$.



We can maybe solve this issue by noting that $langle a-b, f(a)-f(b)rangle geq 0$ holds for all couples of $mathbbR^d$




geometry




share|cite|improve this question























    up vote
    1
    down vote

    favorite












    Let us suppose you have a mapping $~f colon mathbbR^d to mathbbR^d$ such that
    $$forall (a,b) in (mathbbR^d)^2,~langle a-b, f(a)-f(b)rangle geq 0$$



    Is it true that
    $$forall (a,b, c) in (mathbbR^d)^2,~langle a-c, b-crangle geq 0 Rightarrow langle f(a)-f(c), f(b)-(c)rangle geq 0$$



    If no, is there an intuitive counter-example to this property ?



    Discussion



    In general for $(a,b,c) in (mathbbR^d)^3$



    • $langle a-c, f(a)-f(c)rangle > 0$

    • $langle b-c, f(b)-f(c)rangle > 0$

    • $langle a-c, b-crangle > 0$

    does not implies anything about $langle f(a) - f(c), f(b) - f(c) rangle$.



    We can maybe solve this issue by noting that $langle a-b, f(a)-f(b)rangle geq 0$ holds for all couples of $mathbbR^d$




    geometry




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let us suppose you have a mapping $~f colon mathbbR^d to mathbbR^d$ such that
      $$forall (a,b) in (mathbbR^d)^2,~langle a-b, f(a)-f(b)rangle geq 0$$



      Is it true that
      $$forall (a,b, c) in (mathbbR^d)^2,~langle a-c, b-crangle geq 0 Rightarrow langle f(a)-f(c), f(b)-(c)rangle geq 0$$



      If no, is there an intuitive counter-example to this property ?



      Discussion



      In general for $(a,b,c) in (mathbbR^d)^3$



      • $langle a-c, f(a)-f(c)rangle > 0$

      • $langle b-c, f(b)-f(c)rangle > 0$

      • $langle a-c, b-crangle > 0$

      does not implies anything about $langle f(a) - f(c), f(b) - f(c) rangle$.



      We can maybe solve this issue by noting that $langle a-b, f(a)-f(b)rangle geq 0$ holds for all couples of $mathbbR^d$




      geometry




      share|cite|improve this question











      Let us suppose you have a mapping $~f colon mathbbR^d to mathbbR^d$ such that
      $$forall (a,b) in (mathbbR^d)^2,~langle a-b, f(a)-f(b)rangle geq 0$$



      Is it true that
      $$forall (a,b, c) in (mathbbR^d)^2,~langle a-c, b-crangle geq 0 Rightarrow langle f(a)-f(c), f(b)-(c)rangle geq 0$$



      If no, is there an intuitive counter-example to this property ?



      Discussion



      In general for $(a,b,c) in (mathbbR^d)^3$



      • $langle a-c, f(a)-f(c)rangle > 0$

      • $langle b-c, f(b)-f(c)rangle > 0$

      • $langle a-c, b-crangle > 0$

      does not implies anything about $langle f(a) - f(c), f(b) - f(c) rangle$.



      We can maybe solve this issue by noting that $langle a-b, f(a)-f(b)rangle geq 0$ holds for all couples of $mathbbR^d$





      geometry





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 30 at 13:53









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