$h(z)=|z-a|. |z-b|. |z-c|$, max value of $h$ is attained












3












$begingroup$


Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$



Then max value of $h$




  1. is not attained at any point of $Delta$


  2. is attained at an interior point of $Delta$


  3. is attained at the centre of gravity of $Delta$


  4. is attained at the boundary point of $Delta$



$4$ is correct due to Maximum Modulas Principle.Just confirm me please.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:44






  • 4




    $begingroup$
    well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
    $endgroup$
    – Markov
    Jan 6 '14 at 14:47








  • 1




    $begingroup$
    Ah, then you have a correct argument.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:51
















3












$begingroup$


Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$



Then max value of $h$




  1. is not attained at any point of $Delta$


  2. is attained at an interior point of $Delta$


  3. is attained at the centre of gravity of $Delta$


  4. is attained at the boundary point of $Delta$



$4$ is correct due to Maximum Modulas Principle.Just confirm me please.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:44






  • 4




    $begingroup$
    well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
    $endgroup$
    – Markov
    Jan 6 '14 at 14:47








  • 1




    $begingroup$
    Ah, then you have a correct argument.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:51














3












3








3


2



$begingroup$


Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$



Then max value of $h$




  1. is not attained at any point of $Delta$


  2. is attained at an interior point of $Delta$


  3. is attained at the centre of gravity of $Delta$


  4. is attained at the boundary point of $Delta$



$4$ is correct due to Maximum Modulas Principle.Just confirm me please.






complex-analysis







share|cite|improve this question











$endgroup$




Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$



Then max value of $h$




  1. is not attained at any point of $Delta$


  2. is attained at an interior point of $Delta$


  3. is attained at the centre of gravity of $Delta$


  4. is attained at the boundary point of $Delta$



$4$ is correct due to Maximum Modulas Principle.Just confirm me please.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 12 '15 at 20:15









Empty

8,10252660




8,10252660










asked Jan 6 '14 at 14:43









MarkovMarkov

17.2k957178




17.2k957178








  • 1




    $begingroup$
    $h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:44






  • 4




    $begingroup$
    well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
    $endgroup$
    – Markov
    Jan 6 '14 at 14:47








  • 1




    $begingroup$
    Ah, then you have a correct argument.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:51














  • 1




    $begingroup$
    $h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:44






  • 4




    $begingroup$
    well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
    $endgroup$
    – Markov
    Jan 6 '14 at 14:47








  • 1




    $begingroup$
    Ah, then you have a correct argument.
    $endgroup$
    – Daniel Fischer♦
    Jan 6 '14 at 14:51








1




1




$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44




$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44




4




4




$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47






$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47






1




1




$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51




$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

Except your question, we can also answer to the maximum value of the function $h(x)$.



Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$



Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$



If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$



We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.



    The specific triangular shape does not matter; the same holds for any domain.




    I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
      $endgroup$
      – Empty
      Oct 12 '15 at 20:13













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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Except your question, we can also answer to the maximum value of the function $h(x)$.



    Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$



    Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
    On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$



    If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$



    We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Except your question, we can also answer to the maximum value of the function $h(x)$.



      Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$



      Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
      On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$



      If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$



      We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Except your question, we can also answer to the maximum value of the function $h(x)$.



        Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$



        Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
        On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$



        If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$



        We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$






        share|cite|improve this answer









        $endgroup$



        Except your question, we can also answer to the maximum value of the function $h(x)$.



        Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$



        Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
        On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$



        If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$



        We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 15:55









        EminEmin

        1,36021330




        1,36021330























            2












            $begingroup$

            Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.



            The specific triangular shape does not matter; the same holds for any domain.




            I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
              $endgroup$
              – Empty
              Oct 12 '15 at 20:13


















            2












            $begingroup$

            Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.



            The specific triangular shape does not matter; the same holds for any domain.




            I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
              $endgroup$
              – Empty
              Oct 12 '15 at 20:13
















            2












            2








            2





            $begingroup$

            Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.



            The specific triangular shape does not matter; the same holds for any domain.




            I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.






            share|cite|improve this answer









            $endgroup$



            Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.



            The specific triangular shape does not matter; the same holds for any domain.




            I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 '14 at 15:58









            Post No BullsPost No Bulls

            6,42321682




            6,42321682












            • $begingroup$
              @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
              $endgroup$
              – Empty
              Oct 12 '15 at 20:13




















            • $begingroup$
              @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
              $endgroup$
              – Empty
              Oct 12 '15 at 20:13


















            $begingroup$
            @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
            $endgroup$
            – Empty
            Oct 12 '15 at 20:13






            $begingroup$
            @@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
            $endgroup$
            – Empty
            Oct 12 '15 at 20:13




















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