Assuming that $f$ is a bounded linear function on $X$, compute $|f|$.












1












$begingroup$


Let $X$ be a finite dimensional linear space and let ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}
Now, we define $|cdot|$ by begin{align} |x|=left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}.end{align}
Assuming that $f$ is a bounded linear function on $X$, I want compute $|f|$.



MY TRIAL



Since $f$ is a bounded linear function on $X$, then there exists $Kgeq 0$ such that
begin{align} |f(x)|leq K|x|,;;forall;xin X.end{align}
So, taking $sup$ over $|x|leq 1,$ we get
begin{align} |f|=suplimits_{|x|leq 1}|f(x)|leq K,;;forall;xin X.end{align}
I might be wrong. Kindly check if I'm wrong or right. If I'm wrong, can you give me an idea of what to do?






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't use the same definition of a bounded function... Are you sure of yours?
    $endgroup$
    – Damien
    Dec 27 '18 at 8:22










  • $begingroup$
    @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    @Damien: Which do you use?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
    $endgroup$
    – Damien
    Dec 27 '18 at 8:35










  • $begingroup$
    @Damien: No problems! We all learn!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:51
















1












$begingroup$


Let $X$ be a finite dimensional linear space and let ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}
Now, we define $|cdot|$ by begin{align} |x|=left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}.end{align}
Assuming that $f$ is a bounded linear function on $X$, I want compute $|f|$.



MY TRIAL



Since $f$ is a bounded linear function on $X$, then there exists $Kgeq 0$ such that
begin{align} |f(x)|leq K|x|,;;forall;xin X.end{align}
So, taking $sup$ over $|x|leq 1,$ we get
begin{align} |f|=suplimits_{|x|leq 1}|f(x)|leq K,;;forall;xin X.end{align}
I might be wrong. Kindly check if I'm wrong or right. If I'm wrong, can you give me an idea of what to do?






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't use the same definition of a bounded function... Are you sure of yours?
    $endgroup$
    – Damien
    Dec 27 '18 at 8:22










  • $begingroup$
    @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    @Damien: Which do you use?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
    $endgroup$
    – Damien
    Dec 27 '18 at 8:35










  • $begingroup$
    @Damien: No problems! We all learn!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:51














1












1








1


1



$begingroup$


Let $X$ be a finite dimensional linear space and let ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}
Now, we define $|cdot|$ by begin{align} |x|=left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}.end{align}
Assuming that $f$ is a bounded linear function on $X$, I want compute $|f|$.



MY TRIAL



Since $f$ is a bounded linear function on $X$, then there exists $Kgeq 0$ such that
begin{align} |f(x)|leq K|x|,;;forall;xin X.end{align}
So, taking $sup$ over $|x|leq 1,$ we get
begin{align} |f|=suplimits_{|x|leq 1}|f(x)|leq K,;;forall;xin X.end{align}
I might be wrong. Kindly check if I'm wrong or right. If I'm wrong, can you give me an idea of what to do?






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$




Let $X$ be a finite dimensional linear space and let ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}
Now, we define $|cdot|$ by begin{align} |x|=left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}.end{align}
Assuming that $f$ is a bounded linear function on $X$, I want compute $|f|$.



MY TRIAL



Since $f$ is a bounded linear function on $X$, then there exists $Kgeq 0$ such that
begin{align} |f(x)|leq K|x|,;;forall;xin X.end{align}
So, taking $sup$ over $|x|leq 1,$ we get
begin{align} |f|=suplimits_{|x|leq 1}|f(x)|leq K,;;forall;xin X.end{align}
I might be wrong. Kindly check if I'm wrong or right. If I'm wrong, can you give me an idea of what to do?






functional-analysis normed-spaces




functional-analysis normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 8:31







Omojola Micheal

















asked Dec 27 '18 at 8:12









Omojola MichealOmojola Micheal

1,901324




1,901324












  • $begingroup$
    I don't use the same definition of a bounded function... Are you sure of yours?
    $endgroup$
    – Damien
    Dec 27 '18 at 8:22










  • $begingroup$
    @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    @Damien: Which do you use?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
    $endgroup$
    – Damien
    Dec 27 '18 at 8:35










  • $begingroup$
    @Damien: No problems! We all learn!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:51


















  • $begingroup$
    I don't use the same definition of a bounded function... Are you sure of yours?
    $endgroup$
    – Damien
    Dec 27 '18 at 8:22










  • $begingroup$
    @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    @Damien: Which do you use?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:24










  • $begingroup$
    It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
    $endgroup$
    – Damien
    Dec 27 '18 at 8:35










  • $begingroup$
    @Damien: No problems! We all learn!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 8:51
















$begingroup$
I don't use the same definition of a bounded function... Are you sure of yours?
$endgroup$
– Damien
Dec 27 '18 at 8:22




$begingroup$
I don't use the same definition of a bounded function... Are you sure of yours?
$endgroup$
– Damien
Dec 27 '18 at 8:22












$begingroup$
@Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:24




$begingroup$
@Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:24












$begingroup$
@Damien: Which do you use?
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:24




$begingroup$
@Damien: Which do you use?
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:24












$begingroup$
It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
$endgroup$
– Damien
Dec 27 '18 at 8:35




$begingroup$
It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry
$endgroup$
– Damien
Dec 27 '18 at 8:35












$begingroup$
@Damien: No problems! We all learn!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:51




$begingroup$
@Damien: No problems! We all learn!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 8:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

$|f|=sup {frac {|f(sum a_ie_i)|} {|sum a_ie_i|}}leqsqrt {sum |f(e_i)|^{2}}$ by C-S inequality. The exact value of $|f|$ is not easy to write and it is not $sqrt {sum |f(e_i)|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $sqrt {sum |f(e_i)|^{2}}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
    $endgroup$
    – Damien
    Dec 27 '18 at 8:52






  • 1




    $begingroup$
    @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 9:09



















0












$begingroup$

I guess it should be:
begin{align} |f|&=suplimits_{xneq 0}dfrac{|f(x)|}{|x|}\&=suplimits_{alpha_ineq 0}dfrac{left|f(sum^{n}_{i=1}alpha_i e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}\&=suplimits_{alpha_ineq 0}dfrac{left|sum^{n}_{i=1}alpha_i f(e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}.end{align}






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $|f|=sup {frac {|f(sum a_ie_i)|} {|sum a_ie_i|}}leqsqrt {sum |f(e_i)|^{2}}$ by C-S inequality. The exact value of $|f|$ is not easy to write and it is not $sqrt {sum |f(e_i)|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $sqrt {sum |f(e_i)|^{2}}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
      $endgroup$
      – Damien
      Dec 27 '18 at 8:52






    • 1




      $begingroup$
      @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
      $endgroup$
      – Kavi Rama Murthy
      Dec 27 '18 at 9:09
















    1












    $begingroup$

    $|f|=sup {frac {|f(sum a_ie_i)|} {|sum a_ie_i|}}leqsqrt {sum |f(e_i)|^{2}}$ by C-S inequality. The exact value of $|f|$ is not easy to write and it is not $sqrt {sum |f(e_i)|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $sqrt {sum |f(e_i)|^{2}}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
      $endgroup$
      – Damien
      Dec 27 '18 at 8:52






    • 1




      $begingroup$
      @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
      $endgroup$
      – Kavi Rama Murthy
      Dec 27 '18 at 9:09














    1












    1








    1





    $begingroup$

    $|f|=sup {frac {|f(sum a_ie_i)|} {|sum a_ie_i|}}leqsqrt {sum |f(e_i)|^{2}}$ by C-S inequality. The exact value of $|f|$ is not easy to write and it is not $sqrt {sum |f(e_i)|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $sqrt {sum |f(e_i)|^{2}}$






    share|cite|improve this answer











    $endgroup$



    $|f|=sup {frac {|f(sum a_ie_i)|} {|sum a_ie_i|}}leqsqrt {sum |f(e_i)|^{2}}$ by C-S inequality. The exact value of $|f|$ is not easy to write and it is not $sqrt {sum |f(e_i)|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $sqrt {sum |f(e_i)|^{2}}$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 27 '18 at 9:15

























    answered Dec 27 '18 at 8:20









    Kavi Rama MurthyKavi Rama Murthy

    62.5k42262




    62.5k42262












    • $begingroup$
      Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
      $endgroup$
      – Damien
      Dec 27 '18 at 8:52






    • 1




      $begingroup$
      @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
      $endgroup$
      – Kavi Rama Murthy
      Dec 27 '18 at 9:09


















    • $begingroup$
      Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
      $endgroup$
      – Damien
      Dec 27 '18 at 8:52






    • 1




      $begingroup$
      @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
      $endgroup$
      – Kavi Rama Murthy
      Dec 27 '18 at 9:09
















    $begingroup$
    Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
    $endgroup$
    – Damien
    Dec 27 '18 at 8:52




    $begingroup$
    Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$
    $endgroup$
    – Damien
    Dec 27 '18 at 8:52




    1




    1




    $begingroup$
    @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 9:09




    $begingroup$
    @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $|f|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $|f|$. Thanks for your comment.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 9:09











    0












    $begingroup$

    I guess it should be:
    begin{align} |f|&=suplimits_{xneq 0}dfrac{|f(x)|}{|x|}\&=suplimits_{alpha_ineq 0}dfrac{left|f(sum^{n}_{i=1}alpha_i e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}\&=suplimits_{alpha_ineq 0}dfrac{left|sum^{n}_{i=1}alpha_i f(e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}.end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I guess it should be:
      begin{align} |f|&=suplimits_{xneq 0}dfrac{|f(x)|}{|x|}\&=suplimits_{alpha_ineq 0}dfrac{left|f(sum^{n}_{i=1}alpha_i e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}\&=suplimits_{alpha_ineq 0}dfrac{left|sum^{n}_{i=1}alpha_i f(e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}.end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I guess it should be:
        begin{align} |f|&=suplimits_{xneq 0}dfrac{|f(x)|}{|x|}\&=suplimits_{alpha_ineq 0}dfrac{left|f(sum^{n}_{i=1}alpha_i e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}\&=suplimits_{alpha_ineq 0}dfrac{left|sum^{n}_{i=1}alpha_i f(e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}.end{align}






        share|cite|improve this answer









        $endgroup$



        I guess it should be:
        begin{align} |f|&=suplimits_{xneq 0}dfrac{|f(x)|}{|x|}\&=suplimits_{alpha_ineq 0}dfrac{left|f(sum^{n}_{i=1}alpha_i e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}\&=suplimits_{alpha_ineq 0}dfrac{left|sum^{n}_{i=1}alpha_i f(e_i)right|}{left(sum^{n}_{i=1}|alpha_i |^2right)^{1/2}}.end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 14:39









        Omojola MichealOmojola Micheal

        1,901324




        1,901324






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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