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

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    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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