Find the norm of a linear functional in $L^2[0,1]$.












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Define the linear functional $f : L^2[0,1] text{(As completion of $C[0,1]$, all the continuous complex-valued function )} mapsto mathbb{C}$ by
$$f(psi)=3int_{0}^{1}psi(t)dt + iint_{0}^{1} psi(t)sin(pi t)cos(pi t)dt.$$



I have made use of Cauchy Schwarz to set a bound of $|f| leq 3+ frac{1}{sqrt{8}}$, yet I can not figure out what to substitute to obtain the norm. Can I get a small hint? Thanks!










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  • 1




    $begingroup$
    Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
    $endgroup$
    – Prahlad Vaidyanathan
    May 20 '16 at 10:35










  • $begingroup$
    always forget such representation.. thanks and solved ;)
    $endgroup$
    – Kenneth Ng
    May 20 '16 at 10:51








  • 1




    $begingroup$
    @KennethNg: In this case you might want to add your solution as an answer for further reference.
    $endgroup$
    – gerw
    May 20 '16 at 11:23
















1












$begingroup$


Define the linear functional $f : L^2[0,1] text{(As completion of $C[0,1]$, all the continuous complex-valued function )} mapsto mathbb{C}$ by
$$f(psi)=3int_{0}^{1}psi(t)dt + iint_{0}^{1} psi(t)sin(pi t)cos(pi t)dt.$$



I have made use of Cauchy Schwarz to set a bound of $|f| leq 3+ frac{1}{sqrt{8}}$, yet I can not figure out what to substitute to obtain the norm. Can I get a small hint? Thanks!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
    $endgroup$
    – Prahlad Vaidyanathan
    May 20 '16 at 10:35










  • $begingroup$
    always forget such representation.. thanks and solved ;)
    $endgroup$
    – Kenneth Ng
    May 20 '16 at 10:51








  • 1




    $begingroup$
    @KennethNg: In this case you might want to add your solution as an answer for further reference.
    $endgroup$
    – gerw
    May 20 '16 at 11:23














1












1








1


1



$begingroup$


Define the linear functional $f : L^2[0,1] text{(As completion of $C[0,1]$, all the continuous complex-valued function )} mapsto mathbb{C}$ by
$$f(psi)=3int_{0}^{1}psi(t)dt + iint_{0}^{1} psi(t)sin(pi t)cos(pi t)dt.$$



I have made use of Cauchy Schwarz to set a bound of $|f| leq 3+ frac{1}{sqrt{8}}$, yet I can not figure out what to substitute to obtain the norm. Can I get a small hint? Thanks!










share|cite|improve this question









$endgroup$




Define the linear functional $f : L^2[0,1] text{(As completion of $C[0,1]$, all the continuous complex-valued function )} mapsto mathbb{C}$ by
$$f(psi)=3int_{0}^{1}psi(t)dt + iint_{0}^{1} psi(t)sin(pi t)cos(pi t)dt.$$



I have made use of Cauchy Schwarz to set a bound of $|f| leq 3+ frac{1}{sqrt{8}}$, yet I can not figure out what to substitute to obtain the norm. Can I get a small hint? Thanks!







functional-analysis norm






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asked May 20 '16 at 10:10









Kenneth NgKenneth Ng

455




455








  • 1




    $begingroup$
    Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
    $endgroup$
    – Prahlad Vaidyanathan
    May 20 '16 at 10:35










  • $begingroup$
    always forget such representation.. thanks and solved ;)
    $endgroup$
    – Kenneth Ng
    May 20 '16 at 10:51








  • 1




    $begingroup$
    @KennethNg: In this case you might want to add your solution as an answer for further reference.
    $endgroup$
    – gerw
    May 20 '16 at 11:23














  • 1




    $begingroup$
    Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
    $endgroup$
    – Prahlad Vaidyanathan
    May 20 '16 at 10:35










  • $begingroup$
    always forget such representation.. thanks and solved ;)
    $endgroup$
    – Kenneth Ng
    May 20 '16 at 10:51








  • 1




    $begingroup$
    @KennethNg: In this case you might want to add your solution as an answer for further reference.
    $endgroup$
    – gerw
    May 20 '16 at 11:23








1




1




$begingroup$
Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
$endgroup$
– Prahlad Vaidyanathan
May 20 '16 at 10:35




$begingroup$
Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
$endgroup$
– Prahlad Vaidyanathan
May 20 '16 at 10:35












$begingroup$
always forget such representation.. thanks and solved ;)
$endgroup$
– Kenneth Ng
May 20 '16 at 10:51






$begingroup$
always forget such representation.. thanks and solved ;)
$endgroup$
– Kenneth Ng
May 20 '16 at 10:51






1




1




$begingroup$
@KennethNg: In this case you might want to add your solution as an answer for further reference.
$endgroup$
– gerw
May 20 '16 at 11:23




$begingroup$
@KennethNg: In this case you might want to add your solution as an answer for further reference.
$endgroup$
– gerw
May 20 '16 at 11:23










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As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$






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    0












    $begingroup$

    As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$






        share|cite|improve this answer









        $endgroup$



        As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 17:29









        vidyarthividyarthi

        3,0611833




        3,0611833






























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