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!
functional-analysis norm
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add a comment |
$begingroup$
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!
functional-analysis norm
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1
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Try writing it as $f(psi) = langle psi, grangle$ for some function $g$. Then find $|g|_2$.
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– Prahlad Vaidyanathan
May 20 '16 at 10:35
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always forget such representation.. thanks and solved ;)
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– Kenneth Ng
May 20 '16 at 10:51
1
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@KennethNg: In this case you might want to add your solution as an answer for further reference.
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– gerw
May 20 '16 at 11:23
add a comment |
$begingroup$
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!
functional-analysis norm
$endgroup$
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
functional-analysis norm
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
add a comment |
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
add a comment |
1 Answer
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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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$begingroup$
As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$
$endgroup$
add a comment |
$begingroup$
As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$
$endgroup$
add a comment |
$begingroup$
As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$
$endgroup$
As pointed out by @KennethNg, using Riesz representation, and writing $f(psi)$ as $langlepsi,(3+isin(pi t)cos(pi t)rangle$, we get $$|f|=|3+isin(pi t)cos(pi t)|=int_0^13+frac{isin(2pi t)}{2}dt=3$$
answered Jan 26 at 17:29
vidyarthividyarthi
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$begingroup$
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