If $|langle f, g rangle| = | f |_2 |g|_2$ then $exists alpha, beta in mathbb{R} $ s.t. $alpha f = beta g $
$begingroup$
Show the following are equivalent, for $f, g in L^2[a,b]$
a. $|langle f, g rangle| = |f|_2 |g|_2$
b. $exists alpha, beta in mathbb{R} $ s.t. $alpha f = beta g $
where $langle cdot , cdot rangle$ : inner product.
Try
(b $Rightarrow$ a) If $alpha=0$, things are trivial. Otherwise, note $f = frac{beta}{alpha} g$
$|langle f, g rangle| = |frac{beta}{alpha}| langle g, g rangle = |frac{beta}{alpha}| |g|_2 cdot|g|_2 = |f|_2 |g|_2 $
But I'm stuck at (a $Rightarrow$ b).
real-analysis measure-theory
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
$endgroup$
add a comment |
$begingroup$
Show the following are equivalent, for $f, g in L^2[a,b]$
a. $|langle f, g rangle| = |f|_2 |g|_2$
b. $exists alpha, beta in mathbb{R} $ s.t. $alpha f = beta g $
where $langle cdot , cdot rangle$ : inner product.
Try
(b $Rightarrow$ a) If $alpha=0$, things are trivial. Otherwise, note $f = frac{beta}{alpha} g$
$|langle f, g rangle| = |frac{beta}{alpha}| langle g, g rangle = |frac{beta}{alpha}| |g|_2 cdot|g|_2 = |f|_2 |g|_2 $
But I'm stuck at (a $Rightarrow$ b).
real-analysis measure-theory
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
$endgroup$
add a comment |
$begingroup$
Show the following are equivalent, for $f, g in L^2[a,b]$
a. $|langle f, g rangle| = |f|_2 |g|_2$
b. $exists alpha, beta in mathbb{R} $ s.t. $alpha f = beta g $
where $langle cdot , cdot rangle$ : inner product.
Try
(b $Rightarrow$ a) If $alpha=0$, things are trivial. Otherwise, note $f = frac{beta}{alpha} g$
$|langle f, g rangle| = |frac{beta}{alpha}| langle g, g rangle = |frac{beta}{alpha}| |g|_2 cdot|g|_2 = |f|_2 |g|_2 $
But I'm stuck at (a $Rightarrow$ b).
real-analysis measure-theory
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
$endgroup$
Show the following are equivalent, for $f, g in L^2[a,b]$
a. $|langle f, g rangle| = |f|_2 |g|_2$
b. $exists alpha, beta in mathbb{R} $ s.t. $alpha f = beta g $
where $langle cdot , cdot rangle$ : inner product.
Try
(b $Rightarrow$ a) If $alpha=0$, things are trivial. Otherwise, note $f = frac{beta}{alpha} g$
$|langle f, g rangle| = |frac{beta}{alpha}| langle g, g rangle = |frac{beta}{alpha}| |g|_2 cdot|g|_2 = |f|_2 |g|_2 $
But I'm stuck at (a $Rightarrow$ b).
real-analysis measure-theory
real-analysis measure-theory
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
edited Dec 20 '18 at 13:53
Nathanael Skrepek
1,6191515
1,6191515
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
asked Dec 20 '18 at 12:53
MoreblueMoreblue
8791216
8791216
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
One can consider the function $f:mathbb{R}rightarrow mathbb{R},xmapsto ||f-xg||^2$. Since we have $||f-xg||^2=langle f-xg,f-xgrangle=||f||^2-2xlangle f,grangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $left(2langle f,grangleright)^2-4||u||^2||v||^2=4langle f,grangle^2-4langle f,grangle^2=0$.
Hence there is some $xin mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One can consider the function $f:mathbb{R}rightarrow mathbb{R},xmapsto ||f-xg||^2$. Since we have $||f-xg||^2=langle f-xg,f-xgrangle=||f||^2-2xlangle f,grangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $left(2langle f,grangleright)^2-4||u||^2||v||^2=4langle f,grangle^2-4langle f,grangle^2=0$.
Hence there is some $xin mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
$endgroup$
add a comment |
$begingroup$
One can consider the function $f:mathbb{R}rightarrow mathbb{R},xmapsto ||f-xg||^2$. Since we have $||f-xg||^2=langle f-xg,f-xgrangle=||f||^2-2xlangle f,grangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $left(2langle f,grangleright)^2-4||u||^2||v||^2=4langle f,grangle^2-4langle f,grangle^2=0$.
Hence there is some $xin mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
$endgroup$
add a comment |
$begingroup$
One can consider the function $f:mathbb{R}rightarrow mathbb{R},xmapsto ||f-xg||^2$. Since we have $||f-xg||^2=langle f-xg,f-xgrangle=||f||^2-2xlangle f,grangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $left(2langle f,grangleright)^2-4||u||^2||v||^2=4langle f,grangle^2-4langle f,grangle^2=0$.
Hence there is some $xin mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
$endgroup$
One can consider the function $f:mathbb{R}rightarrow mathbb{R},xmapsto ||f-xg||^2$. Since we have $||f-xg||^2=langle f-xg,f-xgrangle=||f||^2-2xlangle f,grangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $left(2langle f,grangleright)^2-4||u||^2||v||^2=4langle f,grangle^2-4langle f,grangle^2=0$.
Hence there is some $xin mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
edited Dec 20 '18 at 15:13
BigbearZzz
8,58221652
8,58221652
answered Dec 20 '18 at 13:11
Student7Student7
2089
answered Dec 20 '18 at 13:11
Student7Student7
2089
answered Dec 20 '18 at 13:11
Student7Student7
2089
2089
add a comment |
add a comment |
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