Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$
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If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ displaystyle langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $displaystyle ;;H_{frac{1}{sqrt{2}}} =left{gin V: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$.
($f text{ has period } 1$)
($g text{ has period } 1 text{ and } frac{1}{sqrt{2}}$)
What can be said about $H_{frac{1}{sqrt{2}}}^perp$?
$H_{frac{1}{sqrt{2}}}^perp = {fin V: langle f | g rangle=0, forall gin H_a }
= left{fin V: langle f | g rangle=0, forall g: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$
$displaystyle 0 = langle f | g rangle = int_0^1 overline{f(t)}g(t)dt
= int_0^1 overline{f(t)}gleft(t+n+frac{m}{sqrt{2}}right)dt, ;; n,m in mathbb{Z}, ;; fin H_{frac{1}{sqrt{2}}}^perp, ;; gin H_{frac{1}{sqrt{2}}}$.
Related post: Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$
inner-product-space orthogonality
asked Nov 20 at 13:44
Filip
427
add a comment |
up vote
0
down vote
favorite
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ displaystyle langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $displaystyle ;;H_{frac{1}{sqrt{2}}} =left{gin V: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$.
($f text{ has period } 1$)
($g text{ has period } 1 text{ and } frac{1}{sqrt{2}}$)
What can be said about $H_{frac{1}{sqrt{2}}}^perp$?
$H_{frac{1}{sqrt{2}}}^perp = {fin V: langle f | g rangle=0, forall gin H_a }
= left{fin V: langle f | g rangle=0, forall g: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$
$displaystyle 0 = langle f | g rangle = int_0^1 overline{f(t)}g(t)dt
= int_0^1 overline{f(t)}gleft(t+n+frac{m}{sqrt{2}}right)dt, ;; n,m in mathbb{Z}, ;; fin H_{frac{1}{sqrt{2}}}^perp, ;; gin H_{frac{1}{sqrt{2}}}$.
Related post: Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$
inner-product-space orthogonality
asked Nov 20 at 13:44
Filip
427
Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ displaystyle langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $displaystyle ;;H_{frac{1}{sqrt{2}}} =left{gin V: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$.
($f text{ has period } 1$)
($g text{ has period } 1 text{ and } frac{1}{sqrt{2}}$)
What can be said about $H_{frac{1}{sqrt{2}}}^perp$?
$H_{frac{1}{sqrt{2}}}^perp = {fin V: langle f | g rangle=0, forall gin H_a }
= left{fin V: langle f | g rangle=0, forall g: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$
$displaystyle 0 = langle f | g rangle = int_0^1 overline{f(t)}g(t)dt
= int_0^1 overline{f(t)}gleft(t+n+frac{m}{sqrt{2}}right)dt, ;; n,m in mathbb{Z}, ;; fin H_{frac{1}{sqrt{2}}}^perp, ;; gin H_{frac{1}{sqrt{2}}}$.
Related post: Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$
inner-product-space orthogonality
asked Nov 20 at 13:44
Filip
427
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ displaystyle langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $displaystyle ;;H_{frac{1}{sqrt{2}}} =left{gin V: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$.
($f text{ has period } 1$)
($g text{ has period } 1 text{ and } frac{1}{sqrt{2}}$)
What can be said about $H_{frac{1}{sqrt{2}}}^perp$?
$H_{frac{1}{sqrt{2}}}^perp = {fin V: langle f | g rangle=0, forall gin H_a }
= left{fin V: langle f | g rangle=0, forall g: gleft(t+frac{1}{sqrt{2}}right)=g(t) right}$
$displaystyle 0 = langle f | g rangle = int_0^1 overline{f(t)}g(t)dt
= int_0^1 overline{f(t)}gleft(t+n+frac{m}{sqrt{2}}right)dt, ;; n,m in mathbb{Z}, ;; fin H_{frac{1}{sqrt{2}}}^perp, ;; gin H_{frac{1}{sqrt{2}}}$.
Related post: Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$
inner-product-space orthogonality
inner-product-space orthogonality
asked Nov 20 at 13:44
Filip
427
asked Nov 20 at 13:44
Filip
427
asked Nov 20 at 13:44
Filip
427
asked Nov 20 at 13:44
Filip
427
asked Nov 20 at 13:44
Filip
427
427
Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53
add a comment |
Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53
Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53
add a comment |
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Hint: $mathbb{Z}+mathbb{Z}frac1{sqrt2}$ is dense in $mathbb{R}$.
– user10354138
Nov 20 at 13:50
@user10354138 What does it mean for a set to be dense in $mathbb{R}$?
– Filip
Nov 20 at 13:53