Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $;;H_a =left{gin V: gleft(t+frac{1}{2}right)=g(t) right};;;$ (period $frac{1}{2}$).
What can be said about $H_a^perp$?
$H_a^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}{2}right)=g(t) right}
$
inner-product-space orthogonality
edited Nov 20 at 11:14
rtybase
10.4k21433
asked Nov 20 at 11:05
Filip
428
add a comment |
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $;;H_a =left{gin V: gleft(t+frac{1}{2}right)=g(t) right};;;$ (period $frac{1}{2}$).
What can be said about $H_a^perp$?
$H_a^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}{2}right)=g(t) right}
$
inner-product-space orthogonality
edited Nov 20 at 11:14
rtybase
10.4k21433
asked Nov 20 at 11:05
Filip
428
You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37
add a comment |
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $;;H_a =left{gin V: gleft(t+frac{1}{2}right)=g(t) right};;;$ (period $frac{1}{2}$).
What can be said about $H_a^perp$?
$H_a^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}{2}right)=g(t) right}
$
inner-product-space orthogonality
edited Nov 20 at 11:14
rtybase
10.4k21433
asked Nov 20 at 11:05
Filip
428
If $;;V={ f:mathbb{R}rightarrow mathbb{C} |; f text{ is continuous and has period }1}$, $;; langle f | g rangle$ is defined as $ langle f | g rangle = int_0^1 overline{f(t)}g(t)dt$, $forall f,g in V;;$ and $;;H_a =left{gin V: gleft(t+frac{1}{2}right)=g(t) right};;;$ (period $frac{1}{2}$).
What can be said about $H_a^perp$?
$H_a^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}{2}right)=g(t) right}
$
inner-product-space orthogonality
inner-product-space orthogonality
edited Nov 20 at 11:14
rtybase
10.4k21433
asked Nov 20 at 11:05
Filip
428
edited Nov 20 at 11:14
rtybase
10.4k21433
asked Nov 20 at 11:05
Filip
428
edited Nov 20 at 11:14
rtybase
10.4k21433
edited Nov 20 at 11:14
rtybase
10.4k21433
edited Nov 20 at 11:14
rtybase
10.4k21433
10.4k21433
asked Nov 20 at 11:05
Filip
428
asked Nov 20 at 11:05
Filip
428
asked Nov 20 at 11:05
Filip
428
428
You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37
add a comment |
You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37
You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37
You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37
add a comment |
1 Answer
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Hints: $$0=int_0^{1/2} f(x)g(x) , dx+int_{1/2}^{1} f(x)g(x) , dx=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y-frac 1 2) , dy$$ $$=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y) , dy$$ for all continuous functions $g$ on $[0,frac 1 2]$ with $g(0)=g(frac 1 2)$ iff $f(y-frac 1 2)=-f(y)$. Thus the orthogonal complement consists of all continuous functions $f$ with period $1$ satisfying $f(x-frac 1 2)=-f(x)$ for all $x$. Some details are missing but I think you can fill in the details.
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
add a comment |
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1 Answer
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1 Answer
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Hints: $$0=int_0^{1/2} f(x)g(x) , dx+int_{1/2}^{1} f(x)g(x) , dx=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y-frac 1 2) , dy$$ $$=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y) , dy$$ for all continuous functions $g$ on $[0,frac 1 2]$ with $g(0)=g(frac 1 2)$ iff $f(y-frac 1 2)=-f(y)$. Thus the orthogonal complement consists of all continuous functions $f$ with period $1$ satisfying $f(x-frac 1 2)=-f(x)$ for all $x$. Some details are missing but I think you can fill in the details.
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
add a comment |
Hints: $$0=int_0^{1/2} f(x)g(x) , dx+int_{1/2}^{1} f(x)g(x) , dx=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y-frac 1 2) , dy$$ $$=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y) , dy$$ for all continuous functions $g$ on $[0,frac 1 2]$ with $g(0)=g(frac 1 2)$ iff $f(y-frac 1 2)=-f(y)$. Thus the orthogonal complement consists of all continuous functions $f$ with period $1$ satisfying $f(x-frac 1 2)=-f(x)$ for all $x$. Some details are missing but I think you can fill in the details.
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
add a comment |
Hints: $$0=int_0^{1/2} f(x)g(x) , dx+int_{1/2}^{1} f(x)g(x) , dx=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y-frac 1 2) , dy$$ $$=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y) , dy$$ for all continuous functions $g$ on $[0,frac 1 2]$ with $g(0)=g(frac 1 2)$ iff $f(y-frac 1 2)=-f(y)$. Thus the orthogonal complement consists of all continuous functions $f$ with period $1$ satisfying $f(x-frac 1 2)=-f(x)$ for all $x$. Some details are missing but I think you can fill in the details.
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
Hints: $$0=int_0^{1/2} f(x)g(x) , dx+int_{1/2}^{1} f(x)g(x) , dx=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y-frac 1 2) , dy$$ $$=int_0^{1/2} f(x)g(x) , dx+int_{0}^{1/2} f(y-frac 1 2)g(y) , dy$$ for all continuous functions $g$ on $[0,frac 1 2]$ with $g(0)=g(frac 1 2)$ iff $f(y-frac 1 2)=-f(y)$. Thus the orthogonal complement consists of all continuous functions $f$ with period $1$ satisfying $f(x-frac 1 2)=-f(x)$ for all $x$. Some details are missing but I think you can fill in the details.
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
answered Nov 20 at 11:54
Kavi Rama Murthy
49.7k31854
49.7k31854
add a comment |
add a comment |
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You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
– астон вілла олоф мэллбэрг
Nov 20 at 11:37