Orthogonal complement of $H_a =left{g in V: gleft(t+frac{1}{2}right)=g(t) right}$












1














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}
$










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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 '18 at 11:37
















1














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}
$










share|cite|improve this question
























  • You can proceed further than just the definition. Try to find some elements of $H_a^perp$.
    – астон вілла олоф мэллбэрг
    Nov 20 '18 at 11:37














1












1








1







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}
$










share|cite|improve this question















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






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edited Nov 20 '18 at 11:14









rtybase

10.4k21433




10.4k21433










asked Nov 20 '18 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 '18 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 '18 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 '18 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 '18 at 11:37










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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.






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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.






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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.






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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.






        share|cite|improve this answer












        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.







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        share|cite|improve this answer










        answered Nov 20 '18 at 11:54









        Kavi Rama Murthy

        50.3k31854




        50.3k31854






























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