Given the subspace of $L^2$ made by constant functions, characterize its orthogonal complement
Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.
I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.
$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$
$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$
So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.
Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?
functional-analysis hilbert-spaces
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Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.
I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.
$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$
$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$
So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.
Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?
functional-analysis hilbert-spaces
1
Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
1
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38
add a comment |
Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.
I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.
$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$
$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$
So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.
Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?
functional-analysis hilbert-spaces
Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.
I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.
$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$
$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$
So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.
Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Nov 30 at 10:15
sound wave
1548
1548
1
Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
1
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38
add a comment |
1
Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
1
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38
1
1
Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
1
1
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38
add a comment |
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While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).
add a comment |
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1 Answer
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While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).
add a comment |
While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).
add a comment |
While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).
While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).
answered Nov 30 at 11:23
астон вілла олоф мэллбэрг
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Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22
1
complete and correct.
– DisintegratingByParts
Nov 30 at 10:38