A complete orthonormal system ${e_i}^infty_{i=1}$ in $H$ is a basis in $H$
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I'm studying about Hilbert space from a book of functional analysis and I just read this theorem (2.1.10) and its' proof.
I cannot understand why $(y-x)perp e_i$? why is it implied?
functional-analysis hilbert-spaces normed-spaces orthonormal complete-spaces
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add a comment |
$begingroup$
I'm studying about Hilbert space from a book of functional analysis and I just read this theorem (2.1.10) and its' proof.
I cannot understand why $(y-x)perp e_i$? why is it implied?
functional-analysis hilbert-spaces normed-spaces orthonormal complete-spaces
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$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43
add a comment |
$begingroup$
I'm studying about Hilbert space from a book of functional analysis and I just read this theorem (2.1.10) and its' proof.
I cannot understand why $(y-x)perp e_i$? why is it implied?
functional-analysis hilbert-spaces normed-spaces orthonormal complete-spaces
$endgroup$
I'm studying about Hilbert space from a book of functional analysis and I just read this theorem (2.1.10) and its' proof.
I cannot understand why $(y-x)perp e_i$? why is it implied?
functional-analysis hilbert-spaces normed-spaces orthonormal complete-spaces
functional-analysis hilbert-spaces normed-spaces orthonormal complete-spaces
asked Jan 4 at 20:36
ChikChakChikChak
764418
764418
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$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43
add a comment |
$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43
$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43
$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43
add a comment |
1 Answer
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In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
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Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
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– ChikChak
Jan 4 at 20:51
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For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
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– Mark
Jan 4 at 21:02
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By the way, how is that book called?
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– Mark
Jan 4 at 21:11
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Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
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– ChikChak
Jan 5 at 13:23
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
$endgroup$
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
add a comment |
$begingroup$
In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
$endgroup$
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
add a comment |
$begingroup$
In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
$endgroup$
In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
answered Jan 4 at 20:44
MarkMark
9,579622
9,579622
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
add a comment |
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
$endgroup$
– ChikChak
Jan 4 at 20:51
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
For each $N$ we have $langle sum_{i=1}^N langle x,e_irangle e_i,e_jrangle$=$sum_{i=1}^Nlangle x,e_iranglelangle e_i,e_jrangle$. Here I'm only using the definition of dot product. And now you can take $Ntoinfty$ on both sides using the continuity of dot product.
$endgroup$
– Mark
Jan 4 at 21:02
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
By the way, how is that book called?
$endgroup$
– Mark
Jan 4 at 21:11
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
$begingroup$
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
$endgroup$
– ChikChak
Jan 5 at 13:23
add a comment |
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$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43