A complete orthonormal system ${e_i}^infty_{i=1}$ in $H$ is a basis in $H$
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
asked yesterday
ChikChak
790418
add a comment |
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
asked yesterday
ChikChak
790418
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday
add a comment |
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
asked yesterday
ChikChak
790418
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 yesterday
ChikChak
790418
asked yesterday
ChikChak
790418
asked yesterday
ChikChak
790418
asked yesterday
ChikChak
790418
asked yesterday
ChikChak
790418
790418
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday
add a comment |
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday
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:
$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$
$=langle x,e_jrangle-langle x,e_jrangle=0$
answered yesterday
Mark
6,080415
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 yesterday
Mark
6,080415
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
add a comment |
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 yesterday
Mark
6,080415
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
add a comment |
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 yesterday
Mark
6,080415
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 yesterday
Mark
6,080415
answered yesterday
Mark
6,080415
answered yesterday
Mark
6,080415
answered yesterday
Mark
6,080415
6,080415
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
add a comment |
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
Could you explain why $langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle$?
– ChikChak
yesterday
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.
– Mark
yesterday
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.
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
By the way, how is that book called?
– Mark
yesterday
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
Milman Eidelman Tsolomitis, Functional analysis An Introduction(T)
– ChikChak
17 hours ago
add a comment |
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$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
– mathworker21
yesterday