A complete orthonormal system ${e_i}^infty_{i=1}$ in $H$ is a basis in $H$












1












$begingroup$


enter image description hereenter image description hereenter image description here



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
    $endgroup$
    – mathworker21
    Jan 4 at 20:43
















1












$begingroup$


enter image description hereenter image description hereenter image description here



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
    $endgroup$
    – mathworker21
    Jan 4 at 20:43














1












1








1


1



$begingroup$


enter image description hereenter image description hereenter image description here



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







share|cite|improve this question









$endgroup$




enter image description hereenter image description hereenter image description here



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 20:36









ChikChakChikChak

764418




764418












  • $begingroup$
    $langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
    $endgroup$
    – 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




$begingroup$
$langle y,e_irangle = langle sum_j langle x,e_j rangle e_j, e_i rangle = sum_j langle x, e_j rangle langle e_j , e_i rangle = langle x, e_i rangle$
$endgroup$
– mathworker21
Jan 4 at 20:43










1 Answer
1






active

oldest

votes


















1












$begingroup$

In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:



$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$



$=langle x,e_jrangle-langle x,e_jrangle=0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:



$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$



$=langle x,e_jrangle-langle x,e_jrangle=0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:



$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$



$=langle x,e_jrangle-langle x,e_jrangle=0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:



$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$



$=langle x,e_jrangle-langle x,e_jrangle=0$






share|cite|improve this answer









$endgroup$



In Hilbert spaces the dot product is closed to infinite sums as well if the series converges. So because the system is orthonormal:



$langle y-x,e_jrangle=langle sum_{i=1}^inftylangle x,e_irangle e_i,e_jrangle-langle x,e_jrangle=sum_{i=1}^inftylangle x,e_iranglelangle e_i,e_jrangle-langle x,e_jrangle=$



$=langle x,e_jrangle-langle x,e_jrangle=0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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