orthonormal subset of $L_2(0,1) $ is complete
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Let ${v_k}_{k=1}^{infty} subset L_2[0,1]$ be orthonormal.
Assuming for each $xin [0,1]$ , $x = sum_{k=1}^{infty} |int_0^xv_k(t) dt|^2$.
I want to show that ${v_k}_{k=1}^{infty}$ is complete.
I showed the reversed statement, and I noticed that if we define $f_x(t) = 1$ for $t le x$ and $f_x(t) = 0$ otherwise we get the $x = ||f_x||^2 = sum_{k=1}^{infty} |(f_x,v_k) |^2$ , so Parseval's equality hold for $f_x in L_2[0,1]$.
So now I want to show Parseval's equality holds for each $fin L_2[0,1]$, but I cant find a way to do that(maybe ${f_x}_{xin [0,1]}$ are dense in $L_2[0,1]$ ? ).
Thanks for helping.
functional-analysis hilbert-spaces
asked Nov 16 at 13:45
Liad
1,215316
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0
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Let ${v_k}_{k=1}^{infty} subset L_2[0,1]$ be orthonormal.
Assuming for each $xin [0,1]$ , $x = sum_{k=1}^{infty} |int_0^xv_k(t) dt|^2$.
I want to show that ${v_k}_{k=1}^{infty}$ is complete.
I showed the reversed statement, and I noticed that if we define $f_x(t) = 1$ for $t le x$ and $f_x(t) = 0$ otherwise we get the $x = ||f_x||^2 = sum_{k=1}^{infty} |(f_x,v_k) |^2$ , so Parseval's equality hold for $f_x in L_2[0,1]$.
So now I want to show Parseval's equality holds for each $fin L_2[0,1]$, but I cant find a way to do that(maybe ${f_x}_{xin [0,1]}$ are dense in $L_2[0,1]$ ? ).
Thanks for helping.
functional-analysis hilbert-spaces
asked Nov 16 at 13:45
Liad
1,215316
Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let ${v_k}_{k=1}^{infty} subset L_2[0,1]$ be orthonormal.
Assuming for each $xin [0,1]$ , $x = sum_{k=1}^{infty} |int_0^xv_k(t) dt|^2$.
I want to show that ${v_k}_{k=1}^{infty}$ is complete.
I showed the reversed statement, and I noticed that if we define $f_x(t) = 1$ for $t le x$ and $f_x(t) = 0$ otherwise we get the $x = ||f_x||^2 = sum_{k=1}^{infty} |(f_x,v_k) |^2$ , so Parseval's equality hold for $f_x in L_2[0,1]$.
So now I want to show Parseval's equality holds for each $fin L_2[0,1]$, but I cant find a way to do that(maybe ${f_x}_{xin [0,1]}$ are dense in $L_2[0,1]$ ? ).
Thanks for helping.
functional-analysis hilbert-spaces
asked Nov 16 at 13:45
Liad
1,215316
Let ${v_k}_{k=1}^{infty} subset L_2[0,1]$ be orthonormal.
Assuming for each $xin [0,1]$ , $x = sum_{k=1}^{infty} |int_0^xv_k(t) dt|^2$.
I want to show that ${v_k}_{k=1}^{infty}$ is complete.
I showed the reversed statement, and I noticed that if we define $f_x(t) = 1$ for $t le x$ and $f_x(t) = 0$ otherwise we get the $x = ||f_x||^2 = sum_{k=1}^{infty} |(f_x,v_k) |^2$ , so Parseval's equality hold for $f_x in L_2[0,1]$.
So now I want to show Parseval's equality holds for each $fin L_2[0,1]$, but I cant find a way to do that(maybe ${f_x}_{xin [0,1]}$ are dense in $L_2[0,1]$ ? ).
Thanks for helping.
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Nov 16 at 13:45
Liad
1,215316
asked Nov 16 at 13:45
Liad
1,215316
edited Nov 16 at 13:58
asked Nov 16 at 13:45
Liad
1,215316
asked Nov 16 at 13:45
Liad
1,215316
asked Nov 16 at 13:45
Liad
1,215316
1,215316
Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26
add a comment |
Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26
Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26
add a comment |
1 Answer
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Indeed ${f_x}_{xin [0, 1]}$ spans a space that is dense in $L_2(0,1)$ :
For $alt b$, we have $f_b - f_a = 1_{]a,b]}$ (where $1_I$ is the function that values $1$ in $I$ and $0$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $L_2(0,1)$.
answered Nov 17 at 12:26
Astyx
2,5801518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Indeed ${f_x}_{xin [0, 1]}$ spans a space that is dense in $L_2(0,1)$ :
For $alt b$, we have $f_b - f_a = 1_{]a,b]}$ (where $1_I$ is the function that values $1$ in $I$ and $0$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $L_2(0,1)$.
answered Nov 17 at 12:26
Astyx
2,5801518
add a comment |
up vote
0
down vote
Indeed ${f_x}_{xin [0, 1]}$ spans a space that is dense in $L_2(0,1)$ :
For $alt b$, we have $f_b - f_a = 1_{]a,b]}$ (where $1_I$ is the function that values $1$ in $I$ and $0$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $L_2(0,1)$.
answered Nov 17 at 12:26
Astyx
2,5801518
add a comment |
up vote
0
down vote
up vote
0
down vote
Indeed ${f_x}_{xin [0, 1]}$ spans a space that is dense in $L_2(0,1)$ :
For $alt b$, we have $f_b - f_a = 1_{]a,b]}$ (where $1_I$ is the function that values $1$ in $I$ and $0$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $L_2(0,1)$.
answered Nov 17 at 12:26
Astyx
2,5801518
Indeed ${f_x}_{xin [0, 1]}$ spans a space that is dense in $L_2(0,1)$ :
For $alt b$, we have $f_b - f_a = 1_{]a,b]}$ (where $1_I$ is the function that values $1$ in $I$ and $0$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $L_2(0,1)$.
answered Nov 17 at 12:26
Astyx
2,5801518
answered Nov 17 at 12:26
Astyx
2,5801518
answered Nov 17 at 12:26
Astyx
2,5801518
answered Nov 17 at 12:26
Astyx
2,5801518
2,5801518
add a comment |
add a comment |
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Why? $||f_x||^2= int_0^1 |f_x|^2 = x$ @nicomezi
– Liad
Nov 16 at 14:24
Hmm, you are right, I computed $(int_0^1 f_x dx )^2$ .
– nicomezi
Nov 16 at 14:26