weighted norm modification
Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
$$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?
I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?
functional-analysis lp-spaces inner-product-space
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
add a comment |
Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
$$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?
I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?
functional-analysis lp-spaces inner-product-space
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
add a comment |
Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
$$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?
I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?
functional-analysis lp-spaces inner-product-space
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
$$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?
I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?
functional-analysis lp-spaces inner-product-space
functional-analysis lp-spaces inner-product-space
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
edited Dec 3 '18 at 9:53
Davide Giraudo
125k16150260
125k16150260
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
asked Dec 2 '18 at 19:18
Chase Sariaslani
805
805
add a comment |
add a comment |
1 Answer
1
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oldest
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For $mathbb{C}^n$:
$$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$
where $overline{y_i}$ denotes the complex conjugate of $y_i$.
For $mathcal{l}^2(mathbb{N})$:
$$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$
To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
For $mathbb{C}^n$:
$$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$
where $overline{y_i}$ denotes the complex conjugate of $y_i$.
For $mathcal{l}^2(mathbb{N})$:
$$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$
To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
add a comment |
For $mathbb{C}^n$:
$$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$
where $overline{y_i}$ denotes the complex conjugate of $y_i$.
For $mathcal{l}^2(mathbb{N})$:
$$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$
To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
add a comment |
For $mathbb{C}^n$:
$$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$
where $overline{y_i}$ denotes the complex conjugate of $y_i$.
For $mathcal{l}^2(mathbb{N})$:
$$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$
To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
For $mathbb{C}^n$:
$$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$
where $overline{y_i}$ denotes the complex conjugate of $y_i$.
For $mathcal{l}^2(mathbb{N})$:
$$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$
To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
answered Dec 2 '18 at 19:25
the_candyman
8,73122044
8,73122044
add a comment |
add a comment |
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