Does the matrix norm inequality or the Cauchy-Schwarz inequality hold for L2,1 norms
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I read here https://statweb.stanford.edu/~souravc/Lecture32.pdf that Cauchy-Schwarz inequality holds for the Hilbert-Schmit or Frobenius norms. I wanted to know if the same holds for other norms too specifically the L2,1 norm.
matrices norm cauchy-schwarz-inequality
asked Jan 9 at 12:14
S. NagS. Nag
1
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add a comment |
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I read here https://statweb.stanford.edu/~souravc/Lecture32.pdf that Cauchy-Schwarz inequality holds for the Hilbert-Schmit or Frobenius norms. I wanted to know if the same holds for other norms too specifically the L2,1 norm.
matrices norm cauchy-schwarz-inequality
asked Jan 9 at 12:14
S. NagS. Nag
1
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$begingroup$
I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
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– Student7
Jan 9 at 13:17
add a comment |
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I read here https://statweb.stanford.edu/~souravc/Lecture32.pdf that Cauchy-Schwarz inequality holds for the Hilbert-Schmit or Frobenius norms. I wanted to know if the same holds for other norms too specifically the L2,1 norm.
matrices norm cauchy-schwarz-inequality
asked Jan 9 at 12:14
S. NagS. Nag
1
$endgroup$
I read here https://statweb.stanford.edu/~souravc/Lecture32.pdf that Cauchy-Schwarz inequality holds for the Hilbert-Schmit or Frobenius norms. I wanted to know if the same holds for other norms too specifically the L2,1 norm.
matrices norm cauchy-schwarz-inequality
matrices norm cauchy-schwarz-inequality
asked Jan 9 at 12:14
S. NagS. Nag
1
asked Jan 9 at 12:14
S. NagS. Nag
1
asked Jan 9 at 12:14
S. NagS. Nag
1
asked Jan 9 at 12:14
S. NagS. Nag
1
asked Jan 9 at 12:14
S. NagS. Nag
1
1
$begingroup$
I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
$endgroup$
– Student7
Jan 9 at 13:17
add a comment |
$begingroup$
I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
$endgroup$
– Student7
Jan 9 at 13:17
$begingroup$
I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
$endgroup$
– Student7
Jan 9 at 13:17
$begingroup$
I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
$endgroup$
– Student7
Jan 9 at 13:17
add a comment |
1 Answer
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Since both Hilbert-Schmidt norm and $L^2$-norm arise from an inner product (more generally, semi-inner product), Cauchy-Schwarz holds in those cases. But if we consider $L^p([0,1])$, $pneq 2$ for example, there does not exist an inner product to apply Cauchy-Schwarz inequality, because $L^p$-norm does not satisfy the parallelogram law
$$
|x+y|_p^2+|x-y|_p^2 = 2(|x|_p^2+|y|_p^2),
$$which should be true for all norms obtained from an inner product.
answered Jan 9 at 13:31
SongSong
15.1k1636
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$begingroup$
Since both Hilbert-Schmidt norm and $L^2$-norm arise from an inner product (more generally, semi-inner product), Cauchy-Schwarz holds in those cases. But if we consider $L^p([0,1])$, $pneq 2$ for example, there does not exist an inner product to apply Cauchy-Schwarz inequality, because $L^p$-norm does not satisfy the parallelogram law
$$
|x+y|_p^2+|x-y|_p^2 = 2(|x|_p^2+|y|_p^2),
$$which should be true for all norms obtained from an inner product.
answered Jan 9 at 13:31
SongSong
15.1k1636
$endgroup$
add a comment |
$begingroup$
Since both Hilbert-Schmidt norm and $L^2$-norm arise from an inner product (more generally, semi-inner product), Cauchy-Schwarz holds in those cases. But if we consider $L^p([0,1])$, $pneq 2$ for example, there does not exist an inner product to apply Cauchy-Schwarz inequality, because $L^p$-norm does not satisfy the parallelogram law
$$
|x+y|_p^2+|x-y|_p^2 = 2(|x|_p^2+|y|_p^2),
$$which should be true for all norms obtained from an inner product.
answered Jan 9 at 13:31
SongSong
15.1k1636
$endgroup$
add a comment |
$begingroup$
Since both Hilbert-Schmidt norm and $L^2$-norm arise from an inner product (more generally, semi-inner product), Cauchy-Schwarz holds in those cases. But if we consider $L^p([0,1])$, $pneq 2$ for example, there does not exist an inner product to apply Cauchy-Schwarz inequality, because $L^p$-norm does not satisfy the parallelogram law
$$
|x+y|_p^2+|x-y|_p^2 = 2(|x|_p^2+|y|_p^2),
$$which should be true for all norms obtained from an inner product.
answered Jan 9 at 13:31
SongSong
15.1k1636
$endgroup$
Since both Hilbert-Schmidt norm and $L^2$-norm arise from an inner product (more generally, semi-inner product), Cauchy-Schwarz holds in those cases. But if we consider $L^p([0,1])$, $pneq 2$ for example, there does not exist an inner product to apply Cauchy-Schwarz inequality, because $L^p$-norm does not satisfy the parallelogram law
$$
|x+y|_p^2+|x-y|_p^2 = 2(|x|_p^2+|y|_p^2),
$$which should be true for all norms obtained from an inner product.
answered Jan 9 at 13:31
SongSong
15.1k1636
answered Jan 9 at 13:31
SongSong
15.1k1636
answered Jan 9 at 13:31
SongSong
15.1k1636
answered Jan 9 at 13:31
SongSong
15.1k1636
15.1k1636
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I don't know the L2,1-norm, but the CSI holds for all norms which are induced by a scalar product. So when you know, that it is induced by a scalar product, the CSI holds for it.
$endgroup$
– Student7
Jan 9 at 13:17