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Confusion about the dual norm
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Consider $mathbbR^2$ with inner product $langle x,yrangle_A = x^T A y$, where $A$ is diagonal positive definite, i.e. $A = beginpmatrix a & 0\ 0 & bendpmatrixquad a,b >0$. To compute the dual norm of this norm, according to The Riesz representation theorem, we can compute $|z|^* = underset_A = 1mathrmmax z^T A x$, this gives $|z|^* = |z|_A = sqrtaz_1^2 + bz_2^2$.
My question is why cannot we compute the dual norm using $|z|^* = underset_A = 1mathrmmax z^T x$. In this case, the "dual norm" would be $|z|^* = sqrtfracz_1^2a+fracz_2^2b$. Specifically, given an element $z$ in the dual space, what is wrong with defining $z(x) = z^Tx$ instead of $z(x) = z^T A x$ ?
functional-analysis hilbert-spaces dual-spaces
asked Jul 16 at 15:54
Sean Ian
545
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Consider $mathbbR^2$ with inner product $langle x,yrangle_A = x^T A y$, where $A$ is diagonal positive definite, i.e. $A = beginpmatrix a & 0\ 0 & bendpmatrixquad a,b >0$. To compute the dual norm of this norm, according to The Riesz representation theorem, we can compute $|z|^* = underset_A = 1mathrmmax z^T A x$, this gives $|z|^* = |z|_A = sqrtaz_1^2 + bz_2^2$.
My question is why cannot we compute the dual norm using $|z|^* = underset_A = 1mathrmmax z^T x$. In this case, the "dual norm" would be $|z|^* = sqrtfracz_1^2a+fracz_2^2b$. Specifically, given an element $z$ in the dual space, what is wrong with defining $z(x) = z^Tx$ instead of $z(x) = z^T A x$ ?
functional-analysis hilbert-spaces dual-spaces
asked Jul 16 at 15:54
Sean Ian
545
Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49
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up vote
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Consider $mathbbR^2$ with inner product $langle x,yrangle_A = x^T A y$, where $A$ is diagonal positive definite, i.e. $A = beginpmatrix a & 0\ 0 & bendpmatrixquad a,b >0$. To compute the dual norm of this norm, according to The Riesz representation theorem, we can compute $|z|^* = underset_A = 1mathrmmax z^T A x$, this gives $|z|^* = |z|_A = sqrtaz_1^2 + bz_2^2$.
My question is why cannot we compute the dual norm using $|z|^* = underset_A = 1mathrmmax z^T x$. In this case, the "dual norm" would be $|z|^* = sqrtfracz_1^2a+fracz_2^2b$. Specifically, given an element $z$ in the dual space, what is wrong with defining $z(x) = z^Tx$ instead of $z(x) = z^T A x$ ?
functional-analysis hilbert-spaces dual-spaces
asked Jul 16 at 15:54
Sean Ian
545
Consider $mathbbR^2$ with inner product $langle x,yrangle_A = x^T A y$, where $A$ is diagonal positive definite, i.e. $A = beginpmatrix a & 0\ 0 & bendpmatrixquad a,b >0$. To compute the dual norm of this norm, according to The Riesz representation theorem, we can compute $|z|^* = underset_A = 1mathrmmax z^T A x$, this gives $|z|^* = |z|_A = sqrtaz_1^2 + bz_2^2$.
My question is why cannot we compute the dual norm using $|z|^* = underset_A = 1mathrmmax z^T x$. In this case, the "dual norm" would be $|z|^* = sqrtfracz_1^2a+fracz_2^2b$. Specifically, given an element $z$ in the dual space, what is wrong with defining $z(x) = z^Tx$ instead of $z(x) = z^T A x$ ?
functional-analysis hilbert-spaces dual-spaces
asked Jul 16 at 15:54
Sean Ian
545
asked Jul 16 at 15:54
Sean Ian
545
asked Jul 16 at 15:54
Sean Ian
545
asked Jul 16 at 15:54
Sean Ian
545
545
Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49
add a comment |Â
Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49
Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49
Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49
add a comment |Â
1 Answer
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Suppose you have some functional $zeta(x) = f^T x$. Suppose $A>0$ is symmetric
and define the $A$ inner product as above. Then $zeta(x) = (A^-1f)^T Ax$, and so the Riesz representation of $zeta$ with respect to the $A$ inner product is
$A^-1f$.
Then $|zeta| = sup__A le 1 zeta(x) = sup__A le 1 langle A^-1 f, xrangle_A = |A^-1 f|_A = sqrtf^T A^-1 f$.
answered Jul 16 at 20:09
copper.hat
122k557156
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose you have some functional $zeta(x) = f^T x$. Suppose $A>0$ is symmetric
and define the $A$ inner product as above. Then $zeta(x) = (A^-1f)^T Ax$, and so the Riesz representation of $zeta$ with respect to the $A$ inner product is
$A^-1f$.
Then $|zeta| = sup__A le 1 zeta(x) = sup__A le 1 langle A^-1 f, xrangle_A = |A^-1 f|_A = sqrtf^T A^-1 f$.
answered Jul 16 at 20:09
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
accepted
Suppose you have some functional $zeta(x) = f^T x$. Suppose $A>0$ is symmetric
and define the $A$ inner product as above. Then $zeta(x) = (A^-1f)^T Ax$, and so the Riesz representation of $zeta$ with respect to the $A$ inner product is
$A^-1f$.
Then $|zeta| = sup__A le 1 zeta(x) = sup__A le 1 langle A^-1 f, xrangle_A = |A^-1 f|_A = sqrtf^T A^-1 f$.
answered Jul 16 at 20:09
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose you have some functional $zeta(x) = f^T x$. Suppose $A>0$ is symmetric
and define the $A$ inner product as above. Then $zeta(x) = (A^-1f)^T Ax$, and so the Riesz representation of $zeta$ with respect to the $A$ inner product is
$A^-1f$.
Then $|zeta| = sup__A le 1 zeta(x) = sup__A le 1 langle A^-1 f, xrangle_A = |A^-1 f|_A = sqrtf^T A^-1 f$.
answered Jul 16 at 20:09
copper.hat
122k557156
Suppose you have some functional $zeta(x) = f^T x$. Suppose $A>0$ is symmetric
and define the $A$ inner product as above. Then $zeta(x) = (A^-1f)^T Ax$, and so the Riesz representation of $zeta$ with respect to the $A$ inner product is
$A^-1f$.
Then $|zeta| = sup__A le 1 zeta(x) = sup__A le 1 langle A^-1 f, xrangle_A = |A^-1 f|_A = sqrtf^T A^-1 f$.
answered Jul 16 at 20:09
copper.hat
122k557156
answered Jul 16 at 20:09
copper.hat
122k557156
answered Jul 16 at 20:09
copper.hat
122k557156
answered Jul 16 at 20:09
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
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Presumably $z$ is an element of the dual space here. Riesz says that there is some $z^* in mathbbR^2$ such that $ z(x) = langle z^*, x rangle_A = (z^*)^T Ax$. Hence $|z| = sup__A le 1 z(x) = sup__A le 1 (z^*)^T Ax$. The functional $z$ and its representation $z^*$ depend on the inner product used
– copper.hat
Jul 16 at 16:49