Dual norm of a product space

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I am wondering how to compute the dual norm of this norm. By the definition of dual norm, we need to find for $z^* in Z^*$, $|z^*|_* = undersetmathrmsup ;z^*(z)$, but what is $z^*(z)$ here? What are the elements in $Z^*$ like?

asked Jul 17 at 14:39

Sean Ian

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up vote
3
down vote

favorite

asked Jul 17 at 14:39

Sean Ian

545

add a commentÂ |Â

up vote
3
down vote

favorite

asked Jul 17 at 14:39

Sean Ian

545

asked Jul 17 at 14:39

Sean Ian

545

asked Jul 17 at 14:39

Sean Ian

545

asked Jul 17 at 14:39

Sean Ian

545

asked Jul 17 at 14:39

Sean Ian

545

add a commentÂ |Â

2 Answers
2

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oldest

votes

up vote
1
down vote

accepted

Your dual space $(X times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* times Y^*$ equipped with the norm $|(f,g)| = sqrtg$.

Proof:

Define $J : X^* times Y^* to (X times Y)^*$ as

$$[J(f,g)](x,y) = af(x) + bg(y)$$

for $(f,g) in X^* times Y^*$.

$J$ is clearly well-defined and linear. Check that its inverse is given by

$$J^-1(F) = left(frac1a F(cdot, 0), frac1b F(0, cdot)right)$$

for $F in (X times Y)^*$ so $J$ is bijective.

We have

$$[J(f,g)](x,y) = |af(x) + bg(y)| le a|f(x)| + b|g(y)| le a|f||x| + b|g||y| le sqrt^2sqrty = |(f,g)||(x,y)|$$

so $|J(f,g)| le |(f,g)|$.

To prove the converse inequality, let $varepsilon > 0$ and pick $x in X$, $y in Y$ such that $f(x) ge (1-varepsilon)|f||x|$ and $g(y) ge (1-varepsilon)|g||y|$.

Consider $left(fracxx|f|, fracy|g|right) in X times Y$. Its norm is $$left|left(fracxx|f|, fracy|g|right)right| = sqrtright = sqrt^2 = |(f,g)|$$

We have

$$J(f,g)(x,y) = af(x) + bg(x) ge a(1-varepsilon)|f|left|fracxx|f|right| + b(1-varepsilon)left|fracy|g|right| = (1-varepsilon)(a|f|^2 + b|g|^2) = (1-varepsilon)|(f,g)|^2 = (1-varepsilon)|(f,g)||(x,y)|$$

so $|J(f,g)| ge (1-varepsilon)|(f,g)|$. Since $varepsilon$ was arbitrary, we conclude $|J(f,g)| ge |(f,g)|$.

Therefore $|J(f,g)| =|(f,g)|$ so $J$ is an isometric isomorphism.

This gives you an explicit formula for the dual norm of $F in (X times Y)^*$:

$$|F| = sqrt^2$$

answered Jul 17 at 17:04

mechanodroid

22.2k52041

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up vote
1
down vote

Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:

$phi(x)+rho(y)$ where $phiin X^*$, $rho in Y^*$.

It is first easy to see that elements of this form are indeed functionals on $Xtimes Y$. Furthermore, all $(x,y)in Xtimes Y$ can be written as $(x,y)=(x,0)+(0,y)$. Hence by linearity of the functionals, we know that for all $varphi in Z$ we have:

$varphi(x,y)= varphi(x,0)+varphi(0,y)$. Where $phi(x):=varphi(x,0)$ and $rho(y):=varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.

answered Jul 17 at 14:56

Keen-ameteur

644213

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

Your dual space $(X times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* times Y^*$ equipped with the norm $|(f,g)| = sqrtg$.

Proof:

Define $J : X^* times Y^* to (X times Y)^*$ as

$$[J(f,g)](x,y) = af(x) + bg(y)$$

for $(f,g) in X^* times Y^*$.

$J$ is clearly well-defined and linear. Check that its inverse is given by

$$J^-1(F) = left(frac1a F(cdot, 0), frac1b F(0, cdot)right)$$

for $F in (X times Y)^*$ so $J$ is bijective.

We have

$$[J(f,g)](x,y) = |af(x) + bg(y)| le a|f(x)| + b|g(y)| le a|f||x| + b|g||y| le sqrt^2sqrty = |(f,g)||(x,y)|$$

so $|J(f,g)| le |(f,g)|$.

To prove the converse inequality, let $varepsilon > 0$ and pick $x in X$, $y in Y$ such that $f(x) ge (1-varepsilon)|f||x|$ and $g(y) ge (1-varepsilon)|g||y|$.

Consider $left(fracxx|f|, fracy|g|right) in X times Y$. Its norm is $$left|left(fracxx|f|, fracy|g|right)right| = sqrtright = sqrt^2 = |(f,g)|$$

We have

so $|J(f,g)| ge (1-varepsilon)|(f,g)|$. Since $varepsilon$ was arbitrary, we conclude $|J(f,g)| ge |(f,g)|$.

Therefore $|J(f,g)| =|(f,g)|$ so $J$ is an isometric isomorphism.

This gives you an explicit formula for the dual norm of $F in (X times Y)^*$:

$$|F| = sqrt^2$$

answered Jul 17 at 17:04

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
1
down vote

accepted

Your dual space $(X times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* times Y^*$ equipped with the norm $|(f,g)| = sqrtg$.

Proof:

Define $J : X^* times Y^* to (X times Y)^*$ as

$$[J(f,g)](x,y) = af(x) + bg(y)$$

for $(f,g) in X^* times Y^*$.

$J$ is clearly well-defined and linear. Check that its inverse is given by

$$J^-1(F) = left(frac1a F(cdot, 0), frac1b F(0, cdot)right)$$

for $F in (X times Y)^*$ so $J$ is bijective.

We have

$$[J(f,g)](x,y) = |af(x) + bg(y)| le a|f(x)| + b|g(y)| le a|f||x| + b|g||y| le sqrt^2sqrty = |(f,g)||(x,y)|$$

so $|J(f,g)| le |(f,g)|$.

To prove the converse inequality, let $varepsilon > 0$ and pick $x in X$, $y in Y$ such that $f(x) ge (1-varepsilon)|f||x|$ and $g(y) ge (1-varepsilon)|g||y|$.

Consider $left(fracxx|f|, fracy|g|right) in X times Y$. Its norm is $$left|left(fracxx|f|, fracy|g|right)right| = sqrtright = sqrt^2 = |(f,g)|$$

We have

so $|J(f,g)| ge (1-varepsilon)|(f,g)|$. Since $varepsilon$ was arbitrary, we conclude $|J(f,g)| ge |(f,g)|$.

Therefore $|J(f,g)| =|(f,g)|$ so $J$ is an isometric isomorphism.

This gives you an explicit formula for the dual norm of $F in (X times Y)^*$:

$$|F| = sqrt^2$$

answered Jul 17 at 17:04

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
1
down vote

accepted

Your dual space $(X times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* times Y^*$ equipped with the norm $|(f,g)| = sqrtg$.

Proof:

Define $J : X^* times Y^* to (X times Y)^*$ as

$$[J(f,g)](x,y) = af(x) + bg(y)$$

for $(f,g) in X^* times Y^*$.

$J$ is clearly well-defined and linear. Check that its inverse is given by

$$J^-1(F) = left(frac1a F(cdot, 0), frac1b F(0, cdot)right)$$

for $F in (X times Y)^*$ so $J$ is bijective.

We have

$$[J(f,g)](x,y) = |af(x) + bg(y)| le a|f(x)| + b|g(y)| le a|f||x| + b|g||y| le sqrt^2sqrty = |(f,g)||(x,y)|$$

so $|J(f,g)| le |(f,g)|$.

To prove the converse inequality, let $varepsilon > 0$ and pick $x in X$, $y in Y$ such that $f(x) ge (1-varepsilon)|f||x|$ and $g(y) ge (1-varepsilon)|g||y|$.

Consider $left(fracxx|f|, fracy|g|right) in X times Y$. Its norm is $$left|left(fracxx|f|, fracy|g|right)right| = sqrtright = sqrt^2 = |(f,g)|$$

We have

so $|J(f,g)| ge (1-varepsilon)|(f,g)|$. Since $varepsilon$ was arbitrary, we conclude $|J(f,g)| ge |(f,g)|$.

Therefore $|J(f,g)| =|(f,g)|$ so $J$ is an isometric isomorphism.

This gives you an explicit formula for the dual norm of $F in (X times Y)^*$:

$$|F| = sqrt^2$$

answered Jul 17 at 17:04

mechanodroid

22.2k52041

Your dual space $(X times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* times Y^*$ equipped with the norm $|(f,g)| = sqrtg$.

Proof:

Define $J : X^* times Y^* to (X times Y)^*$ as

$$[J(f,g)](x,y) = af(x) + bg(y)$$

for $(f,g) in X^* times Y^*$.

$J$ is clearly well-defined and linear. Check that its inverse is given by

$$J^-1(F) = left(frac1a F(cdot, 0), frac1b F(0, cdot)right)$$

for $F in (X times Y)^*$ so $J$ is bijective.

We have

$$[J(f,g)](x,y) = |af(x) + bg(y)| le a|f(x)| + b|g(y)| le a|f||x| + b|g||y| le sqrt^2sqrty = |(f,g)||(x,y)|$$

so $|J(f,g)| le |(f,g)|$.

To prove the converse inequality, let $varepsilon > 0$ and pick $x in X$, $y in Y$ such that $f(x) ge (1-varepsilon)|f||x|$ and $g(y) ge (1-varepsilon)|g||y|$.

Consider $left(fracxx|f|, fracy|g|right) in X times Y$. Its norm is $$left|left(fracxx|f|, fracy|g|right)right| = sqrtright = sqrt^2 = |(f,g)|$$

We have

so $|J(f,g)| ge (1-varepsilon)|(f,g)|$. Since $varepsilon$ was arbitrary, we conclude $|J(f,g)| ge |(f,g)|$.

Therefore $|J(f,g)| =|(f,g)|$ so $J$ is an isometric isomorphism.

This gives you an explicit formula for the dual norm of $F in (X times Y)^*$:

$$|F| = sqrt^2$$

answered Jul 17 at 17:04

mechanodroid

22.2k52041

answered Jul 17 at 17:04

mechanodroid

22.2k52041

answered Jul 17 at 17:04

mechanodroid

22.2k52041

answered Jul 17 at 17:04

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
1
down vote

Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:

$phi(x)+rho(y)$ where $phiin X^*$, $rho in Y^*$.

$varphi(x,y)= varphi(x,0)+varphi(0,y)$. Where $phi(x):=varphi(x,0)$ and $rho(y):=varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.

answered Jul 17 at 14:56

Keen-ameteur

644213

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

add a commentÂ |Â

up vote
1
down vote

Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:

$phi(x)+rho(y)$ where $phiin X^*$, $rho in Y^*$.

$varphi(x,y)= varphi(x,0)+varphi(0,y)$. Where $phi(x):=varphi(x,0)$ and $rho(y):=varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.

answered Jul 17 at 14:56

Keen-ameteur

644213

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

add a commentÂ |Â

up vote
1
down vote

Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:

$phi(x)+rho(y)$ where $phiin X^*$, $rho in Y^*$.

$varphi(x,y)= varphi(x,0)+varphi(0,y)$. Where $phi(x):=varphi(x,0)$ and $rho(y):=varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.

answered Jul 17 at 14:56

Keen-ameteur

644213

Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:

$phi(x)+rho(y)$ where $phiin X^*$, $rho in Y^*$.

$varphi(x,y)= varphi(x,0)+varphi(0,y)$. Where $phi(x):=varphi(x,0)$ and $rho(y):=varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.

answered Jul 17 at 14:56

Keen-ameteur

644213

answered Jul 17 at 14:56

Keen-ameteur

644213

answered Jul 17 at 14:56

Keen-ameteur

644213

answered Jul 17 at 14:56

Keen-ameteur

644213

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

add a commentÂ |Â

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

Seems correct. So the dual of the product space doesn't depend on the defined norm? How would you proceed to compute the dual norm then?
â€“Â Sean Ian
Jul 17 at 15:16

If we agree to denote all functional $varphi in Z^*$ as $varphi =phi+ rho$ as above, I would probably try to see whether $ fracVert phi Vert vert avert geq fracVert phi Vert vert avert$ and observe the expression:
â€“Â Keen-ameteur
Jul 17 at 15:58

$frac vert phi(x)+rho(y)vert sqrt aVert xVert_X^2+ b Vert yVert _Y^2 $
â€“Â Keen-ameteur
Jul 17 at 15:59

add a commentÂ |Â

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