Norm of $f$ in a dual space space $(ell^{infty})^{*}$
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For $x in ell^{infty}$, let $f(x)= sum_{n in mathbb{N}}x_{n}2^{-n}$ determine the norm of $f$ in $(ell^{infty})^{*}$ (the dual space of $(ell^{infty})$.
Notes: I think I need to related this back to the fact that $(ell^{infty})^{*}$ is isometrically isomorphic to $ell^{1}$ so I know there is some map T such that the norm of T$x$ in the dual space is the norm of $x$ in $ell^{1}$, but I'm not sure how to relate this back to find the norm of $f$.
functional-analysis
asked Nov 17 at 22:36
Roger
545
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up vote
1
down vote
favorite
For $x in ell^{infty}$, let $f(x)= sum_{n in mathbb{N}}x_{n}2^{-n}$ determine the norm of $f$ in $(ell^{infty})^{*}$ (the dual space of $(ell^{infty})$.
Notes: I think I need to related this back to the fact that $(ell^{infty})^{*}$ is isometrically isomorphic to $ell^{1}$ so I know there is some map T such that the norm of T$x$ in the dual space is the norm of $x$ in $ell^{1}$, but I'm not sure how to relate this back to find the norm of $f$.
functional-analysis
asked Nov 17 at 22:36
Roger
545
3
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For $x in ell^{infty}$, let $f(x)= sum_{n in mathbb{N}}x_{n}2^{-n}$ determine the norm of $f$ in $(ell^{infty})^{*}$ (the dual space of $(ell^{infty})$.
Notes: I think I need to related this back to the fact that $(ell^{infty})^{*}$ is isometrically isomorphic to $ell^{1}$ so I know there is some map T such that the norm of T$x$ in the dual space is the norm of $x$ in $ell^{1}$, but I'm not sure how to relate this back to find the norm of $f$.
functional-analysis
asked Nov 17 at 22:36
Roger
545
For $x in ell^{infty}$, let $f(x)= sum_{n in mathbb{N}}x_{n}2^{-n}$ determine the norm of $f$ in $(ell^{infty})^{*}$ (the dual space of $(ell^{infty})$.
Notes: I think I need to related this back to the fact that $(ell^{infty})^{*}$ is isometrically isomorphic to $ell^{1}$ so I know there is some map T such that the norm of T$x$ in the dual space is the norm of $x$ in $ell^{1}$, but I'm not sure how to relate this back to find the norm of $f$.
functional-analysis
functional-analysis
asked Nov 17 at 22:36
Roger
545
asked Nov 17 at 22:36
Roger
545
asked Nov 17 at 22:36
Roger
545
asked Nov 17 at 22:36
Roger
545
asked Nov 17 at 22:36
Roger
545
545
3
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36
add a comment |
3
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36
3
3
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36
add a comment |
1 Answer
1
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2
down vote
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We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:ell^1to (ell^infty)^*$, so we simply need to compute $|f|_1$. This is $1$.
Alternatively, we know that $|f|ge |f(1)|=1$, and then you just need an upper bound by $1$.
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:ell^1to (ell^infty)^*$, so we simply need to compute $|f|_1$. This is $1$.
Alternatively, we know that $|f|ge |f(1)|=1$, and then you just need an upper bound by $1$.
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
add a comment |
up vote
2
down vote
accepted
We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:ell^1to (ell^infty)^*$, so we simply need to compute $|f|_1$. This is $1$.
Alternatively, we know that $|f|ge |f(1)|=1$, and then you just need an upper bound by $1$.
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:ell^1to (ell^infty)^*$, so we simply need to compute $|f|_1$. This is $1$.
Alternatively, we know that $|f|ge |f(1)|=1$, and then you just need an upper bound by $1$.
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:ell^1to (ell^infty)^*$, so we simply need to compute $|f|_1$. This is $1$.
Alternatively, we know that $|f|ge |f(1)|=1$, and then you just need an upper bound by $1$.
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
answered Nov 17 at 22:51
Ashwin Trisal
1,0901515
1,0901515
add a comment |
add a comment |
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3
$(ell^infty)^*$ is far from being isomorphic to $ell^1$.
– MisterRiemann
Nov 17 at 22:42
In general X is not isomorphic to $X^{ast ast}$ (its bidual) but $X$ is contained isometrically in it.
– Adrián González-Pérez
Nov 20 at 12:36