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A question regarding bounded linear functional
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Let $(f_n)$ be a bounded sequence of bounded linear functionals on a real normed linear space $X$. Then there exits a linear functional $fin X^*$ such that $liminflimits_{nto infty}f_n(x)leq f(x)leq limsuplimits_{nto infty}f_n(x)$ for all $xin X$.
I proceed as follows: Since for each $xin X$, $(f_n(x))$ is a bounded real sequence, by Bolzano-Weierstrass theorem there exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k}(x))$ converges. We define $f(x)=limlimits_{kto infty}f_{n_k}(x)$ for all $xin X$. The very first problem I face is in showing $f$ to be linear. Once it can be shown to be linear, the problem is solved. Any hint is appreciated.
functional-analysis linear-transformations
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
$endgroup$
add a comment |
$begingroup$
Let $(f_n)$ be a bounded sequence of bounded linear functionals on a real normed linear space $X$. Then there exits a linear functional $fin X^*$ such that $liminflimits_{nto infty}f_n(x)leq f(x)leq limsuplimits_{nto infty}f_n(x)$ for all $xin X$.
I proceed as follows: Since for each $xin X$, $(f_n(x))$ is a bounded real sequence, by Bolzano-Weierstrass theorem there exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k}(x))$ converges. We define $f(x)=limlimits_{kto infty}f_{n_k}(x)$ for all $xin X$. The very first problem I face is in showing $f$ to be linear. Once it can be shown to be linear, the problem is solved. Any hint is appreciated.
functional-analysis linear-transformations
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
$endgroup$
$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
1
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
1
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12
add a comment |
$begingroup$
Let $(f_n)$ be a bounded sequence of bounded linear functionals on a real normed linear space $X$. Then there exits a linear functional $fin X^*$ such that $liminflimits_{nto infty}f_n(x)leq f(x)leq limsuplimits_{nto infty}f_n(x)$ for all $xin X$.
I proceed as follows: Since for each $xin X$, $(f_n(x))$ is a bounded real sequence, by Bolzano-Weierstrass theorem there exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k}(x))$ converges. We define $f(x)=limlimits_{kto infty}f_{n_k}(x)$ for all $xin X$. The very first problem I face is in showing $f$ to be linear. Once it can be shown to be linear, the problem is solved. Any hint is appreciated.
functional-analysis linear-transformations
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
$endgroup$
Let $(f_n)$ be a bounded sequence of bounded linear functionals on a real normed linear space $X$. Then there exits a linear functional $fin X^*$ such that $liminflimits_{nto infty}f_n(x)leq f(x)leq limsuplimits_{nto infty}f_n(x)$ for all $xin X$.
I proceed as follows: Since for each $xin X$, $(f_n(x))$ is a bounded real sequence, by Bolzano-Weierstrass theorem there exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k}(x))$ converges. We define $f(x)=limlimits_{kto infty}f_{n_k}(x)$ for all $xin X$. The very first problem I face is in showing $f$ to be linear. Once it can be shown to be linear, the problem is solved. Any hint is appreciated.
functional-analysis linear-transformations
functional-analysis linear-transformations
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
asked Feb 3 at 3:29
AnupamAnupam
2,5741925
2,5741925
$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
1
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
1
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12
add a comment |
$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
1
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
1
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12
$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
1
1
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
1
1
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12
add a comment |
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$begingroup$
Are you sure you don't have $liminf$ and $limsup$ switched around? Wouldn't $limsup f_n(x) = infty$?
$endgroup$
– Robert Lewis
Feb 3 at 3:33
1
$begingroup$
$(f_n)$ is a bounded sequence and so for each $xin X$, $(f_n(x))$ is a bounded sequence of real numbers. Thus $limsup f_n(x)$ must be finite.
$endgroup$
– Anupam
Feb 3 at 3:40
1
$begingroup$
@Anupam The sequence ${n_k}$ depends on $x$ so there is no hope of proving that $f$ is linear. I think this requires Hahn Banach Theorem. See the arguments used to define Banach limits.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:12