Finding norm of a bounded linear functionalContinuous linear functionalCalculating norm of an operatorShow $F_1$ is a continuous linear functionalShow $T$ is a bounded linear operatorShowing continuity of a linear functionalBounded Linear Functional and Riesz's LemmaCalculating the norm of a specific bounded linear functionalShow that $f$ is a bounded linear functional and find its normBounded linear functional separating a hyperplane and a pointRegarding norm-attaining functional
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Finding norm of a bounded linear functional
Continuous linear functionalCalculating norm of an operatorShow $F_1$ is a continuous linear functionalShow $T$ is a bounded linear operatorShowing continuity of a linear functionalBounded Linear Functional and Riesz's LemmaCalculating the norm of a specific bounded linear functionalShow that $f$ is a bounded linear functional and find its normBounded linear functional separating a hyperplane and a pointRegarding norm-attaining functional
$begingroup$
Let $X=xin C[0,1]: x(0)=0$ with sup norm and let $f:Xto mathbbK$ be defined by $$f(x)=intlimits_0^1 x(t)dt text for all xin X.$$ I want to show that $|f|=1$. It is easy to show that $|f|leq 1$ also I have shown that there does not exist ant $xin B_X$ such that $|f|=|f(x)|$. Thus I have to find a sequence $(x_n)subset B_X$ such that $|f(x_n)|to |f|$. I am stuck here. Any help to find such a sequence is appreciated.
functional-analysis continuity linear-transformations norm normed-spaces
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
$endgroup$
add a comment |
$begingroup$
Let $X=xin C[0,1]: x(0)=0$ with sup norm and let $f:Xto mathbbK$ be defined by $$f(x)=intlimits_0^1 x(t)dt text for all xin X.$$ I want to show that $|f|=1$. It is easy to show that $|f|leq 1$ also I have shown that there does not exist ant $xin B_X$ such that $|f|=|f(x)|$. Thus I have to find a sequence $(x_n)subset B_X$ such that $|f(x_n)|to |f|$. I am stuck here. Any help to find such a sequence is appreciated.
functional-analysis continuity linear-transformations norm normed-spaces
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
$endgroup$
add a comment |
$begingroup$
Let $X=xin C[0,1]: x(0)=0$ with sup norm and let $f:Xto mathbbK$ be defined by $$f(x)=intlimits_0^1 x(t)dt text for all xin X.$$ I want to show that $|f|=1$. It is easy to show that $|f|leq 1$ also I have shown that there does not exist ant $xin B_X$ such that $|f|=|f(x)|$. Thus I have to find a sequence $(x_n)subset B_X$ such that $|f(x_n)|to |f|$. I am stuck here. Any help to find such a sequence is appreciated.
functional-analysis continuity linear-transformations norm normed-spaces
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
$endgroup$
Let $X=xin C[0,1]: x(0)=0$ with sup norm and let $f:Xto mathbbK$ be defined by $$f(x)=intlimits_0^1 x(t)dt text for all xin X.$$ I want to show that $|f|=1$. It is easy to show that $|f|leq 1$ also I have shown that there does not exist ant $xin B_X$ such that $|f|=|f(x)|$. Thus I have to find a sequence $(x_n)subset B_X$ such that $|f(x_n)|to |f|$. I am stuck here. Any help to find such a sequence is appreciated.
functional-analysis continuity linear-transformations norm normed-spaces
functional-analysis continuity linear-transformations norm normed-spaces
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
asked Mar 28 at 15:53
AnupamAnupam
2,5281825
2,5281825
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider $f_n(x)=1,$ if $x>1over n$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,1over n]$, $|f|=1$, and $int_0^1f_n(x)geq 1-1over ngeq|f|$. This implies that $|f|geq 1$. Since you have remarked that $|f|leq 1$, $|f|=1$.
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Consider $f_n(x)=1,$ if $x>1over n$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,1over n]$, $|f|=1$, and $int_0^1f_n(x)geq 1-1over ngeq|f|$. This implies that $|f|geq 1$. Since you have remarked that $|f|leq 1$, $|f|=1$.
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
Consider $f_n(x)=1,$ if $x>1over n$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,1over n]$, $|f|=1$, and $int_0^1f_n(x)geq 1-1over ngeq|f|$. This implies that $|f|geq 1$. Since you have remarked that $|f|leq 1$, $|f|=1$.
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
Consider $f_n(x)=1,$ if $x>1over n$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,1over n]$, $|f|=1$, and $int_0^1f_n(x)geq 1-1over ngeq|f|$. This implies that $|f|geq 1$. Since you have remarked that $|f|leq 1$, $|f|=1$.
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
Consider $f_n(x)=1,$ if $x>1over n$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,1over n]$, $|f|=1$, and $int_0^1f_n(x)geq 1-1over ngeq|f|$. This implies that $|f|geq 1$. Since you have remarked that $|f|leq 1$, $|f|=1$.
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 15:58
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
add a comment |
add a comment |
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