Mixing
Prove that $ |x|=sup{|f(x)|:fin X^*, ,|f|=1},$ where $xin X$ and $X^*$ denotes the dual space of $X$.
$begingroup$
Let $x$ be an element of a normed linear space $X$ and let $X^*$ denote the dual space of $X$. Prove that
begin{align} |x|=sup{|f(x)|:fin X^*, ,|f|=1} end{align}
MY TRIAL
It suffices to show that
begin{align} forall;epsilon>0,;exists;|f(x_{epsilon})|in {|f(x)|:fin X^*, ,|f|=1};;text{such that}end{align}
begin{align} |x|-epsilon< |f(x_{epsilon})|leq |x|.end{align}
Let $xin X$ such that $xneq 0.$ Otherwise, $|f|=0$. Then, by Hanh-Banach Theorem, there exists a linear functional $f$ on $X$ such that
begin{align} |f|=1 ;;;text{and};;;|f(x)|= |x|leq |x|.end{align}
Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at
begin{align} |x|-epsilon< |f(x)|.end{align}
If no, can you help fix my wrong(s)?
functional-analysis analysis
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
$endgroup$
add a comment |
$begingroup$
Let $x$ be an element of a normed linear space $X$ and let $X^*$ denote the dual space of $X$. Prove that
begin{align} |x|=sup{|f(x)|:fin X^*, ,|f|=1} end{align}
MY TRIAL
It suffices to show that
begin{align} forall;epsilon>0,;exists;|f(x_{epsilon})|in {|f(x)|:fin X^*, ,|f|=1};;text{such that}end{align}
begin{align} |x|-epsilon< |f(x_{epsilon})|leq |x|.end{align}
Let $xin X$ such that $xneq 0.$ Otherwise, $|f|=0$. Then, by Hanh-Banach Theorem, there exists a linear functional $f$ on $X$ such that
begin{align} |f|=1 ;;;text{and};;;|f(x)|= |x|leq |x|.end{align}
Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at
begin{align} |x|-epsilon< |f(x)|.end{align}
If no, can you help fix my wrong(s)?
functional-analysis analysis
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
$endgroup$
$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19
add a comment |
$begingroup$
Let $x$ be an element of a normed linear space $X$ and let $X^*$ denote the dual space of $X$. Prove that
begin{align} |x|=sup{|f(x)|:fin X^*, ,|f|=1} end{align}
MY TRIAL
It suffices to show that
begin{align} forall;epsilon>0,;exists;|f(x_{epsilon})|in {|f(x)|:fin X^*, ,|f|=1};;text{such that}end{align}
begin{align} |x|-epsilon< |f(x_{epsilon})|leq |x|.end{align}
Let $xin X$ such that $xneq 0.$ Otherwise, $|f|=0$. Then, by Hanh-Banach Theorem, there exists a linear functional $f$ on $X$ such that
begin{align} |f|=1 ;;;text{and};;;|f(x)|= |x|leq |x|.end{align}
Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at
begin{align} |x|-epsilon< |f(x)|.end{align}
If no, can you help fix my wrong(s)?
functional-analysis analysis
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
$endgroup$
Let $x$ be an element of a normed linear space $X$ and let $X^*$ denote the dual space of $X$. Prove that
begin{align} |x|=sup{|f(x)|:fin X^*, ,|f|=1} end{align}
MY TRIAL
It suffices to show that
begin{align} forall;epsilon>0,;exists;|f(x_{epsilon})|in {|f(x)|:fin X^*, ,|f|=1};;text{such that}end{align}
begin{align} |x|-epsilon< |f(x_{epsilon})|leq |x|.end{align}
Let $xin X$ such that $xneq 0.$ Otherwise, $|f|=0$. Then, by Hanh-Banach Theorem, there exists a linear functional $f$ on $X$ such that
begin{align} |f|=1 ;;;text{and};;;|f(x)|= |x|leq |x|.end{align}
Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at
begin{align} |x|-epsilon< |f(x)|.end{align}
If no, can you help fix my wrong(s)?
functional-analysis analysis
functional-analysis analysis
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
asked Jan 10 at 21:15
Omojola MichealOmojola Micheal
1,853324
1,853324
$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19
add a comment |
$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19
$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your approach is flawed for two reasons that I see: first, you talk about an $x_epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say
"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $|f|=1$ and $|f(x)|=|x|$".
That's not wrong: but it's precisely what you are supposed to prove.
For any $f$ with $|f|leq1$, you have $|f(x)|leq|f|,|x|=|x|$. So $|f(x)|leq|x|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $mathbb C,x$, define $f_0(lambda x)=lambda,|x|$. Then
$$
|f_0(lambda x)|=|lambda,|x|,|=|lambda x|,
$$
so $|f_0|=1$. Now, by Hahn-Banach, there exists $fin X^*$, with $|f|=|f_0|=1$, and $f(x)=f_0(x)=|x|$. So
$$
|x|=max{|f(x)|: fin X^*, |f|=1}.
$$
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Let’s denote $S_x =sup{|f(x)|:fin X^*, ,|f|=1}$.
We have to prove that $Vert x Vert =S_x$.
If $Vert f Vert =1$, then $vert f(x)vert le Vert xVert$. Hence $S_x le Vert xVert$. Conversely for $xneq0$, on the subspace $mathbb Rx$, the linear form defined by $f(y) = lambda Vert x Vert$ for $y = lambda x$ is such that $f(y) le Vert y Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $vert f(y)vert le Vert y Vert$ for $y in X$ and $vert f(x) vert = Vert x Vert$. Hence $S_x ge Vert x Vert$, allowing us to conclude.
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
$endgroup$
add a comment |
$begingroup$
Credits to Martin Argerami. I present the full proof for future readers.
Let $xin X$ be fixed, such that $xneq 0.$ Consider the subspace spanned by $x$, defined by $Z={ alpha x:;alphain Bbb{R} }.$
Define $f$ arbitrarily by
begin{align} f:,&Zto Bbb{R}\& zmapsto alpha | x |. end{align}
Let $zin Z$, then $f(z)= alpha | x |= |alpha x |= |z |.,$ This implies that $|f(z)|=left||z | right|leq left||z |right|=|z |$ which gives boundedness of $f$ and $|f |=1.$
Let $gamma,lambdain Bbb{R}$ and $y,zin Z,$ then $exists,alpha,beta in Bbb{R}$ such that $y=alpha x$ and $z=beta x.$ Then, $f$ is linear, since
begin{align}fleft(gamma y+lambda zright)&= fleft[(gamma alpha+lambda beta)xright] \&= (gamma alpha+lambda beta)| x |\&= gamma (alpha| x |)+lambda( beta| x |)\&= gamma fleft( yright)+lambda fleft(zright).end{align}
Since $f$ is a linear functional, $;alpha fleft( xright)=fleft(alpha xright)=fleft(z right)=alpha | x |$ implies that $fleft( xright)=| x |.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $Fin X^{*}$ s.t. $F(x)=f(x)$ for all $xin Z$ and
begin{align}F(x)=f(x)=| x |;;text{and};;|F |=| f |=1.end{align}
So,
begin{align}| x |=left|| x | right|=left| F(x) right|;;text{for some};;Fin X^{*};;text{and};;|F |=1.end{align}
Taking $sup $ over such $;Fin X^{*};text{and};|F |=1,$ we get
begin{align}| x |=sup{left| F(x) right|:,Fin X^{*},;|F |=1}.end{align}
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your approach is flawed for two reasons that I see: first, you talk about an $x_epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say
"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $|f|=1$ and $|f(x)|=|x|$".
That's not wrong: but it's precisely what you are supposed to prove.
For any $f$ with $|f|leq1$, you have $|f(x)|leq|f|,|x|=|x|$. So $|f(x)|leq|x|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $mathbb C,x$, define $f_0(lambda x)=lambda,|x|$. Then
$$
|f_0(lambda x)|=|lambda,|x|,|=|lambda x|,
$$
so $|f_0|=1$. Now, by Hahn-Banach, there exists $fin X^*$, with $|f|=|f_0|=1$, and $f(x)=f_0(x)=|x|$. So
$$
|x|=max{|f(x)|: fin X^*, |f|=1}.
$$
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Your approach is flawed for two reasons that I see: first, you talk about an $x_epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say
"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $|f|=1$ and $|f(x)|=|x|$".
That's not wrong: but it's precisely what you are supposed to prove.
For any $f$ with $|f|leq1$, you have $|f(x)|leq|f|,|x|=|x|$. So $|f(x)|leq|x|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $mathbb C,x$, define $f_0(lambda x)=lambda,|x|$. Then
$$
|f_0(lambda x)|=|lambda,|x|,|=|lambda x|,
$$
so $|f_0|=1$. Now, by Hahn-Banach, there exists $fin X^*$, with $|f|=|f_0|=1$, and $f(x)=f_0(x)=|x|$. So
$$
|x|=max{|f(x)|: fin X^*, |f|=1}.
$$
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Your approach is flawed for two reasons that I see: first, you talk about an $x_epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say
"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $|f|=1$ and $|f(x)|=|x|$".
That's not wrong: but it's precisely what you are supposed to prove.
For any $f$ with $|f|leq1$, you have $|f(x)|leq|f|,|x|=|x|$. So $|f(x)|leq|x|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $mathbb C,x$, define $f_0(lambda x)=lambda,|x|$. Then
$$
|f_0(lambda x)|=|lambda,|x|,|=|lambda x|,
$$
so $|f_0|=1$. Now, by Hahn-Banach, there exists $fin X^*$, with $|f|=|f_0|=1$, and $f(x)=f_0(x)=|x|$. So
$$
|x|=max{|f(x)|: fin X^*, |f|=1}.
$$
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
Your approach is flawed for two reasons that I see: first, you talk about an $x_epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say
"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $|f|=1$ and $|f(x)|=|x|$".
That's not wrong: but it's precisely what you are supposed to prove.
For any $f$ with $|f|leq1$, you have $|f(x)|leq|f|,|x|=|x|$. So $|f(x)|leq|x|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $mathbb C,x$, define $f_0(lambda x)=lambda,|x|$. Then
$$
|f_0(lambda x)|=|lambda,|x|,|=|lambda x|,
$$
so $|f_0|=1$. Now, by Hahn-Banach, there exists $fin X^*$, with $|f|=|f_0|=1$, and $f(x)=f_0(x)=|x|$. So
$$
|x|=max{|f(x)|: fin X^*, |f|=1}.
$$
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:09
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
$begingroup$
Let’s denote $S_x =sup{|f(x)|:fin X^*, ,|f|=1}$.
We have to prove that $Vert x Vert =S_x$.
If $Vert f Vert =1$, then $vert f(x)vert le Vert xVert$. Hence $S_x le Vert xVert$. Conversely for $xneq0$, on the subspace $mathbb Rx$, the linear form defined by $f(y) = lambda Vert x Vert$ for $y = lambda x$ is such that $f(y) le Vert y Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $vert f(y)vert le Vert y Vert$ for $y in X$ and $vert f(x) vert = Vert x Vert$. Hence $S_x ge Vert x Vert$, allowing us to conclude.
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
$endgroup$
add a comment |
$begingroup$
Let’s denote $S_x =sup{|f(x)|:fin X^*, ,|f|=1}$.
We have to prove that $Vert x Vert =S_x$.
If $Vert f Vert =1$, then $vert f(x)vert le Vert xVert$. Hence $S_x le Vert xVert$. Conversely for $xneq0$, on the subspace $mathbb Rx$, the linear form defined by $f(y) = lambda Vert x Vert$ for $y = lambda x$ is such that $f(y) le Vert y Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $vert f(y)vert le Vert y Vert$ for $y in X$ and $vert f(x) vert = Vert x Vert$. Hence $S_x ge Vert x Vert$, allowing us to conclude.
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
$endgroup$
add a comment |
$begingroup$
Let’s denote $S_x =sup{|f(x)|:fin X^*, ,|f|=1}$.
We have to prove that $Vert x Vert =S_x$.
If $Vert f Vert =1$, then $vert f(x)vert le Vert xVert$. Hence $S_x le Vert xVert$. Conversely for $xneq0$, on the subspace $mathbb Rx$, the linear form defined by $f(y) = lambda Vert x Vert$ for $y = lambda x$ is such that $f(y) le Vert y Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $vert f(y)vert le Vert y Vert$ for $y in X$ and $vert f(x) vert = Vert x Vert$. Hence $S_x ge Vert x Vert$, allowing us to conclude.
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
$endgroup$
Let’s denote $S_x =sup{|f(x)|:fin X^*, ,|f|=1}$.
We have to prove that $Vert x Vert =S_x$.
If $Vert f Vert =1$, then $vert f(x)vert le Vert xVert$. Hence $S_x le Vert xVert$. Conversely for $xneq0$, on the subspace $mathbb Rx$, the linear form defined by $f(y) = lambda Vert x Vert$ for $y = lambda x$ is such that $f(y) le Vert y Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $vert f(y)vert le Vert y Vert$ for $y in X$ and $vert f(x) vert = Vert x Vert$. Hence $S_x ge Vert x Vert$, allowing us to conclude.
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
answered Jan 10 at 22:08
mathcounterexamples.netmathcounterexamples.net
26.6k22157
26.6k22157
add a comment |
add a comment |
$begingroup$
Credits to Martin Argerami. I present the full proof for future readers.
Let $xin X$ be fixed, such that $xneq 0.$ Consider the subspace spanned by $x$, defined by $Z={ alpha x:;alphain Bbb{R} }.$
Define $f$ arbitrarily by
begin{align} f:,&Zto Bbb{R}\& zmapsto alpha | x |. end{align}
Let $zin Z$, then $f(z)= alpha | x |= |alpha x |= |z |.,$ This implies that $|f(z)|=left||z | right|leq left||z |right|=|z |$ which gives boundedness of $f$ and $|f |=1.$
Let $gamma,lambdain Bbb{R}$ and $y,zin Z,$ then $exists,alpha,beta in Bbb{R}$ such that $y=alpha x$ and $z=beta x.$ Then, $f$ is linear, since
begin{align}fleft(gamma y+lambda zright)&= fleft[(gamma alpha+lambda beta)xright] \&= (gamma alpha+lambda beta)| x |\&= gamma (alpha| x |)+lambda( beta| x |)\&= gamma fleft( yright)+lambda fleft(zright).end{align}
Since $f$ is a linear functional, $;alpha fleft( xright)=fleft(alpha xright)=fleft(z right)=alpha | x |$ implies that $fleft( xright)=| x |.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $Fin X^{*}$ s.t. $F(x)=f(x)$ for all $xin Z$ and
begin{align}F(x)=f(x)=| x |;;text{and};;|F |=| f |=1.end{align}
So,
begin{align}| x |=left|| x | right|=left| F(x) right|;;text{for some};;Fin X^{*};;text{and};;|F |=1.end{align}
Taking $sup $ over such $;Fin X^{*};text{and};|F |=1,$ we get
begin{align}| x |=sup{left| F(x) right|:,Fin X^{*},;|F |=1}.end{align}
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
$endgroup$
add a comment |
$begingroup$
Credits to Martin Argerami. I present the full proof for future readers.
Let $xin X$ be fixed, such that $xneq 0.$ Consider the subspace spanned by $x$, defined by $Z={ alpha x:;alphain Bbb{R} }.$
Define $f$ arbitrarily by
begin{align} f:,&Zto Bbb{R}\& zmapsto alpha | x |. end{align}
Let $zin Z$, then $f(z)= alpha | x |= |alpha x |= |z |.,$ This implies that $|f(z)|=left||z | right|leq left||z |right|=|z |$ which gives boundedness of $f$ and $|f |=1.$
Let $gamma,lambdain Bbb{R}$ and $y,zin Z,$ then $exists,alpha,beta in Bbb{R}$ such that $y=alpha x$ and $z=beta x.$ Then, $f$ is linear, since
begin{align}fleft(gamma y+lambda zright)&= fleft[(gamma alpha+lambda beta)xright] \&= (gamma alpha+lambda beta)| x |\&= gamma (alpha| x |)+lambda( beta| x |)\&= gamma fleft( yright)+lambda fleft(zright).end{align}
Since $f$ is a linear functional, $;alpha fleft( xright)=fleft(alpha xright)=fleft(z right)=alpha | x |$ implies that $fleft( xright)=| x |.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $Fin X^{*}$ s.t. $F(x)=f(x)$ for all $xin Z$ and
begin{align}F(x)=f(x)=| x |;;text{and};;|F |=| f |=1.end{align}
So,
begin{align}| x |=left|| x | right|=left| F(x) right|;;text{for some};;Fin X^{*};;text{and};;|F |=1.end{align}
Taking $sup $ over such $;Fin X^{*};text{and};|F |=1,$ we get
begin{align}| x |=sup{left| F(x) right|:,Fin X^{*},;|F |=1}.end{align}
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
$endgroup$
add a comment |
$begingroup$
Credits to Martin Argerami. I present the full proof for future readers.
Let $xin X$ be fixed, such that $xneq 0.$ Consider the subspace spanned by $x$, defined by $Z={ alpha x:;alphain Bbb{R} }.$
Define $f$ arbitrarily by
begin{align} f:,&Zto Bbb{R}\& zmapsto alpha | x |. end{align}
Let $zin Z$, then $f(z)= alpha | x |= |alpha x |= |z |.,$ This implies that $|f(z)|=left||z | right|leq left||z |right|=|z |$ which gives boundedness of $f$ and $|f |=1.$
Let $gamma,lambdain Bbb{R}$ and $y,zin Z,$ then $exists,alpha,beta in Bbb{R}$ such that $y=alpha x$ and $z=beta x.$ Then, $f$ is linear, since
begin{align}fleft(gamma y+lambda zright)&= fleft[(gamma alpha+lambda beta)xright] \&= (gamma alpha+lambda beta)| x |\&= gamma (alpha| x |)+lambda( beta| x |)\&= gamma fleft( yright)+lambda fleft(zright).end{align}
Since $f$ is a linear functional, $;alpha fleft( xright)=fleft(alpha xright)=fleft(z right)=alpha | x |$ implies that $fleft( xright)=| x |.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $Fin X^{*}$ s.t. $F(x)=f(x)$ for all $xin Z$ and
begin{align}F(x)=f(x)=| x |;;text{and};;|F |=| f |=1.end{align}
So,
begin{align}| x |=left|| x | right|=left| F(x) right|;;text{for some};;Fin X^{*};;text{and};;|F |=1.end{align}
Taking $sup $ over such $;Fin X^{*};text{and};|F |=1,$ we get
begin{align}| x |=sup{left| F(x) right|:,Fin X^{*},;|F |=1}.end{align}
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
$endgroup$
Credits to Martin Argerami. I present the full proof for future readers.
Let $xin X$ be fixed, such that $xneq 0.$ Consider the subspace spanned by $x$, defined by $Z={ alpha x:;alphain Bbb{R} }.$
Define $f$ arbitrarily by
begin{align} f:,&Zto Bbb{R}\& zmapsto alpha | x |. end{align}
Let $zin Z$, then $f(z)= alpha | x |= |alpha x |= |z |.,$ This implies that $|f(z)|=left||z | right|leq left||z |right|=|z |$ which gives boundedness of $f$ and $|f |=1.$
Let $gamma,lambdain Bbb{R}$ and $y,zin Z,$ then $exists,alpha,beta in Bbb{R}$ such that $y=alpha x$ and $z=beta x.$ Then, $f$ is linear, since
begin{align}fleft(gamma y+lambda zright)&= fleft[(gamma alpha+lambda beta)xright] \&= (gamma alpha+lambda beta)| x |\&= gamma (alpha| x |)+lambda( beta| x |)\&= gamma fleft( yright)+lambda fleft(zright).end{align}
Since $f$ is a linear functional, $;alpha fleft( xright)=fleft(alpha xright)=fleft(z right)=alpha | x |$ implies that $fleft( xright)=| x |.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $Fin X^{*}$ s.t. $F(x)=f(x)$ for all $xin Z$ and
begin{align}F(x)=f(x)=| x |;;text{and};;|F |=| f |=1.end{align}
So,
begin{align}| x |=left|| x | right|=left| F(x) right|;;text{for some};;Fin X^{*};;text{and};;|F |=1.end{align}
Taking $sup $ over such $;Fin X^{*};text{and};|F |=1,$ we get
begin{align}| x |=sup{left| F(x) right|:,Fin X^{*},;|F |=1}.end{align}
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
answered Jan 11 at 6:09
Omojola MichealOmojola Micheal
1,853324
1,853324
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$begingroup$
Well $|x|-epsilon<|x|$
$endgroup$
– SmileyCraft
Jan 10 at 21:18
$begingroup$
@SmileyCraft: Yes, I agree with you! Okay, do you mean $| x|-epsilon<| x|=|f(x)|$?
$endgroup$
– Omojola Micheal
Jan 10 at 21:19