A linear functional $l$ on a Banach space $B$ is continuous if and only if ${f in B : l(f)=0 }$ is closed .
$begingroup$
Suppose $B$ is a Banach space and $S$ a closed proper subspace , and assume $f_0 notin S$ .
(a) Show that there is a continuous linear functional $l$ on $B$ , so that $l(f)=0$ for $f in S$ , and $l(f_0)=1$ . The linear functional $l$ can be chosen so that $vertvert l vertvert = frac{1}{d}$ where $d$ is the distance from $f_0$ to $S$ .
(b) A linear functional $l$ on a Banach space $B$ is continuous iff ${f in B: l(f)=0 }$ is closed . [Hint: (b) is a consequence of (a) ]
My attempt:
$$d(f,S)= inf {| f-f_s | , f_s in S }$$ $$p(f)=frac{d(f,S)}{d(f_0,S)}$$
Let $S_1 = span(S,f_0)$ , and $l(f_0)=1$ , $l(f_s)=0$ $$l(af_0 +bf_s) =al(f_0) +b l(f_s)$$ Then it is obvious that $p(af)=|a|p(f) $ , $p(f_1 + f_2) le p(f_1) + p(f_2)$ , $|l(f)| le p(f) $for $a in C $ ,$f in S_1$ and $f_1,f_2 in S_1$
So By Hahn-Banach , $l$ can be extended from $S_1$ to $B$ , with $|l(f)| le p(f) le frac{|f|_B}{d}$ .
The proof of (a) is complete .
To prove (b) , let $S= {f in B: l(f) =0 }$ , if $S=B$ ,then there's nothing to prove . If $S neq B$ , WLOG we choose $l(f_0)=1 , f_0 notin S$ then we denote $l_s$ the functional $l $ restrict on $S$ , and by (a) , $l_s$ has an extention to a continuous functional $l'$ . But I think $l' neq l$ , and I don't know how to prove (b) then .
functional-analysis banach-spaces
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
$endgroup$
add a comment |
$begingroup$
Suppose $B$ is a Banach space and $S$ a closed proper subspace , and assume $f_0 notin S$ .
(a) Show that there is a continuous linear functional $l$ on $B$ , so that $l(f)=0$ for $f in S$ , and $l(f_0)=1$ . The linear functional $l$ can be chosen so that $vertvert l vertvert = frac{1}{d}$ where $d$ is the distance from $f_0$ to $S$ .
(b) A linear functional $l$ on a Banach space $B$ is continuous iff ${f in B: l(f)=0 }$ is closed . [Hint: (b) is a consequence of (a) ]
My attempt:
$$d(f,S)= inf {| f-f_s | , f_s in S }$$ $$p(f)=frac{d(f,S)}{d(f_0,S)}$$
Let $S_1 = span(S,f_0)$ , and $l(f_0)=1$ , $l(f_s)=0$ $$l(af_0 +bf_s) =al(f_0) +b l(f_s)$$ Then it is obvious that $p(af)=|a|p(f) $ , $p(f_1 + f_2) le p(f_1) + p(f_2)$ , $|l(f)| le p(f) $for $a in C $ ,$f in S_1$ and $f_1,f_2 in S_1$
So By Hahn-Banach , $l$ can be extended from $S_1$ to $B$ , with $|l(f)| le p(f) le frac{|f|_B}{d}$ .
The proof of (a) is complete .
To prove (b) , let $S= {f in B: l(f) =0 }$ , if $S=B$ ,then there's nothing to prove . If $S neq B$ , WLOG we choose $l(f_0)=1 , f_0 notin S$ then we denote $l_s$ the functional $l $ restrict on $S$ , and by (a) , $l_s$ has an extention to a continuous functional $l'$ . But I think $l' neq l$ , and I don't know how to prove (b) then .
functional-analysis banach-spaces
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
$endgroup$
add a comment |
$begingroup$
Suppose $B$ is a Banach space and $S$ a closed proper subspace , and assume $f_0 notin S$ .
(a) Show that there is a continuous linear functional $l$ on $B$ , so that $l(f)=0$ for $f in S$ , and $l(f_0)=1$ . The linear functional $l$ can be chosen so that $vertvert l vertvert = frac{1}{d}$ where $d$ is the distance from $f_0$ to $S$ .
(b) A linear functional $l$ on a Banach space $B$ is continuous iff ${f in B: l(f)=0 }$ is closed . [Hint: (b) is a consequence of (a) ]
My attempt:
$$d(f,S)= inf {| f-f_s | , f_s in S }$$ $$p(f)=frac{d(f,S)}{d(f_0,S)}$$
Let $S_1 = span(S,f_0)$ , and $l(f_0)=1$ , $l(f_s)=0$ $$l(af_0 +bf_s) =al(f_0) +b l(f_s)$$ Then it is obvious that $p(af)=|a|p(f) $ , $p(f_1 + f_2) le p(f_1) + p(f_2)$ , $|l(f)| le p(f) $for $a in C $ ,$f in S_1$ and $f_1,f_2 in S_1$
So By Hahn-Banach , $l$ can be extended from $S_1$ to $B$ , with $|l(f)| le p(f) le frac{|f|_B}{d}$ .
The proof of (a) is complete .
To prove (b) , let $S= {f in B: l(f) =0 }$ , if $S=B$ ,then there's nothing to prove . If $S neq B$ , WLOG we choose $l(f_0)=1 , f_0 notin S$ then we denote $l_s$ the functional $l $ restrict on $S$ , and by (a) , $l_s$ has an extention to a continuous functional $l'$ . But I think $l' neq l$ , and I don't know how to prove (b) then .
functional-analysis banach-spaces
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
$endgroup$
Suppose $B$ is a Banach space and $S$ a closed proper subspace , and assume $f_0 notin S$ .
(a) Show that there is a continuous linear functional $l$ on $B$ , so that $l(f)=0$ for $f in S$ , and $l(f_0)=1$ . The linear functional $l$ can be chosen so that $vertvert l vertvert = frac{1}{d}$ where $d$ is the distance from $f_0$ to $S$ .
(b) A linear functional $l$ on a Banach space $B$ is continuous iff ${f in B: l(f)=0 }$ is closed . [Hint: (b) is a consequence of (a) ]
My attempt:
$$d(f,S)= inf {| f-f_s | , f_s in S }$$ $$p(f)=frac{d(f,S)}{d(f_0,S)}$$
Let $S_1 = span(S,f_0)$ , and $l(f_0)=1$ , $l(f_s)=0$ $$l(af_0 +bf_s) =al(f_0) +b l(f_s)$$ Then it is obvious that $p(af)=|a|p(f) $ , $p(f_1 + f_2) le p(f_1) + p(f_2)$ , $|l(f)| le p(f) $for $a in C $ ,$f in S_1$ and $f_1,f_2 in S_1$
So By Hahn-Banach , $l$ can be extended from $S_1$ to $B$ , with $|l(f)| le p(f) le frac{|f|_B}{d}$ .
The proof of (a) is complete .
To prove (b) , let $S= {f in B: l(f) =0 }$ , if $S=B$ ,then there's nothing to prove . If $S neq B$ , WLOG we choose $l(f_0)=1 , f_0 notin S$ then we denote $l_s$ the functional $l $ restrict on $S$ , and by (a) , $l_s$ has an extention to a continuous functional $l'$ . But I think $l' neq l$ , and I don't know how to prove (b) then .
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
asked Dec 8 '18 at 11:05
J.GuoJ.Guo
3759
3759
add a comment |
add a comment |
1 Answer
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$begingroup$
No, you can't have $l'neq l$. Note that the restrictions of $l$ and $l'$ to $S$ are the same and that $l(f_0)=l'(f_0)=1$. But $B=Sbigoplusmathbb{R}f_0$, and therefore $l=l'$.
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
add a comment |
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$begingroup$
No, you can't have $l'neq l$. Note that the restrictions of $l$ and $l'$ to $S$ are the same and that $l(f_0)=l'(f_0)=1$. But $B=Sbigoplusmathbb{R}f_0$, and therefore $l=l'$.
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
add a comment |
$begingroup$
No, you can't have $l'neq l$. Note that the restrictions of $l$ and $l'$ to $S$ are the same and that $l(f_0)=l'(f_0)=1$. But $B=Sbigoplusmathbb{R}f_0$, and therefore $l=l'$.
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
add a comment |
$begingroup$
No, you can't have $l'neq l$. Note that the restrictions of $l$ and $l'$ to $S$ are the same and that $l(f_0)=l'(f_0)=1$. But $B=Sbigoplusmathbb{R}f_0$, and therefore $l=l'$.
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
No, you can't have $l'neq l$. Note that the restrictions of $l$ and $l'$ to $S$ are the same and that $l(f_0)=l'(f_0)=1$. But $B=Sbigoplusmathbb{R}f_0$, and therefore $l=l'$.
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 8 '18 at 11:31
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
add a comment |
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
$begingroup$
Do you mean that $Sbigoplusmathbb{R}f_0 = Span (S,f_0)$ ? if so , I can't see that $B=Sbigoplusmathbb{R}f_0$ .
$endgroup$
– J.Guo
Dec 8 '18 at 11:37
1
1
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
If $xin B$, $x=x-l(x)f_0+l(x)f_0$. Now, note that $x-l(x)f_0inker l$ and that $l(x)f_0inmathbb{R}f_0$.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 11:47
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
$begingroup$
Ah , I see it ! Thank you .
$endgroup$
– J.Guo
Dec 8 '18 at 11:49
add a comment |
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