A linear functional $l$ on a Banach space $B$ is continuous if and only if ${f in B : l(f)=0 }$ is closed .












2












$begingroup$



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







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$endgroup$

















    2












    $begingroup$



    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







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      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







      share|cite|improve this question









      $endgroup$





      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






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      asked Dec 8 '18 at 11:05









      J.GuoJ.Guo

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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 $l=l'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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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 $l=l'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $l=l'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $l=l'$.






          share|cite|improve this answer









          $endgroup$



          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 $l=l'$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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