Let $E$ be Banach, $F$ a proper subspace, take $y$ outside $F$, define $f(x+ty)=t$ for $x$ in $F$ and $t$ real, it is a linear functional and the kernel is $F$. Then $F$ not dense iff $F$ closed in $E$ iff $f$ is bounded. Is it correct to say I can't have a proper not dense not closed subspace? Thank you.
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1Its not true , consider $Ε = l^{\infty}$ and $F = c_{00}$ then $E$ is a banach space but $F$ its neither closed nor dense in $E$. The problem in your proof is that $F \subsetneq $ kernel. – ChrisNick92 Apr 19 '18 at 15:17
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Thank you for the comment but why is that the kernel of $f$ isn't $F$, I have to find $z$ such that $f(z)=0$, writing $z=x+ty$ for some $t$ i get that this happens iff $z=x$ with $x$ in $F$. Where is the mistake? – lucmobz Apr 19 '18 at 15:27
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1Ohh okay , i thought that you extended your functional in the whole space $E$ by using Hahn-Banach. If you extend the functional in the whole space $E$ then you might have $F \subsetneq $ kernel. If you dont extend your functional is defined only in $span(F \cup {y})$ which might not be the whole $E$ so you cant use the preposition which says $F$ not dense in $E$ iff $F$ closed in $E$ iff f is bounded.This prepositions holds for the space $span(F\cup {y})$ . I hope this is better ! – ChrisNick92 Apr 19 '18 at 16:52
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Ohh okay , i thought that you extended your functional in the whole space E by using Hahn-Banach. If you extend the functional in the whole space E then you might have F⊊ kernel. If you dont extend your functional is defined only in span(F∪{y}) which might not be the whole E so you cant use the preposition which says F not dense in E iff F closed in E iff f is bounded.This prepositions holds for the space span(F∪{y}) . I hope this is better !
lucmobz
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