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If $F$ is nonzero continuous linear functional on the Banach space $E$. Can I find an element $x \in E$ such that $F(x)=1$?

I know $F(0)=0$, but don't think this holds for $1$ as well. Since $ F \ne 0$ then there is $x$ such that $F(x) \ne 0$ but then is there an element say $y=x+\alpha z$ such that $F(y)=1$?

My guess: $z \in Ker(F)$ but this does not imply my claim.

math_for_ever
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  • If $F$ is nonzero there is some $v$ somewhere with $F(v)$ nonzero. What is $F(\frac{1}{F(v)} v)$?. – leslie townes Mar 10 '21 at 04:00
  • Do you have constraints on the field over which you have your Banach space. $\Bbb{Z}/2\Bbb{Z}$ is a perfectly good field which should lead to any number of discrete counterexamples. For many more details, see https://math.stackexchange.com/questions/2151779/normed-vector-spaces-over-finite-fields – Eric Towers Mar 10 '21 at 04:05
  • Note that the range of $F$ is a field of scalars. What $F$ does to the identity operator on $E$ is not that related to the question. – leslie townes Mar 10 '21 at 04:12

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