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Let $E$ be a Banach space and $F \subsetneq E$ a closed subspace. For $e \in E\backslash F$, we can construct a map on $\text{span} \{F \cup \{e\}\}$ as $$ \tilde{\psi}(\lambda e+\sum_jc_j g_j)=\lambda,\ \ g_1,\ldots,g_n\in F. $$ By Hahn-Banach theorem, we can construct $\psi$ on $E$ such that $\psi(e) = 1$ and $\psi|_{F} = 0$. Now, by this contruction we cannot conclude that $F = \text{ker } \psi$ since the kernel of this map may be bigger than $F$. Is it possible to construct $\psi$ such that $F = \text{ker } \psi$ ?

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  • $\begingroup$ You should be writng just $\lambda e+g$. $F$ is a subspace and $\sum c_ig_i \in F$. $\endgroup$ May 19, 2021 at 10:04

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Not possible. The kernel of any non-zero continuous linear functional has co-dimension $1$ so we cannot do this unless your $F$ has this property.

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  • $\begingroup$ If we assume that $F$ has codimension one, is the result true? I guess it is $\endgroup$ May 19, 2021 at 10:34
  • $\begingroup$ @GiuseppeNegro Any subspace of co-dimension $1$ is the kernel of a scalar valued linear map and if the subspace is closed then the linear map is continuous. $\endgroup$ May 19, 2021 at 11:26

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