Let $F \subsetneq E$ a closed subspace, then $\exists \psi \in E^*\backslash\{0\}$ such that $F = \text{ker } \psi$.

Question

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 $F = \text{ker } \psi$ ?

Answer 1

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.