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

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

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

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.
• If we assume that $F$ has codimension one, is the result true? I guess it is Commented May 19, 2021 at 10:34
• @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. Commented May 19, 2021 at 11:26