Here is a functional analysis exercise I've not managed to figure out.
Assume that $Y \subset X$ is a closed subspace of the normed space $X$ and that there exists $m_1,m_2,\dots,m_n \in X'$ such that for every $\ell \in Y'$ and any two extensions $\ell_1,\ell_2 \in X'$ we have that $\ell_1 - \ell_2 \in \text{span}\{m_1,m_2,\dots, m_n \}$. Prove that $Y$ has codimension at most $n$.
It seems like one should consider the quotient space $X'/\text{span}\{m_1,m_2,\dots, m_n \}$ since in this space $[\ell_1] = [\ell_2]$ as congruence classes. Also, the fact that the kernel of a non-zero linear functional has codimension 1 might be useful. I would be very grateful if someone could point me in the right direction.
For every $k\in(X/Y)',$ canonically identified to $k\in X'$ such that $k$ is $0$ on $Y,$ we have that $k\in \text{span}\{m_1,m_2,\dots, m_n \}$ (by applying the hypothesis to $\ell=0,\ell_1=k,\ell_2=0$). Hence $\dim(X/Y)'\le n,$ which proves that $X/Y$ is finite dimensional and of dimension $\le n.$