Let $\{x_1,\dots,x_n\}$ be a basis of a normed linear space $X$. Use Hahn-Banach to prove that, there exist $n$ linear functionals $f_1,\dots,f_n$ such that $f_i(x_j)=\delta_{ij}$.
I can prove the statement using basic linear algebra fact (a linear map is well defined, defining it on a basis, that's it). But how can I prove the statement using Hahn-Banach and which version?
I was thinking using the following version: For all $x\in X$ there is a linear map $\psi\in X^*$ such that $\psi(x)=\|x\|$ and $\|\psi\|=1$ but I couldn't figure out how to use it.
I would use the following consequence of Hahn Banach (for reference, Folland 5.8(a)): If $\mathcal{M}$ is a closed subspace of normed vector space $\mathcal{X}$ and $x\in \mathcal{X}\backslash \mathcal{M}$, then there exists $f\in X^{*}$ so that $f(x) = \delta = \inf_{y\in \mathcal{M}}\|{x - y}\|$, $\|f\| = 1$, and $f\big|_{M} = 0$.
Then if $\mathcal{M} = \text{span}(x_{2},...,x_{n})$, we may find $f$ from the previous theorem so that $f(x_{1}) = \inf \{\|x_{1} - y\|: y\in \text{span}(x_{2},...,x_{n})\} = 1$ and $f(x_{k}) = 0$ for all $k > 1$. This is $f_{1}$.
Then repeat this for every $k=1,...,n$ to get $f_{1},...,f_{n}$.