How do I show that for linearly independent set in dual is a dual of a linearly independent set?

Related: How do I prove that a bilinear map is symmetric in the given condition?

Let $X$ be a Banach space over $\mathbb{R}$ and $X^*$ be the dual of $X$. Let $\lambda_1,....\lambda_n$ be linearly independent elements in $X^*$. Then, how do I prove that there are linearly independent elements $x_1,...,x_n$ in $X$ such that $\lambda_i(x_j)=\delta_{ij}$?

Solution 1:

Linear independence of such elements $x_j$ is automatic, since if $\sum c_jx_j=0$ then for all $i$ $0=\lambda_i(\sum c_jx_j)=c_i$.

For existence, consider the map $\Lambda:X\to\mathbb{R}^n$ given by $\Lambda(x)=(\lambda_1(x),\dots,\lambda_n(x))$. If $\Lambda$ were not surjective, there would be a nonzero functional $f:\mathbb{R}^n\to\mathbb{R}$ which vanishes on the image of $\Lambda$. Such an $f$ would have the form $(a_1,\dots,a_n)\mapsto \sum c_ja_j$ for some $c_j\in\mathbb{R}$, and the fact that $f$ vanishes on the image of $\Lambda$ says exactly that $\sum c_j\lambda_j=0$. Since the $\lambda_j$ are linearly independent, this can't happen unless $c_j=0$ for all $j$, i.e. $f=0$.

Thus $\Lambda$ is surjective. In particular, there exist $x_j\in X$ whose images under $\Lambda$ are the standard basis vectors of $\mathbb{R}^n$. This means exactly that $\lambda_i(x_j)=\delta_{ij}$.