Hahn-Banach needed to show equality?

Question

Let $X$ be a normed space and $x_1, x_2 \in X$. Suppose $x^{\ast}(x_1) = x^{\ast}(x_2)$ for all $x^{\ast} \in X^{\ast}$. Then $x_1 = x_2$.

Do we need Hahn-Banach (hence, equivalently some sort of axiom of choice) to prove this?

Answer 1

For finite dimensional vector spaces we do not need Hahn Banach. Let $X$ be a finite dimensional space with basis $\{e_1,\dots,e_n\}$, and $\{f_1,\dots,f_n\}$ be the usual basis for $X^*$. Let $x_2=\sum_{i=1}^n\alpha_ie_n$ and $x_1=\sum_{i=1}^n\beta_ie_n$, and suppose $f(x_1)=f(x_2)$ for all $f\in X^*$. In particular that means that $f_i(x_1)=f_i(x_2)$ for each $i\in \{1,\dots,n\}$, which means that $\alpha_i=\beta_i$ for each $i\in \{1,\dots,n\}$, so $x_1=x_2$.

We can generalise this argument to any vector space with a Schauder basis, using the continuity of elements in the dual, but in an infinite vector space without a Schauder basis I believe we will need to use the Hahn-Banach theorem. Although I do not rule out the possibility that there may be intermediate spaces where we don't need the full Hahn-Banach theorem, and I hope more learned members of the community will be able to discuss such spaces.