As a corollary of the Hahn-Banach theorem, we proved that if $M$ is a closed subspace of a normed linear space $X$, $0\neq x_0\notin M$, then $\exists f \in X^*$ such that $f(x_0)\neq 0$ and $f(y)=0$ $\forall y\in M$.
This looked like overkill to me and I tried to give an alternative proof. I may be wrong: so, we can extend $x_0$ to a Hamel basis of $X$, then define $f(x_0)=1,$ zero on the other basis elements and extend linearly, this basically gives $f(\alpha x_0)=\alpha$ on $\text{span}\{x_0\}$ and zero elsewhere. This should give us a bounded linear functional on $X$. Is this correct? Please point out if I am wrong.
Best Answer
The problem with your proof is that linear functionals defined by specifying their value on a Hamel basis have no reason to be bounded in general.
For example, in this case, if you extend $\{x_0\}$ to a Hamel basis $\{x_0\} \cup \{x_i: i \in \Lambda\}$ then there could e.g. be linear combinations of the type $x_0 + \sum_{i \in I} \lambda_i x_i$ with very small norm (where $I \subseteq \Lambda$ is finite). This is a problem since $f(x_0 + \sum_{i \in I} \lambda_i x_i) = f(x_0) = 1$ and so $\|f\| \geq \|x_0 + \sum_{i \in I} \lambda_i x_i\|^{-1}$ which gets very large as $\|x_0 + \sum_{i \in I} \lambda_i x_i\|$ gets small.