# An alternative proof of an application of Hahn-Banach

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.

I think you will run into a problem with $$f$$ being continuous.
In case you have a sequence of linear combinations of Hamel basis which do not have $$x_0$$ as summand but converge to $$x_0$$ then the limit of the $$f$$ values of these linear combinations will be zero and not one.

As an example that a functional constructed using a Hamel Basis for its definition and which is not continuous, let us consider an infinite dimensional Banach space $$X$$ having a Hamel Basis $$mathbb{H}$$. Let us consider the family of projections $$pi_h:Xtomathbb{R}$$ where $$hin mathbb{H}$$. For fixed $$h_0$$ these are defined for any $$x$$ which will be represented as unique finite sum $$x=sum_{hinmathbb{H}}alpha_hh$$ with appropriate $$alpha_hinmathbb{R}$$ as $$pi_{h_0}(x)= alpha_{h_0}$$.
Then there is at least one $$pi_{h}$$ which is not continous.
You can see that by taking a countable subset of $$mathbb{H}$$, say $$h_0, h_1,…$$ and consider
$$x=sum_{k=0}^infty frac{h_k}{2^k||h_k||}$$
By construction $$x$$ is not a finite linear combination of the $${h_k}$$.
So for all $$kinmathbb{N}$$ we have $$pi_{h_k}(x)=0$$. On the other hand, if all $$pi_{h_k}$$ would be continuous then we would have for all $$kinmathbb{N}$$ : $$pi_{h_k}(x)=frac{1}{2^k||h_k||} >0$$, which is a contradiction. So at least one projection will be discontinuous.
Rhys and Maksim have both given excellent answers pertaining to the issue of continuity presented by your construction, but I believe it should be pointed out that, even if your construction worked, it would in fact be no less “overkill” than using Hahn-Banach. In the context of foundations your method is even more “overkill”. The statement that every vector space has a Hamel basis is in fact equivalent to the axiom of choice over $$mathbf{ZF}$$, whereas Hahn-Banach can be proven in $$mathbf{ZF}$$ + the ultrafilter theorem, which is a strictly weaker axiom than choice.