# An alternative proof of an application of Hahn-Banach

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.

• Thanks for the answers, I see the problem lies in continuity, can you please give an explicit example may be.. that would be very helpful. – Larsson Mar 11 at 11:29
• Tried to provide an example in my answer - however maybe not as explicit as you might have wished. – Maksim Mar 11 at 12:28
• It should make you worry that you have not used that $M$ is closed. – Carsten S Mar 11 at 14:47
• For an explicit counter-example, consider the set of continuous functions on $\{1/n : n \in \Bbb N^+\}\cup\{0\}$, with the sup norm. Continuity here means $\lim_n f(1/n) = f(0)$. But if $x_0 = 0$, your construction will have $f(0) = 1$ and $\lim_n f(1/n) = 0$. – Paul Sinclair Mar 11 at 16:30

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.
Addendum: 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 $$\pi_h:X\to\mathbb{R}$$ where $$h\in \mathbb{H}$$. For fixed $$h_0$$ these are defined for any $$x$$ which will be represented as unique finite sum $$x=\sum_{h\in\mathbb{H}}\alpha_hh$$ with appropriate $$\alpha_h\in\mathbb{R}$$ 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 $$k\in\mathbb{N}$$ 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 $$k\in\mathbb{N}$$ : $$\pi_{h_k}(x)=\frac{1}{2^k||h_k||} >0$$, which is a contradiction. So at least one projection will be discontinuous.
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.
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.