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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.

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  • $\begingroup$ 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. $\endgroup$ – Larsson Mar 11 at 11:29
  • $\begingroup$ Tried to provide an example in my answer - however maybe not as explicit as you might have wished. $\endgroup$ – Maksim Mar 11 at 12:28
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    $\begingroup$ It should make you worry that you have not used that $M$ is closed. $\endgroup$ – Carsten S Mar 11 at 14:47
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    $\begingroup$ 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 $\lim_n f(1/n) = 0$. $\endgroup$ – Paul Sinclair Mar 11 at 16:30
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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.

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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.

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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.

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