There must be a mistake in my reasoning, but I couldn't find it and that's what I ask you to do:

Let $E_0 \subset E$ be a subspace in a normed space $E$ and $f: E_0 \rightarrow \mathbb{R}$ is continous functional on $E_0$

Let $\{e_0^1, e_0^2, ...\}$ be a basis in $E_0$. We can extend it to the basis of entire $E$ using the Zorn's lemma: let $\{e_1^1, e_1^2,...\}$ be a system that $span \{e_0^1, e_0^2, ..., e_1^1, e_2^2, ...\}=E$

Let's define $f(e_1^i)=0$ for all $i$

So we've extended continous functional from subspace $E_0$ to the entire space $E$. But it's pretty much easier than how we do it in Hahn–Banach theorem so it must be a mistake somewhere

Maybe the extenstion will not be continious or it's not correct. But I can't understand. Need help to find a mistake

  • $\begingroup$ What is your definition of a basis for a normed space? $\endgroup$ – uniquesolution Dec 27 '18 at 21:19
  • $\begingroup$ I meant Hamel basis of course. You know any space that doesn't have Hamel basis? $\endgroup$ – Anton Zagrivin Dec 27 '18 at 21:20
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    $\begingroup$ No, I don't, but I could not see "Hamel" anywhere. Of course the extension does not preserve norm. I believe you will find the problem when you'll try to prove that your extension is bounded. $\endgroup$ – uniquesolution Dec 27 '18 at 21:22
  • $\begingroup$ I beg a pardon fot being unclear $\endgroup$ – Anton Zagrivin Dec 28 '18 at 9:14

Such an extension isn't necessarily continuous.

For an example, consider the canonical (topological) basis $(e_n)_n$ of $\ell^2$ and let $B$ be a Hamel basis for $\ell^2$ which contains all $e_n$.

Pick $b_0 \in B$ different from all $e_n$ and define a continuous functional $f : \operatorname{span}\{b_0\} \to \mathbb{R}$ as $f(\alpha b_0) = \alpha$.

Extend $f$ to a functional $\ell^2 \to \mathbb{R}$ by setting $f(b) = 0, \forall b \in B\setminus \{b_0\}$. In particular $f(e_n) = 0, \forall n\in\mathbb{N}$.

If $f$ were continuous, we would have $f \equiv 0$, but clearly $f(b_0) = 1$. Hence $f$ is not continuous.

To be more explicit, you could take $b_0 = \sum_{n=1}^\infty \frac1n e_n$ as it is linearly independent with all $e_n$, and then let $B$ be a Hamel basis containing $\{b_0\} \cup \{e_n\}_{n\in\mathbb{N}}$.

  • $\begingroup$ Okay, thank you very much. I really had some doubts about continuity of my extention but I needed a counterexample to be 100% sure. Thanks for help $\endgroup$ – Anton Zagrivin Dec 28 '18 at 9:13

Your extension is in general not continuous, let alone norm preserving, which is what Hahn-Banach gives us.

Consider $f:\mathbb{R}\times\{0\}\to\mathbb{R}$, $f(r,0)=r$ and then the unfortunate choice of basis ${(1,0),(1,\varepsilon)}$. Let $g$ denote your extension. Then $g(0,-n\varepsilon)=n$.

This is clearly not norm preserving. In infinite dimensions this can get so bad that the function is not even bounded.


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