# Easy way to extend continous functional

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

• What is your definition of a basis for a normed space? – uniquesolution Dec 27 '18 at 21:19
• I meant Hamel basis of course. You know any space that doesn't have Hamel basis? – Anton Zagrivin Dec 27 '18 at 21:20
• 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. – uniquesolution Dec 27 '18 at 21:22
• I beg a pardon fot being unclear – 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 $$\{b_0\} \cup \{e_n\}_{n\in\mathbb{N}}$$.

• 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 – 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.