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Hahn Banach Theorem states that given a linear continuous functional $f$ on a subspace $N$ of a normed space $M$, it can be extended to a linear functional $F$ on the whole space $M$ and the norm of the extension is the same as the one of $f$.

Can someone tell me if something is wrong in the following lines:

Fact 1: Every vector space has a basis (use axiom of choice if it is infinite dimensional). Fact 2: A linear map is defined if we give the image of a basis.

Take $\{e_i\}_{i\in I}$ a basis of $M$ extending one of $N$, and define $F(e_i):=f(e_i)$ if $e_i$ is in $N$ and $0$ otherwise. This is clearly an extension of $f$, plus the norm of $F$ is the same as that of $f$ because sup $|F(x)|/||x||$ is $0$ for $x \in M \setminus N$ and $||f||$ if $x \in N$.

In case is correct this seems like a very simple proof of this theorem!

inquisitor
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    Hahn Banach Theorem provides a continuous linear extension $F$ of your continuous linear functional $f$. Your "proof" does not take into account continuity of the alleged extension. – Marco Vergura Dec 10 '14 at 13:55
  • @Marco Vergura Well if $F$ is bounded and linear then it is also continuous. And since $f$ is bounded $F$ it is as well by the last observation – inquisitor Dec 10 '14 at 13:59
  • @DavidMitra That's even better. I was a little bit too fast in reading the question and I thought the OP had taken a basis of $N$, then completed it to a basis of $M$ and then extended $f$ by zero out of $N$. This process, I guess, tends to make continuity of the extension impossible to control. – Marco Vergura Dec 10 '14 at 14:04
  • Yes I was thinking to do the extension actually, so I'm going to update the OP. Thanks @David Mitra – inquisitor Dec 10 '14 at 14:07
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    Marco's comment is on point. If $x\in M\setminus N$, how do you know that the expansion of $x$ with respect to the basis is a linear combination of $e_i$'s in $M\setminus N$? It seems this is what you're assuming... – David Mitra Dec 10 '14 at 14:20
  • Yes, now I see it. Thanks Marco Vergura and DavidMitra – inquisitor Dec 10 '14 at 14:29

2 Answers2

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I hope this can serve as a counterexample.

Let $X$ be an infinite dimensional Banach space and $\mathcal{B}=\{e_\gamma: \gamma\in \Gamma \}$ be a Hamel basis of $X$. By the comment of Mike F to this post, we know there exists an unbounded coordinate functional $g_{\gamma_0}$ associated to $e_{\gamma_0}$.

Now let $Y=\text{span}\{ e_{\gamma_0} \}$, a one-dimensional subspace of X. Clearly $g_{\gamma_0}|Y$ is continuous on $Y$ and $g_{\gamma_0}$ is your extension of $g_{\gamma_0}|Y$. But $g_{\gamma_0}$ is unbounded by the first paragraph.

Community
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No One
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Take the space $\mathbb R^2$ with the usual norm. Choose a non-orthogonal basis. Apply your method to extend from $1$ dimension to $2$. See that the norm is not preserved.

GEdgar
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