# Uniqueness of Hahn-Banach extensions for $c_0 \subseteq \ell^\infty$

Let $$c_0$$ be the space of real sequences converging to zero with supremum norm. $$c_0$$ is a (closed) subspace of $$\ell^\infty$$, the space of bounded real sequences. A $$f \in {c_0}^*$$ corresponds to a $$\hat{f} \in \ell^1$$ via

$$f(x) = \sum_{k=0}^\infty \hat{f_k} \, x_k \tag{1} ~,$$ and $$\|f\| = \|\hat{f}\|_1$$. It it obvious that map $$F \in (\ell^\infty)^*$$ defined again by $$(1)$$ is a Hahn-Banach extension of $$f$$, meaning that $$F|_{c_0} = f$$ and $$\|F\| = \|f\|$$.

I have read that in this case the extension $$F$$ is unique. Why?

For simplicity lets assume $$\|\hat f\|_1 = 1$$.

Let $$F_0 \in (l^\infty)^*$$ be defined as $$F_0(x) = \sum \hat f_k x_k$$ and $$F$$ be any extension of $$f$$ that disagrees with $$F_0$$. Let $$x = (x_1, x_2, \ldots) \in l^\infty$$ have norm of $$1$$ be a vector on which $$F_0$$ and $$F$$ disagree: $$F(x) - F_0(x) > \varepsilon > 0$$ (if difference is less then $$0$$ - replace $$x$$ with $$-x$$). Let $$N$$ be such that $$\sum\limits_{k=1}^N |\hat f_k| > 1 - \frac{\varepsilon}{2}$$.

Let $$y = (0, 0, \ldots, 0, x_{N + 1}, \ldots)$$. We have $$\|y\| \leqslant 1$$, $$F(y) - F_0(y) > \varepsilon$$ and $$|F_0(y)| < \frac{\varepsilon}{2}$$, so $$F(y) > \frac\varepsilon 2$$.

Let $$z = (\operatorname{sgn}\hat f_1, \operatorname{sgn}\hat f_2, \ldots, \operatorname{sgn}\hat f_{N}, 0, 0, \ldots)$$. As $$F$$ is extension of $$f$$, we have $$F(z) = \sum\limits_{k=1}^N |\hat f_k| > 1 - \frac{\varepsilon}{2}$$

Then we have $$\|y + z\| = 1$$. At the other hand, we have $$F(y + z) = F(y) + F(z) > \frac{\varepsilon}{2} + 1 - \frac{\varepsilon}{2} = 1$$. This implies $$\|F\| > 1$$.

So any extension that disagrees with $$F_0$$ has norm greater then $$1$$ - so $$F_0$$ is the only extension of $$f$$ that has the same norm.

• You seem to be assuming that $x$ not only hast $\infty$-norm 1 but that there actually is a $n > N$ s. t. $|x_n| > 1$, right? (Although this is not a problem...) Apr 16, 2019 at 23:03
• Probably no. The only place we use norm of $x$ is to get $\|y\| \leqslant 1$ ($\|y + z\| = 1$ as $z$ itself has ones on some positions where $y$ is $0$) Apr 16, 2019 at 23:07
• argh, you're right, of course. I forgot that $|\operatorname{sgn} x| = 1$ :/ Apr 16, 2019 at 23:41