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?


1 Answer 1


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.

  • $\begingroup$ 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...) $\endgroup$ Apr 16, 2019 at 23:03
  • $\begingroup$ 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 $0$) $\endgroup$
    – mihaild
    Apr 16, 2019 at 23:07
  • $\begingroup$ argh, you're right, of course. I forgot that $|\operatorname{sgn} x| = 1$ :/ $\endgroup$ Apr 16, 2019 at 23:41

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