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I'm trying to show that there exists a functional in $l^\infty$ whose value lies between the infimum and supremum of a sequence. That is, there exists a functional $\phi:l^\infty\longrightarrow \mathbb R$ such that $\forall x=(x_n)\in l^\infty$, $\liminf x_n\leq\phi(x)\leq \limsup x_n$. My argument is as follows:

Since $l^1\subset l^\infty$, we first define $\phi(x)=\lim_{N\rightarrow \infty}\frac{x_1+x_2+...+x_N}{N}$ for $x=(x_1,x_2,...x_n)\in l^1$. Since $x\in l^1$, the partial sums converge and hence this is basically the zero functional. Now, the Hahn-Banach theorem (in some version) asserts that any continuous linear functional dominated by a sublinear function $p$ (i.e. $p(x+y)\leq p(x)+p(y)$ defined on a subspace can be extended to the whole space. Since $\limsup$ and $-\liminf$ are sublinear, the functional on $l^1$ can be extended to $l^\infty$ without violating the inequalities. Is this solution correct? If not, what modifications are necessary?

Martin Sleziak
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adrija
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1 Answers1

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Yes, I think your argument works.

Here is another approach: Define, for any $x\in\ell^\infty$, functionals $\phi_n$ given by $$ \phi_n(x)=\frac{x_1+\cdots+x_n}n. $$ All these functionals satisfy your property, and moreover $\|\phi_n\|=1$ for all $n$. As the unit ball is weak$^*$-compact, the sequence $\{\phi_n\}$ has cluster points. Any such cluster point will be one of the functionals you are looking for.

The argument above is sometimes referred as a Banach Limit. The same can be achieved using free ultrafilters. Deep down, it is the same.

As Nate mentions, the functionals $\phi_n (x)=x_n $ can be used.

Martin Argerami
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    Another approach that would work here is to consider the functionals $\pi_n(x) = x_n$. A weak-* cluster point $\ell$ of this sequence will do the job. Unlike the Banach limit, $\ell$ will not be shift invariant, but it does have the interesting property that $\ell(x)$ is always a subsequential limit of the sequence ${x_n}$. – Nate Eldredge Oct 09 '15 at 03:35
  • Good point. That's certainly simpler. – Martin Argerami Oct 09 '15 at 04:06
  • I have a doubt. Why are you saying that the functional I define for $l^1$ will be zero on $l^\infty$? It's zero on $l^1$ and does satisfy the inequalities. The version of Hahn-Banach I'm using is this: any functional on a subspace dominated by a sublinear map can be extended to the whole space such that the extended functional is dominated by the same sublinear map. With this result, the functional I define (zero on $l^1$) can be extended to $l^\infty$ and it'll continue to satisfy the inequalities-I'm not picking any extension of the functional, but one which satisfies the inequality. – adrija Oct 09 '15 at 18:42
  • In fact, why can't I just say this? The zero functional is bounded above by $limsup$ in $l^1$ and so, by above version, it has an extension to $l^\infty$ which is again bounded above by $limsup$. Once that is achieved, the $liminf$ inequality is just a consequence of that. – adrija Oct 09 '15 at 18:51
  • In fact, the extension can't be the zero functional. Note that if $\phi=0$, then your example reads $\phi(1,1,...)=0\leq 1=limsup(1,1,...)$. But then $\phi(-1,-1,...)=0>limsup(-1,-1,...)$. So the zero functional is not an extension that's dominated by the $limsup$ sublinear map and so won't be chosen at all if I apply above version of Hahn-Banach. – adrija Oct 09 '15 at 19:27
  • Yes, I think you are right. – Martin Argerami Oct 10 '15 at 11:00