Let $(\ell^{\infty})'$ be the $\mathbb{F}$-vector space of linear and continuous (bounded) functionals $\ell^{\infty}\rightarrow \mathbb{F}$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$ (but we can assume $\mathbb{F}=\mathbb{R}$, if needed) and $\ell^{\infty}$ has the sup norm $\parallel\cdot\parallel_{\infty}$. Let also $c$ be the subspace of $\ell^{\infty}$ consisting of convergent sequences. Then the limit functional $\lim\colon c\rightarrow \mathbb{F}$ sending a convergent sequence to its limit is a continuous, linear functional with operator norm $1$ ($c$ has the sup norm as well).

I am asked to prove or disprove that there exist distinct elements $f,g\in(\ell^{\infty})'$ which extend the limit functional on $c$. I think the claim is true, but, up to now, I have been able to prove only the following fact (for $\mathbb{F}=\mathbb{R}$), using Hahn-Banach's extension Theorem: for every real number $\lambda$ with $-1\leq \lambda \leq 1$, there exists a linear extension $h_{\lambda}$ of the limit functional to the whole $\ell^{\infty}$ such that, for all $\alpha\in\mathbb{R}$ and any convergent sequence $x\in c$, if $y$ is the sequence $((-1)^{n})_{n\in\mathbb{N}}$, then $$h_{\lambda}(ay+x)=a\lambda +\lim(x)\leq \limsup(ay+x).$$ In particular, there are uncountably many linear extension of the limit functional. I can not prove that at least two of these are continuous though.

Can someone help me solving this problem with a worked solution? (I have looked for Banach limits around, but I have not found an explicit proof of non uniqueness of such continuous extensions of the limit extension).

Thanks in advance.

  • $\begingroup$ Would it be an idea to extend not from $c$ but from a slightly larger space? I.e. take $x_0 \in l^\infty\setminus c$. Define $L = c + span(x_0)$. Then define $f$ as $f(y + a\cdot x_0) = \lim y + a$, $g(y+a\cdot x_0)=\lim y$ for $y\in c$. Then $f\ne g$ on $L$, and their extensions would also be not equal. Are these $f,g$ continuous? $\endgroup$ – daw Nov 29 '14 at 15:55
  • $\begingroup$ Dear @daw, this is precisely what I did to get the $h_{\lambda}$ above and indeed, if you take $x_{0}$ to be the sequence $((-1)^{n})_{n\in\mathbf{N}}$, then your $f$ is my $h_{1}$ on $L$ and your $g$ is my $h_{0}$ on $L$. Still, the problem is that I do not see how to show that there is at least one $h_{\lambda}$ with $\lambda\neq 0$ such that the restriction of $h_{\lambda}$ to $L$ is continuous... $\endgroup$ – Marco Vergura Nov 29 '14 at 16:09
  • $\begingroup$ Sorry, was too quick when reading your question. Does my answer help? This should show continuity of $h_\lambda$ on $c+span(y)$. $\endgroup$ – daw Nov 29 '14 at 16:49
  • $\begingroup$ @MarcoVergura: if you don't know that your functionals are bounded, I don't see how you applied Hahn-Banach. $\endgroup$ – Martin Argerami Nov 30 '14 at 2:44

The idea is to define two functionals on a larger space than $c$, but which are equal to the limit-functional on $c$. Then extend these two functionals to $l^\infty$.

Take $x:=(1,-1,1,-1,\dots)$ the alternating sequence. Define $L:=c + span(x)$. Construct the functionals $f_1,f_2:L\to \mathbb F$ as $$ f_1(y + ax):=\lim(y) + 2a, \quad f_2(y + ax):=\lim(y) - 2a\quad y\in c, a\in \mathbb F. $$ It remains to show that $f_1$ and $f_2$ are continuous with respect to the $l^\infty$-norm.

Let $L$ and $R$ denote the left and right-shift on $l^\infty$. Then $f_2(z)=f_1(Lz)$ for all $z\in L$. Thus, it suffices to show boundedness of $f_1$.

Now let $z=y+ax$, $y\in c$, $a\in\mathbb F$ be given. Then it holds $$ \lim(z + Lz) = \lim y, \quad \lim(x*(z-Lz)) = 2a, $$ where $x*z$ denotes the elementwise multiplication of $x$ and $z$. hence $$ f_1(z) = \lim(z+Lz) + \lim(x*(z-Lz)). $$ Then we can estimate $$ \begin{split} |f_1(z)|&\le |\lim(z+Lz)| + |\lim(x*(z-Lz))| \\ &\le \|\lim\|\cdot( \|I+L\| + \|I-L\|)\|z\|. \end{split} $$ Hence $f_1$ and consequently $f_2$ are bounded. Extending both functionals to $l^\infty$ yields two different extensions of the limit functional.

This construction seems to be too complicated. Anyone aware of a simpler example?


The key observation is that, in the OP's notation, $$\tag{1} \|ay+x\|_\infty\geq\max\{|a|,|\lim x|\}. $$ (proof below). Then $$ |h_\lambda(ay+x)|=|\lambda a+\lim x|\leq|\lambda|\,|a|+|\lim x|\leq(|\lambda|+1)\|\lambda a y + x\|_\infty. $$ Thus $h_\lambda$ is bounded and now you can extend by Hahn-Banach. By construction $h_\lambda\ne h_\lambda'$ if $\lambda\ne\lambda'$.

Proof of (1). This follows from the fact that, for $n$ big enough, $ay_n+ x_n$ is alternatively very near $a+\lim x$ and $-a+\lim x$ (because $x_n$ is very near $\lim x$). More explicitly:

Claim. $$ \max\{|\lambda+\mu|,|\lambda-\mu|\}\geq\sqrt{|\lambda|^2+|\mu|^2}\geq\max\{|\lambda|,|\mu|\}. $$ To prove the claim, assume first that $\lambda\in\mathbb R+$, $\mu=re^{it}$. Then $$ |\lambda\pm re^{it}|^2=(\lambda+r\cos t)^2+r^2\sin^2t=\lambda^2+r^2\pm 2\lambda r\cos t. $$ Since either $\cos t\geq0$ or $\cos t<0$, one of the two choices of sign puts the expression above $\lambda^2+r^2=|\lambda|^2+|\mu|^2$.

When $\lambda=se^{iv}$, $|\lambda+r^{it}|=|s+re^{i(t-v)}|$ and we can proceed as above.


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