Continuous extension of the limit functional

Question

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.

Answer 1

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?

Answer 2

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.