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Let $(x_j) \in X=l^{\infty}(\mathbb{N})$ and define the left shift $T((x_j))=(x_{j+1})$. A linear functional $l:X \to \mathbb{R}$ is called a Banach limit if the following holds

(a) $l$ is positive: if $x_1,x_2, \dots \geq 0$, then $l((x_j))\geq 0$.

(b) $l \circ T=l$

(c) $l((1,1,\dots))=1$.

Establish the existence of a Banach limit using Hahn-Banach.

I considered the subspace $Y\in X$ of constant sequences $x=(a,a,\dots )$ where $a\in \mathbb{R}$. Define $\tilde{l}(x)=a$. Then obviously $\tilde{l}:Y\to \mathbb{R}$ satisfies (a)-(c). By Hahn-Banach there exists a linear functional $l:X\to \mathbb{R}$ such that $l_Y=\tilde{l}$ and $||l||_{X^{*}}= ||\tilde{l}||_{Y^{*}}$. Then certainly condition (c) is satisfied but I am not sure how to show the other ones.

Community
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user30523
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  • You can't. The only thing you know about the functional you constructed is that it maps a constant sequence to the constant element. Many functionals do that, that are not Banach Limits, e.g. projection to first coordinate. In particular, you chose a $Y$ such that $T_Y$ is the identity, so any functional on $Y$ satisfies (b). – Shai Deshe Jan 16 '20 at 20:11

1 Answers1

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Let $X= l^\infty(\mathbb N)$, let $T: X \to X$, given by $T((x_j)_{j \in \mathbb N}) = (x_{j+r})_{j \in \mathbb N}$, where $r \in \mathbb N_+$

Take $p:X \to \mathbb R$ be given by $p(x) = \limsup \frac{1}{n} \sum_{k=1}^n x_k$, that is Cesaro-limit

Check that $p$ is a banach functional $($ that is $p(x+y) \le p(x) + p(y)$ and $p(tx) = tp(x)$ for $t \ge 0$ $)$, hence ( I don't know which version of H-B Theorem you are familiar with, but $p$ is both convex and banach functional. )

Take $\phi : c \to X$, given by $\phi(x) = \lim x_n$, where $c \subset l^\infty(\mathbb N)$ is subspace of convergent sequences.

Clearly $\phi \le p$, since it is known fact from analysis, that for convergent sequences we have $\lim x_n = \lim \frac{1}{n} \sum_{k=1}^n x_k$.

By Hanh Banach theorem ($\phi$ is a linear functional ), we have extension $\psi: X \to \mathbb R$, sch that $\psi(x) = \phi(x)$ for $x \in c$.

Now we want to establish those conditions, firstly note that:

$\psi(x) \le p(x) \le \limsup x_n$

and similarly (since $\psi(-x) = -\psi(x)$)

$\psi(-x) = -\psi(x) \ge -p(x) \ge \liminf(-x)$, so $\psi(x) \ge \liminf(x)$

1) Take $x \ge 0$ (that is $x_k \ge 0$, for all $k \in \mathbb N$). Then both $\liminf, \limsup$ are between $0$ and $\infty$, so $\psi(x) \ge 0$

3) It's easy, we have $\psi((1,1,...)) = \phi((1,1,...)) = \lim (1,1,...) = 1$

2) Note that $\psi(x) - \psi(T(x)) = \psi(x - T(x)) \le p(x-T(x)) = \limsup \frac{1}{n}( \sum_{k=1}^r x_k - \sum_{k=n+1}^{n+r} x_k)$

And since we're in $l^\infty(\mathbb N)$, we have $|x| \le M$, so $p(x-T(x)) \le \limsup \frac{1}{n} \cdot 2rM$ which tends to $0$.

And similarly:

$ \psi(T(x)) - \psi(x) \le p(T(x) - x) \le \limsup \frac{1}{n} \cdot 2rM$, so that

$ \psi(T(x)) = \psi(x)$, which is what we needed.

Presage
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  • Why don't we just extend the limit operator $\ell:c_0\rightarrow \lim_nx_n$ to $\ell^{\infty}$? I think the subset of convergent sequences in $\ell^{\infty}$ is a subspace. –  Feb 13 '22 at 10:27
  • In fact, we did just extend the limit operator. But simply from the definition of the extension it wasn't so obvious (at least for me) that such extension will satisfy both a) and b) conditions. That's the reason of introducing majorizing Cesaro-limit. – Presage Feb 13 '22 at 10:34
  • Oh yes, I think a) can be ensured but it may hard to get b). –  Feb 13 '22 at 10:41