10:38
In another room there was a conversation related to extensions of the usual limit to $\ell_\finty$ (with some additional conditions). I am not sure to which extent this is generally accepted terminology, but we used the name generalized limit.
There was also some additional questions regarding
Banach limits.
In mathematical analysis, a Banach limit is a continuous linear functional
ϕ
:
ℓ
∞
→
C
{\displaystyle \phi :\ell ^{\infty }\to \mathbb {C} }
defined on the Banach space
ℓ
∞
{\displaystyle \ell ^{\infty }}
of all bounded complex-valued sequences
such that for all sequences
x
=
(
x...
Let me just copy-paste the messages from there. (With slight edits - in order to make MathJax properly rendered.)
> Another question, if I consider the Banach sequence which is defined by a generalized limit with shift-invariant. Is $B(\{a_n\})=\operatorname{LIM}(1/n \sum_{j=1}^n x_{j})$ a Banach limit?
> Do you think if it is right?
> Here is the key point: $\operatorname{LIM}((\frac{x_1-x_{n+1}}{n})+B(x_2,x_3,x_4,\dots)=B(x_2,x_3,x_4,\dots)$ since $\Vert x_1-x_{n+1}\Vert \leq 2\Vert x\Vert_{\infty}$ which is bounded. Then $\operatorname{LIM} (\frac{x_1-x_{n+1}}{n})=0$
> Then I can show that $B(\{x_{n+1}\})=B(\{x_{n}\})$.
> Remark: I use $B(\{x_{n}\})=B(x_1,x_2,x_3,\dots)=B(x_1-x_2,x_2-x_3,x_3-x_4,\dots)+B(x_2,x_3,x_4,\dots)=LIM(\frac{1}{n}\sum_{i=1}^n(x_i-x_{i+1}))+B(x_2,x_3,x_4,\dots)$
> Denote the shift operator $T: l_{\infty}\to l_{\infty}$ by $$T: (x_1, x_2, x_3, \dots)\to (x_2, x_3, x_4, \dots).$$ So we have $B(T\{x_{n}\})=B(\{x_{n+1}\})$. > Suppose for $k-1$ we have $B(T^{k-1}\{x_{n}\})=B(\{x_{n+k-1}\})=B(\{x_{n}\})$. Then for $k$, we have $$B(T^{k}\{x_{n}\})=B(T(T^{k-1}\{x_{n}\}))=B(T\{x_{n}\})=B(\{x_{n}\}).$$
> Hence $B(\{x_{n}\})=B(T^{k}\{x_{n}\})=B(\{x_{n+k}\})$.
Here are some short comments on the above proof about Banach limits.
> I consider the Banach sequence which is defined by a generalized limit with shift-invariant. Is $B(\{a_n\})=\operatorname{LIM}(1/n \sum_{j=1}^n x_{j})$ a Banach limit?
Yes, that should work. However, you still need some proof that a shift-invariant generalized limit exists.
Limit along an ultrafilter is an example of such generalized limit.
Possibly you could use the version of Hahn-Banach theorem which deals with extending functionals which are invariant w.r.t. such maps. (Such version is, for example, in Lax's book which you mentioned in your question. And also in Edwards' functional analysis.)
So we will assume that we know that a shift-invariant generalized limit $\operatorname{LIM}$ exists.
> Here is the key point: $\operatorname{LIM}((\frac{x_1-x_{n+1}}{n})+B(x_2,x_3,x_4,\dots)=B(x_2,x_3,x_4,\dots)$ since $\Vert x_1-x_{n+1}\Vert \leq 2\Vert x\Vert_{\infty}$ which is bounded. Then $\operatorname{LIM} (\frac{x_1-x_{n+1}}{n})=0$
It is true that you want to show $$\operatorname{LIM} (\frac{x_1-x_{n+1}}{n})=0.$$
The fact that this sequence is bounded is not enough to get this. However, it's quite easy to see that this sequence converges to zero in the usual sense. (Since the sequence $(x_n)$ is bounded.
So we have $$\lim\limits_{n\to\infty} (\frac{x_1-x_{n+1}}{n})=0$$ and this implies also $\operatorname{LIM} (\frac{x_1-x_{n+1}}{n})=0$.
> Then I can show that $B(\{x_{n+1}\})=B(\{x_{n}\})$.
> Remark: I use $B(\{x_{n}\})=B(x_1,x_2,x_3,\dots)=B(x_1-x_2,x_2-x_3,x_3-x_4,\dots)+B(x_2,x_3,x_4,\dots)=LIM(\frac{1}{n}\sum_{i=1}^n(x_i-x_{i+1}))+B(x_2,x_3,x_4,\dots)$
You have $B(Tx-x)=\operatorname{LIM} (\frac{x_1-x_{n+1}}{n})=0$ and thus $B(Tx)=B(x)$.
