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I'm stuck with the following problem. Show that the map: $ r(x)=\inf\limits_{k\in\mathbb{N}}\limsup\limits_{m\to\infty}\frac{1}{k}\sum\limits_{j=0}^{k-1}S^j(x)(m) $ is subadditive on $\ell_\infty(\mathbb{N})$. Here $ S:\ell_\infty(\mathbb{N})\to\ell_\infty(\mathbb{N}): (x(1),x(2),x(3),\ldots)\mapsto(0,x(1),x(2),x(3),\ldots) $

Any help greatly appreciated!

2 Answers 2

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I will use slightly different notation so that I can copy part of the text from an older answer. In that answer you can read how this is related to the existence and possible values of Banach limits.


Let $T:\ell_\infty\to\ell_\infty$ be shift-operator $T:{(x_n)}\mapsto{(x_{n+1})}$.

For any bounded sequence $x$ we define $T_n(x)=\frac{x+Tx+\dots+T^{n-1}x}n$. I.e., $T_n(x)$ is the sequence $\left(\frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n\right)_{k=1}^\infty$. Let us denote $ \begin{gather*} M(x)=\lim_{n\to\infty} \limsup T_n(x) = \inf_{n\in\mathbb N} \limsup T_n(x),\\ m(x)=\lim_{n\to\infty} \liminf T_n(x) = \inf_{n\in\mathbb N} \liminf T_n(x). \end{gather*} $ The fact that the the above limits exist and that they are equal to infima can be shown using Fekete's lemma - a proof of this lemma can be found in this answer. I've added details below.

Note that $M(x)$ is the same thing as what you denoted $r(x)$ in your question.


It is easy to see, that for every $n\in\mathbb N$ and for every $x,y\in\ell_\infty$ we have $T_n(x+y)=T_n(x)+T_n(y)$. Now we get $\limsup T_n(x+y) = \limsup(T_n(x)+T_n(y)) \le \limsup T_n(x)+\limsup T_n(y)$ from the subadditivity of limit superior.

Now from the basic properties of limit you get $M(x+y) = \lim_{n\to\infty} \limsup T_n(x+y) \le \lim_{n\to\infty} (\limsup T_n(x) +\limsup T_n(y))= \lim_{n\to\infty} \limsup T_n(x) + \lim_{n\to\infty} \limsup T_n(y) = M(x)+M(y).$

(Probably it would be possible to get the required result with $\inf$ instead of $\lim$, but I think this way the solution is nicer.)


Now I get back to the fact that the both expressions (the one using $\lim$ and the one using $\inf$ are the same.)

A sequence $(a_n)$ is called subadditive if for any $m,n\in\mathbb N$ $a_{n+m}\leq a_n+a_m.$

Fekete's lemma. For every subadditive sequence $(a_n)$, the limit $\lim\limits_{n \to \infty} \frac{a_n}{n}$ exists and is equal to $\inf \frac{a_n}{n}$. (The limit may be $-\infty$.)

So to apply Fekete's lemma we need to show that $a_n=\limsup_k (x_k+x_{k+1}+\dots+x_{k+n-1})$ is a subadditive sequence. It suffices to notice that $$a_{m+n} = \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m-1})\le \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1})+ \limsup_{k\to\infty}(x_{k+n}+\dots+x_{k+n+m-1}) = a_n+a_m.$$

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    I still think the both thing are (more-or-less) the same. The expression $\sum\limits_{j=0}^k S^j(x)(m) = S^0(x)(m)+S^1(x)(m)+\dots+S^{k-1}(x)(m)= x_m+x_{m-1}+\dots+x_{m-k+1}$ is the sum of elements of sequence $(x_n)$ in the interval of the length $k$, where $m$ is the endpoint of this interval. I use interval of some length where the beginning of the interval is changing. The only different terms are the terms where m, but they don't influence $\limsup$. $A$nyway, thanks for pointing this out (I did not notice that) and also for correcting a typo in the post.2012-09-30
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Denote $T_k:=\displaystyle\frac1k\cdot (I+S+S^2+\dots+S^{k-1} $), then $\ T_k{\bf x} = \left( \frac{x_0+..+x_{k-1}}{k}, \frac{x_1+..+x_k}{k}, \frac{x_2+..+x_{k+1}}{k} ,\dots\right)$.

For given $\bf x$, $\bf y$, by the definition of $\inf$, for each $\varepsilon>0$ there are $k$ and $l$ indices such that $\limsup (T_k{\bf x}) < r({\bf x})+\varepsilon/2 \ $ and $\ \limsup(T_l{\bf y}) < r({\bf y})+\varepsilon/2$.

We have to provide an $m\in\mathbb N$ such that $\limsup(T_m({\bf x+y})) \le \limsup(T_k{\bf x}) + \limsup(T_l{\bf y}) $...

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    ..neither am I.. what was in my head when I started typing, is just not true.. I was thinking something like $m:=kl$ or their least comon multiplier..2012-09-28