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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!

Norbert
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Tim
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2 Answers2

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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 $$a_{m+n} = \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m-1})\le \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1})+ \limsup_{k\to\infty}(x_{k+n}+\dots+x_{k+n+m-1}) = a_n+a_m.$$

Martin Sleziak
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  • OP used right shift operator instead of left shift – Norbert Sep 30 '12 at 12:18
  • 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 $m<k$, but they don't influence $\limsup$. Anyway, thanks for pointing this out (I did not notice that) and also for correcting a typo in the post. – Martin Sleziak Sep 30 '12 at 12:30
  • It is clear to me that the expression $$\limsup (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m_1})\le \limsup (x_k+x_{k+1}+\dots+x_{k+n-1})+\limsup(x_{k+n}+\dots+x_{k+n+m-1})$$ is true. But to prove that the sequence $(a_n)n$ defined by $a_n=\limsup\limits{n\to\infty}(x_k+x_{k+1}+\dots+x_{k+n-1})$ is subadditive, we have to show that $$\limsup (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m_1})\le \limsup (x_k+x_{k+1}+\dots+x_{k+n-1})+\limsup(x_{k}+\dots+x_{k+m-1})$$ – MAS Aug 08 '19 at 08:53
  • Sorry. $(a_n)n$ is defined by $a_n=\limsup\limits{k\to\infty}(x_k+x_{k+1}+\dots+x_{k+n-1})$. – MAS Aug 08 '19 at 09:03
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    @BijanDatta These two values are the same (for any fixed $m$ and $n$):: $a_m=\limsup\limits_{k\to\infty} (x_{k}+\dots+x_{k+m-1})=\limsup\limits_{k\to\infty} (x_{k+n}+\dots+x_{k+n+m-1})$. – Martin Sleziak Aug 08 '19 at 09:09
  • Oooh yes. Because $\limsup\limits_{k\to\infty}z_k$ does not depend on first finitely many terms of $(z_k)_k$. – MAS Aug 08 '19 at 09:20
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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}) $...

Berci
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    I'm not sure to see how you provide such a m... – Tim Sep 28 '12 at 13:55
  • ..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.. – Berci Sep 28 '12 at 13:59