Help showing subadditivity of a map 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!
 A: 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
$$a_{m+n} = \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m-1})\le
\limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1})+ \limsup_{k\to\infty}(x_{k+n}+\dots+x_{k+n+m-1}) = a_n+a_m.$$
A: 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}) $...
