I'm working on an exercise consisting of several questions that I can't quite figure out. If someone could give me any tips I'd be very happy!
For $\xi = (x_1, x_2,...) \in \ell^\infty$ define \begin{align*} a_n(\xi) = \sup_{i \in \mathbb{N}} \frac{1}{n} \sum_{j = 0}^{n - 1} x_{i + j}. \end{align*}
The first question was to prove that for any $\xi = (x_1, x_2, ...) \in \ell^\infty$ and $n,m,i \in \mathbb{N}$ such that $n > m$ \begin{align*} \frac{1}{n} \sum_{j = 0}^{n - 1} x_{i + j} \leq (1 - \frac{r}{n}) a_m (\xi) + \frac{m}{n} \| \xi \|_\infty, \end{align*} when $n = km + r$ with $0 \leq r < m$ and $k,r \in \mathbb{N}_0$.
The second question is to conclude, using the first question that \begin{align*} \limsup_{n \to \infty} a_n( \xi) \leq a_m (\xi) \end{align*} for all $m \in \mathbb{N}$ and $\xi \in \ell^\infty$.
The final question is to use the second question to prove that $\lim_{m \to \infty} a_m(\xi)$ exists for all $\xi \in \ell^\infty$.
What I have done so far:
Proving the first question isn't that difficult I think:
We can split $\frac{1}{n} \sum_{j = 0}^{n-1} x_{i + j}$ into \begin{align*} \frac{x_{i} + x_{i +1} + ... + x_{i + km}}{n} + \frac{x_{i + km + 1} + ... + x_{i + n - 1}}{n}, \end{align*} where the first sum is $km$ terms long, and the second sum is $r$ terms long.
Rewriting $(1 - \frac{1}{n}) a_m (\xi) + \frac{m}{n} \| \xi \|_\infty$ gives \begin{align*} (1 - \frac{1}{n}) a_m (\xi) + \frac{m}{n} \| \xi \|_\infty &= \frac{n - r}{n} \sup_{i \in \mathbb{N}} \frac{1}{m} \sum_{j = 0}^{m - 1} x_{i + j} + \frac{m}{n} \| \xi \|_\infty \\ &= k \frac{ \sup_{i \in \mathbb{N}} \sum_{j=0}^{m-1} x_{i+j}}{n} + m\frac{\| \xi \|_\infty}{n}. \end{align*} Compare this with the first sum and make two inequalities and you're mostly done.
I'm kind of stumped on the second and third question though...
We see that \begin{align*} \limsup_{n \to \infty} a_n (\xi) = \lim_{n \to \infty} \sup_{k \geq n, i \in \mathbb{N}} \frac{1}{n} \sum_{j = 0}^{k - 1} x_{i + j}. \end{align*}
Now, I'd like to write \begin{align*} \lim_{n \to \infty} \sup_{k \geq n, i \in \mathbb{N}} \frac{1}{n} \sum_{j = 0}^{k - 1} x_{i + j} &\leq \lim_{n \to \infty}\sup_{k \geq n, i \in \mathbb{N}} (1 - \frac{r}{n}) a_m (\xi) + \frac{m}{n} \| \xi \|_\infty \\ &= \lim_{n \to \infty} (1 - \frac{r}{n}) a_m (\xi) + \frac{m}{n} \| \xi \|_\infty \\ &= a_m(\xi). \end{align*}
but I just don't see this holding for all $m \in \mathbb{N}$: we have $n = km + r$, so taking the limit $n \to \infty$, shouldn't $m$ increase as well?
Moving on then, I have no idea how to show that if the $\limsup$ exists, the $\lim$ exists as well. Should I calculate $\liminf$ to show that $\limsup = \liminf$?
For the second question, fix $m$. Then you have for every $n$ from the first question $$ a_n(\xi)\leq (1-\frac{1}{n})a_m(\xi) + \frac{m}{n}||\xi||_\infty. $$ From this you conlude that $\limsup_n a_n(\xi)$ exists and it must be less or equal of the $\lim_n$ of the right side of the inequality. This answers the second question.
The third question is answered similarly, you were on the right track with the $\liminf$. Do you get it?