Question:
$E= \mathbb{R}$ or $\mathbb{C}$, and $E$ is a normed vector space$(E; ||.||)$. Let $a$ be a linear continuous mapping and we writte $a^p = a \circ a \circ ... \circ a$ $p$ times (with $"\circ"$ composition law). Moreover, we define $$b_p=\frac{\sum_{k=0}^{p-1}a^k}{p}$$
Let a point $x \in E$ s.t. the sequence $\left \{ a^p(x) \right \} $ is bounded. Show that every point in $E$ that is a possible sub-sequential limit of the sequence $\left \{ b_p(x) \right \}$ is a fixed point of the linear mapping $a$.
My best attempt: see a more details answer in the answer
My best attempt until now it the one below (just take it as a try and any feedback will be great).
First of all note that $a \circ b_p(x)=\frac{\sum_{k=1}^{p}a^k}{p}=b_p \circ a(x)$.
Now let note $b_{p_k}(x)$ a sub sequence with limit $l$, hence we have
(i) $a \circ b_{p_k}(x)\underset{k \to \infty}{\rightarrow} a(l) $
(ii) $b_{p_k} \circ a(x)\underset{k \to \infty}{\rightarrow} l \circ a(x)$
But as (i)=(ii) and because $l \circ a(x) = l$ we have that $a(l)=l$
In all the case I don't fell enough good with this answer (not rigorous enough...)
Moreover, I have tried different ways to solve this question, but I am really stuck any help will be great. Thank you.
As i didn't get yet any help bellow my best answer for the moment.
First of all let remember that in any metric space we have that for any continuous function $f$ if $\lim_{n \to \infty } x_n $ exists then $\lim_{n \to \infty } f(x_n ) = f(\lim_{n \to \infty } x_n)$.
Now let take $l$ a sub sequence limit of $\left \{ b_p(x) \right \}$, hence we have $\lim_{k \to \infty }b_{p_k}(x)=l$ and as obvisouly $b_p(x)$ is a continuous fuction (as a finit sum of linear continuous mapping) we get that: $\lim_{k \to \infty }a [b_{p_k}(x)]=a[\lim_{k \to \infty }b_{p_k}(x)]=a(l)$.
On a other side, we can writte $ b_{pk} \circ a = \frac{\sum_{i=1}^{p_k}a^i}{p_k}=\frac{p_{k+1}}{p_k}b_{pk+1}-\frac{1}{p_k}Id $. Now if $p_k \to \infty$ by limit arithmetic and by the unicity of the limit of the sub sequence we get that: $\lim_{k \to \infty}{b_{pk} \circ a} = l$.
As we know that $a \circ b_{p} = b_{p} \circ a$ (easy to prove it as it is only continuous linear mapping), hence by applying the limit on the considering sub sequence with limit $l$ we get: $a(l)=l$.
I will be happy to know if it is correct and to read your advices.
EDIT: I get a feed back and it is correct