Help regarding an iteration formula Consider the mapping $$ (a,b,c,d) \mapsto  \left(\frac{a+b}{2}, \frac{b+c}{2},\frac{c+d}{2},\frac{d+a}{2}\right );a,b,c,d \in \mathcal{R^+}$$
Intuitively it is obvious that the if the above mapping is applied  iteratively, some that    is halving the sum of two neighbouring    quantities while keeping the overall  total equal which is $a+b+c+d.$  So , it is expected that the iteration process should  in the limit in the result in the tuple  $(av,av,av,av)$ where $$ av=\frac{\sum a}{4}.$$ However, unfortunately ,I am not able to put this into a formal argument.Would somebody kindly help?ANy help will be greatly appreciated.
PS:-There is nothing peculiar with 4.Any other number of quntities can be taken in the tuple
 A: Let $(a_n, b_n, c_n, d_n)$ be the quadruple after the $n$-th iteration, where $(a_0, b_0, c_0, d_0)$ are the initial values and can be arbitrary real numbers, the restriction to positive numbers is not necessary.
Without loss of generality we can assume that $a_0+b_0+c_0+d_0 = 0$, and consequently $a_n+b_n+c_n+d_n = 0$ for all $n$.
Then
$$
\begin{align}
 a_{n+1}^2+b_{n+1}^2+c_{n+1}^2+d_{n+1}^2
 &= \frac 12 \left( a_n^2+b_n^2+c_n^2+d_n^2 + a_nb_n+b_nc_n+c_nd_n + d_n a_n\right) \\
&= \frac 12 \left( a_n^2+b_n^2+c_n^2+d_n^2 + (a_n+c_n)(b_n+d_n)\right) \\
 &= \frac 12 \left( a_n^2+b_n^2+c_n^2+d_n^2 - (a_n+c_n)^2\right) \\
 &\le \frac 12 \left( a_n^2+b_n^2+c_n^2+d_n^2  \right) \, .
\end{align}
$$
It follows that
$$
 \lim_{n\to \infty} a_n^2+b_n^2+c_n^2+d_n^2 = 0
$$
and that implies that the four sequences $(a_n)$, $(b_n)$, $(c_n)$, $(d_n)$ converge to zero.

Generalization for iterations of tuples with arbitrary length:
For fixed $m \ge 2$, let $(x_{n,1}, x_{n,2}, \ldots, x_{n,m})$ be the tuple after the $n$-th iteration of the initial tuple $(x_{0,1}, x_{0,2}, \ldots, x_{0,m})$. Again we assume that $x_{0,1} + x_{0,2} +\cdots + x_{0,m} = 0$.
In order to prove that
$$
 \lim_{n \to \infty} x_{n,1}^2 + x_{n,2}^2 +\cdots + x_{n,m}^2 = 0
$$
and consequently $\lim_{n \to \infty} x_{n, j} = 0$ for all $1 \le j \le m$, it suffices to prove the following

Lemma: Let $m \ge 2$ and $x_1, x_2, \ldots, x_m \in \Bbb R$  with $x_1 + x_2 + \ldots + x_m = 0$. Set $x_{m+1} = x_1$. Then
$$
 \sum_{j=1}^m \left( \frac{x_j + x_{j+1}}{2}\right)^2 \le C_m \sum_{j=1}^m x_j^2
$$
with a constant $0 < C_m < 1$ which depends only on $m$.

Proof: Set $S=\sum_{j=1}^m x_j^2$ and $S' = \sum_{j=1}^m \left( \frac{x_j + x_{j+1}}{2}\right)^2$. If $S=0$ then $S' = 0$, so from now on we assume that $S > 0$.
$$ \tag{*}
 S' - S = \sum_{j=1}^m \left( \frac{x_j + x_{j+1}}{2}\right)^2
 - \sum_{j=1}^m\frac{x_j^2+x_{j+1}^2}{2}
 = -\frac 14 \sum_{j=1}^m (x_j - x_{j+1})^2
$$
which already shows that $S' \le S$. But we need a slightly better estimate.
There must be (at least) one index $j$ with $|x_k| \ge \sqrt{\frac Sm}$, and since the sum of all $x_j$ is zero, there must be another index $l$ such that $x_k$ and $x_l$ have opposite sign. It follows that there is an index $j$ with $|x_j - x_{j+1}| \ge \frac 1m \sqrt{\frac Sm}$. This gives a negative upper bound for the right-hand side in $(*)$, and we get
$$
 S \le S - \frac 14 \frac{S}{m^3} = \left( 1 - \frac{1}{4m^3}\right) S \, .
$$
This proves the lemma.
A: Proof
Beginning by defining the iterating sequences
\begin{split}
X_n& \triangleq(w_n,x_n,y_n,z_n) \to \left(\frac{w_n+x_n}{2},\frac{x_n+y_n}{2},\frac{y_n+z_n}{2},\frac{z_n+w_n}{2}\right) \\
X_0 & = (a,b,c,d)
\end{split}
So considering :
$$ f (w,x,y,z) \to \left(\frac{w+x}{2},\frac{x+y}{2},\frac{y+z}{2},\frac{z+w}{2}\right) $$
$$ X_{n+1} =f(X_n) $$
Yet, for all $(w,x,y,z)$
$$ Df(w,x,y,z) : H=(h_1,h_2,h_3,h_4) \to \begin{pmatrix} \dfrac{1}{2} & 0  &0 & \dfrac{1}{2}\\ \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 \\ 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 \\ 0 & 0 & \dfrac{1}{2} & \dfrac{1}{2}\end{pmatrix}H $$
It follows that for all $(w,x,y,z)$
$$ ||| Df(w,x,y,z) |||_{\infty}\triangleq\sup_{||H||\neq 0} \dfrac{||Df(w,x,y,z)(H)||}{||H||}=1/2<1 $$
So by Banach-Picard theorem, it exists one and only one fixed point for $X_n$ sequence whatever $X_0$ (finite) is. Then by component combination every $w_n,x_n,y_n,z_n$ sequences converge.
I let you end the proof to show the unique limit for every of them.
