Show that $\lim_{m \rightarrow \infty} z_k^m = \frac 1 n \sum_{j=1}^n z_j^0$. Let $\{z_k^0\}_{k=1}^n$ be $n$ points in the complex plane and consider the following iteration: $$z_k^{m+1}=\frac 1 2 (z_k^m +z_{k+1}^m)$$ for $k=1,2,3,\ldots,n-1$.
$$z_k^{m+1}=\frac 1 2 (z_n^m+z_1^m)$$ for $k=n$.
Show that $\lim_{m \rightarrow \infty} z_k^m = \frac 1 n \sum_{j=1}^n z_j^0$.
I tried to iterate the iteration but I can't see the magic in there; besides, when I iterated, I got the coefficients like $\frac 1 {2^n}$, rather than $\frac 1 n$.
Thanks.
 A: Sketch:


*

*Consider $z_k^m$ as a sum of $z_i^0$'s.  In particular,
$$
z_k^m=\sum a_i z_i^0.
$$
The $a_i$'s are nonnegative and sum to $1$ since every expansion preserves the sum of the coefficients.

*Consider $z_{k+1}^m$ as a sum of $z_i^0$'s (or $z_1^m$ if $k=n$).  In particular,
$$
z_{k+1}^m=\sum b_i z_i^0.
$$
Then, $a_i=b_{i+1}$ for $i<n$ and $a_n=b_1$.  This can be proved by induction.

*Claim: Let $z_k^{m+1}$ be written as a sum of $z_i^0$'s as
$$
z_k^{m+1}=\sum c_iz_i^0.
$$
If the $a_i$'s are not all zero, then either $\max\{a_i\}>\max\{c_i\}$ or the number of $a_i$'s which attain the maximum is greater than the number of $c_i$'s which attain the maximum.

*I'm going to get the indices wrong, but $a_i$ is something like 
$$
\frac{1}{2^m}\sum_{\substack{j=0\\j\equiv i+k\pmod n}}^m\binom{m}{j}.
$$
You can use this to add coefficients together (along with binomial rules).
A: The denominator in question is indeed $n$. The iteration can be rewritten as $\mathbf z^{m+1}=C\mathbf z^m$, where
$$
C=\pmatrix{\frac12&\frac12\\ &\ddots&\ddots\\ &&\frac12&\frac12\\ \frac12&&&\frac12}.
$$
Since $C$ is primitive and doubly stochastic, $\lim_{m\to\infty}C^m$ exists and the limit is a doubly stochastic matrix with identical rows and identical columns. Hence $\lim_{m\to\infty}C^m=\frac1n\mathbf1\mathbf1^T$ (where $\mathbf1$ is all-one vector) and $\lim_{m\to\infty}\mathbf z^m=\lim_{m\to\infty}C^m\mathbf z^0=\frac1n\mathbf1\mathbf1^T\mathbf z^0=\frac{\sum_k z_k^0}{n}\mathbf 1$.
Alternatively, you may explicitly calculate $C^m$ using the properties of circulant matries.
