Suppose that $(f^n)$ is a sequence of continuous functions from a compact metric space $A$ into a convex compact subset of the reals $B$. Is there a subsequence, $(f^{n_k})$, such that the sequence
$f^{n_1},(1/2)f^{n_1}+(1/2)f^{n_2},(1/3)f^{n_1}+(1/3)f^{n_2}+(1/3)f^{n_3},…$
converges pointwise to some map $f:A\to B$?
Best Answer
The answer depends on the sequence $(f^n)$.
Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà –Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $f\in C(A)$. Then a sequence $\left(\frac 1k\sum_{i=1}^k f^{n_i}\right)$ uniformly converges to $f$ too.
But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space $\{0,1\}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:A\to [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $D\subset\Bbb N$ with undefined density, that is such that the limit $d(D)=\lim_{k\to\infty} \frac 1k|\{ a\in D: a\le k\}|$ does not exist. Pick any point $x=(x_n)\in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $k\in D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|\{ a\in D: a\le k\}|=\sum_{i=1}^k f^{n_i}(x)$, the limit $\frac 1k\sum_{i=1}^k f^{n_i}(x)$ does not exist.