Pointwise convergence of average of continuous functions

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 $$f:A\to B$$?

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.
• Thanks for your answer. I am little confused with the construction. Do you mean a set $D\subseteq\{0,1\}^N$ such that $\lim_{k\to\infty}\frac{1}{k}(d_1+\dots+d_k)$ does not exist? And then pick $x\in A$ such that $x_{n_k}=d_k$ for each $k$? – mo15 Dec 8 '18 at 22:02
• @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected. – Alex Ravsky Dec 8 '18 at 23:03
• But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think. – mo15 Dec 8 '18 at 23:15
• In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $\mathbb R^n$. – mo15 Dec 9 '18 at 0:42
• No, $D$ is a subset of $\Bbb N$. – Alex Ravsky Dec 9 '18 at 3:48