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


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.

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  • $\begingroup$ 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 $k$? $\endgroup$ – mo15 Dec 8 '18 at 22:02
  • $\begingroup$ @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected. $\endgroup$ – Alex Ravsky Dec 8 '18 at 23:03
  • $\begingroup$ But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think. $\endgroup$ – mo15 Dec 8 '18 at 23:15
  • $\begingroup$ In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $\mathbb R^n$. $\endgroup$ – mo15 Dec 9 '18 at 0:42
  • $\begingroup$ No, $D$ is a subset of $\Bbb N$. $\endgroup$ – Alex Ravsky Dec 9 '18 at 3:48

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