I have seen in a paper to claim the fact that due to Arzela-Ascoli theorem pointwise convergence with equicontinuity and uniform boundedness of a sequence of functions implies uniform convergence on compacts.

This is not correct. The statement is true, but it is not due to arzela-ascoli as this theorem talks about existence of some subsequence.

Am I correct ?


It is due to Arzela Ascoli (in a sense).

We assume that the claim is false. Then there is a compact subset $K$ and (check this!) Some $\epsilon>0$ and some sub sequence $(f_{n_k})_k$ so that for each $k$, there is some $x_k \in K$ with $$|f_{n_k}(x_k)-f(x_k)| >\epsilon \qquad (\dagger).$$

But Arzela Ascoli yields a further sub sequence $(f_{n_{k_l}})_l$ which converges uniformly in $K$ to some function $g$. But we assume that the sequence $(f_n)_n$ converges pointwise to $f$, hence $f=g$.

One can now easily derive a contradiction to $(\dagger)$.

The principle used here is sometimes called the sub sequence principle, which states that $x_n \to x$ is equivalent to the fact that every sub sequence $(x_{n_k})_k$ has a further sub sequence which converges to $x$, as long as the notion of convergence is given by a topology.

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  • $\begingroup$ It looks like to me you are claiming that pointwise convergence and uniform convergence along a sub-sequence implies uniform convergence for the whole sequence which is not true. Am I correct ? $\endgroup$ – Anonymous Jan 2 '15 at 12:19
  • $\begingroup$ No, I am claiming that pointwise convergence together with uniform convergence of some (sub)subsequence of every subsequence implies uniform convergence of the whole sequence. $\endgroup$ – PhoemueX Jan 2 '15 at 12:45
  • $\begingroup$ Sorry. Got it now. $\endgroup$ – Anonymous Jan 2 '15 at 12:49

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