Note: A corollary of Arzela-Ascoli Theorem asks a similar question, but this question assumes pointwise convergence. Here we do not assume this.

(A version of) the Arzela-Ascoli theorem reads

If $\{f_n\}_n$ is a sequence of functions on a metric space $C$ which is uniformly bounded and uniformly equicontinuous, then there exists a uniformly convergent subsequence $\{f_{n_k}\}_k$.

Does this not mean that the entire sequence of functions $\{f_n\}_n$ in the Arzela-Ascoli theorem, not just a subsequence? Here is why I think this might be plausible:

Let $\{f_n\}$ satisfy the hypotheses of Arzela-Ascoli. Uniform convergence of $\{f_n\}_n$ is equivalent to the convergence \begin{equation*} g_n := \sup_{x \in C} |f_n(x) - f(x)| \to 0 \qquad (n \to \infty) \, . \end{equation*}

Let $\{g_{n_k}\}_k$ be a subsequence achieving $\limsup g_n$, that is \begin{equation*} g_{n_k} \to \limsup g_n =: \ell \qquad (k \to \infty) \, . \end{equation*}

The corresponding subsequence $\{f_{n_k}\}_k$ remains uniformly bounded and uniformly equicontinuous, so Arzela-Ascoli provides a further subsequence, denoted $\{f_{n_{k'}}\}_{k'}$, which converges uniformly, i.e., \begin{equation*} g_{n_{k'}} \to 0 \, . \end{equation*}

Since $\{g_{n_{k'}}\}_{k'}$ is a subsequence of the convergent sequence $\{g_{n_k}\}_k$ (with limit $\ell$), \begin{equation*} \ell = \lim_{k' \to \infty} g_{n_{k'}} = 0 \, . \end{equation*}

Now $\{g_n\}_n$ is nonnegative and its $\limsup$ is zero, so $g_n \to 0$, which means the entire sequence $\{f_n\}$ uniformly converges to $f$.

I feel like I'm missing something since I've never seen Arzela-Ascoli stated as ``$\{f_n\}$ converges uniformly,'' merely that a subsequence converges. If that's the case, what am I missing? Do we need some sort of pointwise convergence of $\{f_n\}_n$ as well, as mentioned in the note?

  • 2
    $\begingroup$ This is about (sequential) compactness or pre-compactness in a certain topology. Then the issue is easily illustrated in familiar (locally compact) metrics spaces such as closed balls in finite-dimensional Euclidean spaces. $\endgroup$ – paul garrett Feb 27 at 16:50
  • $\begingroup$ I see. Do you mean to say that the issue is that in writing the limit $f$ above, that $f$ is merely a subsequential limit? Thus, $f_{n_{k'}}$ does not necessarily converge to $f$. It seems also that if we have some sort of convergence on $\{f_n\}$ (not just pointwise), such as weak, $L^p$, or in measure, that is implied by uniform convergence, then one can replace the pointwise-convergence assumption with that. If so, then I'm happy with that; I had originally asked this question to understand Lemma 1 of jstor.org/stable/2241171?seq=1#metadata_info_tab_contents. Thank you! $\endgroup$ – grogTheFrog Feb 27 at 18:52

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