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Let $~f_{n}: [0,1] \rightarrow \mathbb{R}$ be a sequence of smooth functions that are uniformly bounded and equicontinuous. By Arzela Ascoli theorem we know that a subsequence $\{ f_{n_k} \} $ converges uniformly.

Is there is any additional condition under which one can say that the sequence $\{ f_{n} \} $ converges uniformly? In my case, I have sequence of functions that are uniformly bounded and the derivatives $f_{n}^{\prime}$ are also uniformly bounded. By fundamnetal theorem of calculus this implies the sequence is equicontinuous. Under what additional hypothesis can one conclude this sequence converges uniformly?

For example is this a sufficient criteria: $$ f_{n+1}(x) \geq f_{n}(x) \qquad \forall ~~x, ~~n $$ ?

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    $\begingroup$ Perhaps you are looking for en.wikipedia.org/wiki/Dini%27s_theorem ? $\endgroup$ – Prahlad Vaidyanathan Sep 4 '13 at 17:29
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    $\begingroup$ If the sequence converges pointwise on some dense subset, it converges uniformly. Under the additional hypothesis that the sequence is monotonic, it converges niformly if and only if it is bounded at one point. $\endgroup$ – Daniel Fischer Sep 4 '13 at 18:36
  • $\begingroup$ @Ritwik : in the first sentence of your question, you don't need the word "smooth" $\endgroup$ – Stefan Smith Nov 1 '13 at 1:07
  • $\begingroup$ Yes you are right, the word smooth is not needed. $\endgroup$ – Ritwik Nov 2 '13 at 4:54
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A condition on the derivatives cannot guarantee the whole series (take $f_n(x):=(-1)^n$).

However, in the case where $(f_n(x),n\geqslant 1)$ is non-increasing for all $x$, and the sequence is uniformly bounded, this is true (and called Dini's theorem).

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    $\begingroup$ Also note that it is enough that $⟨f_n(x)⟩_n$ is monotone for every $x$ but the monotonicity doesn't have to be same for all $x$. $\endgroup$ – user87690 Sep 4 '13 at 17:53
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Here is one additional assumption that allows you to conclude: if there is a dense set $D\subset [0,1]$ such that the sequence $(f_n(z))$ is convergent for each point $z\in D$, then $(f_n)$ is uniformly convergent.

To prove this, first use equicontinuity to show that in fact the sequence $(f_n(x))$ is convergent for every $x\in [0,1]$ to some $f(x)$. This implies that the sequence $(f_n)$ has at most one limit point in $\mathcal C([0,1])$. Then apply Ascoli to conclude that $(f_n)$ converges uniformly to this only possible candidate.

For example, this works if the sequence $(f_n)$ is monotonic on a dense set of points (where the monotonicity is allowed to depend on the point). However, in this case the assumption of equicontinuity is not a priori needed (even though you do have equicontinuity a posteriori): it is not difficult to prove that monotonicity on a dense set of points implies monotonicity at each point, and then one can apply Dini's theorem separately to each of the two compact sets $K_1=\{ x;\; (f_n(x))\; \hbox{is nondecreasing}\}$ and $K_2=\{ x;\; (f_n(x))\; \hbox{is nonincreasing}\}$

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