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If $\{f_{n}\}_{n\geq 1}$ is a sequence of continuous functions on $\mathbb R$, with $|f_{n}(x)|\leq 1$ for all $x\in \mathbb{R}$, and all $n\geq 1$. Does there exist a subsequence which converges uniformly (or pointwise) to some continuous function $f$?

As I know the Arzelà–Ascoli theorem works for closed intervals $[a,b]$, I don't know if there is something in case of $\mathbb{R}$?

EDIT: If this assumption help we can consider it: the sequence $\{f'_{n}\}$ is uniformly bounded on $\mathbb R$.

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3 Answers

This fails even on a closed interval. You also need equicontinuity to apply Arzelà–Ascoli theorem. For example, let us consider

$$f_n(x) = \tanh (nx).$$

Each $f_n$ is smooth and $|f_n| \leq 1$, while we have $f_n(x) \to \mathrm{sgn}(x)$ pointwise. This is because this sequence lacks equicontinuity.

Considering the problem of generalizing AA, we have several options.

  1. Assume further that $f_n$ is continuous on $\mathbb{R}^{\ast} = [-\infty, \infty]$, or in other words, both $$f_n(+\infty) := \lim_{x\to\infty} f_n(x) \quad \text{and} \quad f_n(-\infty) := \lim_{x\to-\infty} f_n(x),$$ and also assume that we can find a homeomorphism $\phi : [0, 1] \to \mathbb{R}^{\ast}$ such that $f_n \circ \phi$ is equicontinuous. Then we can apply AA to $(f_n \circ \phi)$ to obtain a uniformly convergent subsequence $(f_{n_k} \circ \phi)$. Now it is clear that $(f_{n_k})$ itself is also uniformly convergent.

  2. We can drop globalness. Let $(f_n)$ be locally uniformly bounded and locally equicontinuous. That is, it is uniformly bounded and equicontinuous on every finite closed intervals. Then by diagonal argument, we can extract a subsequence $(f_{n_k})$ of which converges uniformly on every compact subset of $\mathbb{R}$. In particular, the resulting $(f_{n_k})$ converges pointwise to some continuous function $f$ on $\mathbb{R}$.

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But since my sequence is uniformly bounded then it is equicontinuous! –  ConfusedMath Jun 14 '12 at 4:25
@ConfusedMath, if their Lipschitz constants are uniformly bounded, then it implies equicontinuity. But a mere uniform-boundedness does not imply equicontinuity, as you can confirm at my example. My example is also uniformly bounded! –  sos440 Jun 14 '12 at 4:31
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Take $f_n(x) = x^n 1_{(0,1)}(x)+1_{[1,\infty)}(x)$. Let $x_n = \frac{1}{\sqrt[n]{2}}$, and note that $x_n \to 1$. Also note that $f_n(x_n) = \frac{1}{2}$, $f_n(1) = 1$, and for a fixed $x$, $f_n(x)$ is non-increasing, so for $m\geq n$, $f_m(x_n) \leq \frac{1}{2}$.

Consequently no subsequence can converge to a continuous function. To get a contradiction, suppose $f_n(x) \to f(x)$, where $f$ is continuous. We must have $f(1) = 1$, so choose $\delta>0$ such that if $|x-1|<\delta$, then $|f(x)-1|<\frac{1}{4}$. Now choose $n$ such that $|x_n-1| < \delta$. Then if $m \geq n$ we have $f_m(x_n) \leq \frac{1}{2}$, hence $f(x_n) \leq \frac{1}{2}$, which is a contradiction.

Just noticed that there was an update to the question.

Answer to modified question:

If the derivatives are uniformly bounded, then the functions are uniformly Lipschitz, hence equicontinous. Since the functions themselves are uniformly bounded, we can apply Arzela-Ascoli on nested closed intervals to get the desired result:

Let $I_1 = [-1,1]$, and apply Arzela-Ascoli to get a subsequence (ie, an infinite subset) $N_1 \subset \mathbb{N}$ along which $\{f_n\}_{n \in N_1}$ converges uniformly to $f:I_1 \to \mathbb{R}$. (Since the convergence is uniform, $f$ is continuous.)

Now suppose we have a subsequence $N_k \subset \mathbb{N}$, an interval $I_k=[-k,k]$, and a continuous limit function $f:I_k \to \mathbb{R}$. Consider the functions $\{f_n\}_{n \in N_k}$ on the interval $I_{k+1} = [-(k+1),k+1]$. Again, by Arzela-Ascoli, there is a subsequence $\mathbb{N}_{k+1} \subset \mathbb{N}_k$ along which the functions converge uniformly to a continuous function $\phi: I_{k+1} \to \mathbb{R}$. Since the functions converge to $f$ on the subsequence $\mathbb{N}_k$, it follows that $\phi(x) = f(x)$, $\forall x \in I_k$. Hence we may abuse notation slightly by dropping the symbol $\phi$ and using $f$ to denote the new limit function, ie, $f:I_{k+1} \to \mathbb{R}$.

Continuing this way, we define a continuous function $f:\mathbb{R} \to \mathbb{R}$. If $x \in \mathbb{R}$, then for sufficiently large $k$, we have $x \in I_k$, and $\{f_n(x)\}_{n \in N_k}$ converges to $f(x)$.

The convergence is uniform on any bounded set (since a bounded set will be contained in $I_k$ for some $k$). The convergence need not be uniform on all of $\mathbb{R}$, however. For example, take $f_n(x) = 1_{(-2 n \pi, 2 n \pi)}(x) + \cos(x) 1_{(-\infty,-2 n \pi] \cup [2 n \pi, \infty)}(x)$. Both $f_n$ and $f_n'$ are uniformly bounded. Clearly, $f_n$ converges pointwise to $f(x) = 1$, but $\max_{x \in \mathbb{R}} |f_n(x)-f(x)| = 2$, $\forall n$, and so the convergence in not uniform.

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No, it doesn't exist. Something very bad could happen in the infinity.

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but the functions are bounded on all $x$? –  ConfusedMath Jun 14 '12 at 3:05
This example of yours gives non-continuous functions $\,f_n\,$, @Caozhu... –  DonAntonio Jun 14 '12 at 3:24
Your sequence converges pointwise to the zero function. –  David Mitra Jun 14 '12 at 3:42
So what about uniform convergence on compact subsets of $\mathbb R$? Do we have such result? –  ConfusedMath Jun 14 '12 at 3:44
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