# Subsequence Convergence

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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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.

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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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! –  Sangchul Lee Jun 14 '12 at 4:31

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