Uniform Convergence to find out limit of sequence of real numbers.

Question

Assume that $f_{n} \to f$ uniformly on $\Bbb R$ for a sequence $\{f_{n}\}$ of functions from $\Bbb R$ to $\Bbb R$. Suppose that, for some $x_{0}\in \Bbb R$ and each $n \in \Bbb N$, there exists a limit $a_{n} := \lim_{x\to x_{0}}f_{n}(x)$. Prove that the limit $\lim _{n\to \infty} a_{n}$ exists.

My solution:

Since $f_{n}$ converges uniformly to $f$, then it also converges pointwise to $f$. Now for any given $x_{0}$, $f_{n}(x_{0})$ is a real number for any $n \in \Bbb N$. Given that $a_{n} := \lim_{x\to x_{0}}f_{n}(x)$, $a_{n}$ is a sequence of real numbers. For $n$ large enough, $a_{n}$ becomes a stationary sequence by the pointwise convergence of $f_{n}(x)$ and $\lim _{n\to \infty} a_{n}=f(x_{0})$ So, it converges. So, $\lim _{n\to +\infty} a_{n}$ exists.

Please confirm if it is ok. I am skeptic about missing something.

Answer 1

"For $n$ large enough, $a_{n}$ becomes a stationary sequence by the pointwise convergence of $f_{n}(x)$ and $lim _{n→∞} a_{n}=f(x_{0})$" is false. Why would $a_n$ become stationary for $n$ large enough ?...

A proof goes as follows if $f$ has a limit $l$ when $x \to x_0$ : by uniform convergence you can find an $N\in\mathbf{N}$ such that for all $n\geq N$ and for all $x\in\mathbf{R}$ we have $|f_n(x)-f(x)|\leq \varepsilon$. As $f$ has a limit $l$ when $x \to x_0$ as $x\to x_0$, passing to the limit when $x\to x_0$ in what preceeds ensures that for each $n\in\mathbf{N}$ such that $n\geq N$, we have $|a_n - l|\leq\varepsilon$. We have therefore shown that for every $\varepsilon\in\mathbf{R}_{+}^{\times}$ there is an $N\in\mathbf{N}$ such that for all $n\geq N$ we have $|a_n - l|\leq\varepsilon$. This is the exact quantification of $\lim_{n\to+\infty} a_n = l$, so we are done.

Remark. We have show that we can permute the passage to the limit as $n\to+\infty$ and the passage to the limit as $x\to x_0$.

Answer 2

Your proof used only the fact that $f_n$ converges to $f$ pointwisely, thus is incorrect. Indeed, if we define $f_n : \mathbb R \to \mathbb R$ so that $f_n(x)=0$ whenever $n$ is even and for $n$ odd:

• $f_n$ is symmetric along $x=1$,

• $f_n(y) = 0$ for all $n$ and $y\in (-\infty, 0) \cup\{1\}$, and

• when $x\in [0, 1)$, $f_n(x) =x^n$.

Then $f_n$ converges pointwisely but not uniformly to $f(x) = 0$ on $[0,1]$. We also have $a_n = \lim_{x\to 1}f_n(x)$ exists for all $n$ and

$$a_n =\begin{cases} 0\ \ \ \text{if n is even} \\ 1 \ \ \ \text{if n is odd.}\end{cases}$$

Thus $\{a_n\}$ does not converges.

In order to show your assertion, it suffices to show (argue as in Robert Green's answer) $\lim_{x\to x_0} f(x) $ exists.

To show that, let $\epsilon >0$. Then there is $n \in \mathbb N$ so that $|f(x) - f_n(x)| < \epsilon/3$ for all $x\in \mathbb R$. Now let $n$ be fixed. As $\lim_{x\to x_0} f_n(x)$ exists, there is an $\delta$ such that $|f_n(x) - f_n(y)|<\epsilon/3$ for all $x, y \in (x_0 - \delta, x_0 + \epsilon)\setminus \{x_0\}$. Thus

$$|f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y| + + |f_n(y) - f(y)|< \epsilon$$

whenever $x, y \in (x_0 - \delta, x_0 + \epsilon)\setminus \{x_0\}$. Thus $\lim_{x\to x_0} f(x)$ exists.