Pointwise convergence of sequence $(f_n)_n$ of functions to $f$ and changing limits

Question

My analysis notes contains the following question: if $(f_n)_n$ is a sequence of functions of $A \subset \mathbb{R} \to \mathbb{R}$ and $a \in \mathbb{R} \cup \{-\infty, +\infty\}$ an accumulation point in $A$. Assume that for all $n$, $\lim_{ x \to a}f_n(x) = L_n$ exists and is finite. Suppose $(f_n)_n$ converges pointwise to $f:A \to \mathbb{R}$.

1) Does $\lim_{x \to a} f(x)$ exists?

2) Does $\lim_{n \to \infty} L_n$ exists?

3) Can we change limits (in case both limits exist)?

---

If know all questions should be no. I have found examples for 1) and 3). However, I can not find an example for 2). I know I should look for a sequence of functions which does not converge uniformly. Any hints would be appreciated.

---

My solutions to 1) Define $$f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases} -1 &\text{ if } x \leq -1/n\\ nx & \text{ if } -1/n < x < 1/n\\ 1 &\text{ if } x \geq 1/n\end{cases}.$$ This functions converges to the function $f$ which equals to -1 for $x > 0$, 0 for $x = 0$ and $1$ for $x > 0$. The limit in zero does not exist.

and 3): define $$f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(nx)^2}{1 + (nx)^2}.$$ This sequence converges to $f$ which equals 1 everywhere, except for $x = 0$, where it equals to $0$. We have that $$1 = \lim_{x \to 0} \lim_{n \to \infty} f_n(x) \neq \lim_{n \to \infty} \lim_{x \to 0} f_n(x) = 0$$

Answer 1

Let $(f_n)_{n \geq 1} : \mathbb{R} \to \mathbb{R}$ be defined as $f_{2n} = \chi_{[2n,+\infty)}$ and $f_{2n-1} = \chi_{(-\infty, 2n-1]}$. Now, if $x \in \mathbb{R}$, there exists $k \in \mathbb{N}$ such that $x \in [-k,k]$ and so if $n \geq k$, we have that $f_n(x) = 0$. Therefore we have pointwise convergence.

Moreover, if $a = \infty$ and $n \in \mathbb{N}$, then $\lim_{x \to \infty}f_n(x)$ always exists,

$$ \lim_{x \to \infty}f_n(x) = \cases{0 \quad n \text{ is odd} \\ 1 \quad n \text{ is even}} $$

However, it is clear from here that $(L_n)_{n\geq 1}$ does not converge, singe taking odd and even terms we have two subsequences converging to different values.

As a side note, a sufficient condition for $(2)$ to hold is that $a \in \mathbb{R}$ and $f_n$ continuous at $a$ for all $n$, in which case $L_n \to f(a)$. If so, we would have that

$$ |f(a) - L_n| \leq |f(a) - f_n(a)| + |f_n(a)-f_n(x)| + |f_n(x) - L_n| $$

Taking limit of $x \to a$, we have that

$$ |f(a) - L_n| \leq |f(a) - f_n(a)| \xrightarrow{n \to \infty} 0 $$

as claimed.

Answer 2

Hint for 2: consider $f_n$ to be tent-shaped around $a=0$, where the slope of the tent gets higher with $n$. Spoiler below:

Explicitly, take $f_n$ to be zero except for $x\in (0,\frac1n)$, where $f_n$ has slope $-n^2$, and $x\in(-\frac1n,0)$, where $f_n$ has slope $n^2$. Let $f_n(0)=0$ for every $n$, so $f_n$ has a removable discontinuity at $x=0$. Argue that $f_n(x)\to0$ for every $x$. But for each $n$ we have $\lim_{x\to0}f_n(x)=n=:L_n$, and $L_n$ doesn't converge.

Answer 3

Let $\{q_1,q_2,\dots\}$ be an ordering of $\mathbb{Q}$. Define $f : \mathbb{Q} \to \mathbb{R}$ as follows: $$f_n(x)=\begin{cases} 0 &: x\in \{q_i : 1 \leq i \leq n\}\\ n &: \text{otherwise}\end{cases}$$ Then, $\{f_n\}$ converges to $0$ pointwise and for $a=0$, $L_n=n \, \forall n \in \mathbb{N}$. Clearly, it contradicts $2$.