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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 $$1 = \lim_{x \to 0} \lim_{n \to \infty} f_n(x) \neq \lim_{n \to \infty} \lim_{x \to 0} f_n(x) = 0$$

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    $\begingroup$ Let $A = (0,1]$ and $f_n(x) = (-1)^n g_n(x)$ where $g_n$ is similar to the $f_n$ you used for 1). $\endgroup$ Aug 20 '18 at 20:31
  • $\begingroup$ @DanielFischer thanks for your answer! I have troubles though figuring out what the pointwise limit should be... $\endgroup$
    – Student
    Aug 20 '18 at 20:37
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    $\begingroup$ It's always nice if one can take it as simple as possible, say $f(x) = 0$ for all $x \in A$. $\endgroup$ Aug 20 '18 at 20:40
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    $\begingroup$ You make it so that the limit is $0$. Have $f_n(x) = 0$ for $x \geqslant 1/n$, and on $(0,1/n)$, wiggle the tail. $\endgroup$ Aug 20 '18 at 20:51
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    $\begingroup$ That depends on how you wiggle. You can take an arbitrary sequence $(L_n)$, and define $f_n$ on $(0,1/n)$ so that you get a continuous function with $\lim_{x \to 0} f_n(x) = L_n$. Making them piecewise linear as in your example for 1) is the simplest way, but you can do it arbitrarily. $\endgroup$ Aug 20 '18 at 21:01
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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.

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  • $\begingroup$ Thank you for your answer! I can really visualize it, which makes it nice! $\endgroup$
    – Student
    Aug 20 '18 at 21:30
  • $\begingroup$ @Student no problem! Glad I could help :) $\endgroup$
    – qualcuno
    Aug 20 '18 at 21:32
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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.

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  • $\begingroup$ Thank you for your answer, I can really visualize it. My course notes however distinguish between finite and infinite limits and sequences which have no limit. Yours converges to infinity (according to my coursenotes), so I'm not sure if this is what the lecturer intended. I'll have to ask him... Thanks anyway, since I really like this answer! $\endgroup$
    – Student
    Aug 20 '18 at 21:33
  • $\begingroup$ Realized my answer is pretty similar to yours. Keeping it nonetheless. $\endgroup$ Aug 20 '18 at 21:50
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    $\begingroup$ @Student You can modify my answer to obtain $L_n=(-1)^n$. For $n$ odd the tents are "negative" ($f_n\le0$) and for $n$ even they're "positive". $\endgroup$
    – grand_chat
    Aug 21 '18 at 0:45
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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 $2$.

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  • $\begingroup$ Thanks for your answer. Less straightforward than the others (imo), but another exqmple of how creative I should try to be when looking for counterexamples :) $\endgroup$
    – Student
    Aug 20 '18 at 21:58
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    $\begingroup$ @Student The countability of $\mathbb{Q}$ is very instrumental for making counterexamples in these kinds of problems in my experience. Counterexamples against all three statements can be easily brewed using this fact. $\endgroup$ Aug 21 '18 at 4:29

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