convergence-divergencereal-analysis
Define a sequence of functions $f_n: (0,1)\rightarrow\mathbb{R}$ by
$\
f(x) =
\begin{cases}
1/q^n
& \text{if } x =p/q \space(\space\mathrm{nonzero})\\
0 & \text{otherwise}
\end{cases}
$
Find the pointwise limit $f$ of $\{f_n\}$ and show $\{f_n\}$ converges uniformly.
$f$ looks like a modification of Thomae's function to me, but I can't see how a function that converges uniformly can also have a pointwise limit — I thought uniform convergence was a stronger type of convergence?
Your reasoning that your sequence $(f_n)$ converges pointwise to the zero function $f(x)=0$ on $[0,2]$ is correct.
Considering uniform convergence now, let's first recall the definition of uniform convergence of a sequence of functions:
The sequence $(g_n)$ converges uniformly to $g$ on $I$ if for every $\epsilon>0$ there is a positive integer $N$ such that $$\tag{1} |g_n(x)-g(x)|<\epsilon,\quad\text{for all } n\ge N\ \text{and}\ x\in I. $$
Note that $(1)$ must hold for each $n\ge N$ and every $x\in I$. (It does have something to do with the supremum $\sup\limits_{x\in I}|g_n(x)-g(x)|$: this must be be able to be made as small as desired by taking $n$ sufficiently large.)
Back to your sequence. Note that
for any positive integer $n$, we have $$\tag{2}f_n(1/n)=n.$$
Now let $\epsilon=1 $ and let $N$ be any fixed positive integer. By $(2)$, we have $$|f_N(1/N) -f(1/N)|=f_N(1/N)=N\ge {1 }=\epsilon.$$ This shows that for $\epsilon=1$ there is no positive integer $N$ such that $(1)$ holds (with $g_n=f_n$ and $g=f$). Consequently, $(f_n)$ does not converge uniformly on $[0,2]$ (or in fact, on any interval containing $0$).
The graphs of the $f_n$ prove illuminating.
The sequence resembles the "witch hat" sequence: it consists of spikes near $x=0$ whose widths approach zero but whose heights grow large. The graphs of the first few terms of the sequence are shown below:
For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:
$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$
The supmemum occurs at $x$ such that:
$$\frac{df_n(x)}{dx}=0$$
Solving the above for $x$, you get:
$$x=\pm\frac{1}{n}$$
$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:
$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$
And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.
Addendum for comment:
I am adding a graphic, so you can see what's happening as a response to your second question.
This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.
Best Answer
For every irrational $x \in (0,1)$, $$\lim_{n \to \infty} f_n(x) = 0$$ is obvious. For a rational $x = \frac{p}{q}$, its again easy to see that $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{p}{q^n} = 0$$
Thus the pointwise limit is $0$ at all $x \in (0,1)$.
To show uniform convergence, note that $f_1(x) \leq 1$ for all $x$. Also $f_2(x) \leq \frac{1}{2}$ and so on, its easy to see (since for any rational $\frac{p}{q}$, $q \geq 2$) that $$f_n(x) \leq \frac{1}{2^n}$$
Now given $\epsilon > 0$, choose $N$ such that $$\frac{1}{2^N} < \epsilon$$ and you're done.