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Recall that Thomae's function (also called Popcorn function) $f\colon\mathbb{R}\to\mathbb{R}$ is defined as $$ f(x) = \begin{cases} \frac{1}{q} & \text{ if } x=\frac{p}{q} \neq 0 \text{ is rational, } \gcd(p,q)=1 \text{ and } q> 0\\ 0 &\text{ otherwise.} \end{cases} $$ In particular, it is a standard exercise to show that $f$ is continuous at every irrational, and discontinuous at every rational.

Find a sequence of continuous functions $(f_n)_{n\in\mathbb{N}}$ that converges pointwise to $f$ on $[0,1]$.

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  • $\begingroup$ I had this exercise buried in my folders, which my solution (posted answer below). Figured I may as well post it somewhere — other approaches more than welcome! $\endgroup$ – Clement C. Mar 13 '16 at 16:25
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Detailed sketch of the proof:

Since $\mathbb{Q}\cap[0,1]$ is countable, we can write it $\mathbb{Q}\cap[0,1]=(x_n)_{n\in\mathbb{N}}$, where each $x_n$ has a canonical form $x_n=\frac{p_n}{q_n}$, $p_n\wedge q_n=1$. We then define $f_n$ by considering only the $n+1$ first terms of the sequence $(x_k)$, and putting the "right" value on them (i.e, $f_n(x_k)=\frac{1}{q_n}$, $0\leq k\leq n$).

To make $f_n$ continuous, we then make it "affine by parts": each $x_k$ is surrounded by two elements of $\{0,x_1,\dots,x_n,1\}$ (finite number of elements), we can pick the max $\underline{x}_k$ of those strictly less, the min $\bar{x}_k$ of those strictly greater, and then go from 0 to $\frac{1}{q_n}$ on the interval $\left[x_k-\frac{x_k-\underline{x}_k}{2^n}, x_k\right]$, then from $\frac{1}{q_n}$ to 0 on the interval $\left[x_k, x_k+\frac{\bar{x}_k-x_k}{2^n}\right]$ (cf. the very bad picture below).

Illustration

Since each rational $q$ is one of the $x_n$ and thus will, for $k\geq n$, be be "considered" by $f_k$ and its image put to the desired value, $f_k(q)\to f(q)$ for all rationals in $[0,1]$. The problem is for the irrationals. Pick $r\in\mathbb{R}\setminus\mathbb{Q}$ arbitrary; in order to prove that $f_k(r)\to0$, we can, by contradiction, suppose it doesn't.\ It means that there is an $\varepsilon>0$ such that, for an infinity of $k$, $f_k(r)>\varepsilon$. Pick the subsequence of such $k$: it defines a sequence $(f_{\varphi(k)}(r))_k$ of values always strictly greater that $\varepsilon$.

But that means that there is a corresponding sequence of rationals always very close to $r$ (one for each $f_{\varphi(k)}$, otherwise $f_{\varphi(k)}(r)=0$). By very close, it actually means (closer than $\frac{\bar{q}-q}{2^n}\leq 2^{-\varphi(k)}$ for some rational $x$. In other terms, we have a sequence of rationals converging (since $2^{-\varphi(k)}\to0$, because $\varphi(k)\to\infty$ (it is a subsequence, so $\varphi(k)\geq k$ for every $k$).

Then, we can use the fact that if a sequence of rationals $\frac{p_n}{q_n}\to r$ irrational, necessarily $p_n\to \infty$, $q_n\to \infty$ (it is rather simple to prove*). And thus in particular $\frac{1}{q_n}\to 0$. Since $f_{\varphi(k)}(r)$ is always less than the "peak", which is the $\frac{1}{q}$ of the "close rational associated with $f_{\varphi(k)}$" mentioned earlier, $f_{\varphi(k)}(r)\to0$ when $k\to\infty$. Which is a contradiction, because it is supposed not to go under $\varepsilon>0$.

Hence, $f_k(r)\to 0$, and $f_n$ converges pointwise to $f$.


${}$* for example, by contradiction, supposing some subsequence of $p_n$ (or $q_n$) is bounded, applying Bolzano—Weierstrass' theorem to it to have a converging subsequence, and then considering the corresponding subsequence of $q_n$ which must be itself, then, bounded because of the convergence to $r$, and again apply Bolzano—Weiestrass, and find finally a sub-sub-sequence of $p_n$ and $q_n$ such that both converge to some integers $p$ and $q$, which in turn implies that $r=\frac{p}{q}\in\mathbb{Q}$, which can be seen as a problem for an irrational.

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