| |||||||||||||

## IMC2021: Day 1, Problem 4
\(\displaystyle \text{if} \quad |x-y| < \min\big\{g(x),g(y)\big\}, \quad\text{then}\quad \big|f(x) - f(y)\big| < \varepsilon. \) Prove that \(\displaystyle f\) is the pointwise limit of a sequence of continuous \(\displaystyle \RR\to\RR\) functions, i.e., there is a sequence \(\displaystyle h_1,h_2,\ldots\) of continuous \(\displaystyle \RR\to\RR\) functions such that \(\displaystyle \lim\limits_{n\to\infty}h_n(x)=f(x)\) for every \(\displaystyle x\in\RR\). Camille Mau, Nanyang Technological University, Singapore
For any \(\displaystyle x\in\mathbb R\), choose \(\displaystyle \delta_n(x) = \min\{ 1/n, g(x, 1/n)\}.\) Of the \(\displaystyle \delta_n(x)\)-neighborhoods of all \(\displaystyle x\) select (using local compactness of the reals) an inclusion-minimal locally finite covering \(\displaystyle \{U_i\}\). From its inclusion-minimality it follows that we may enumerate \(\displaystyle U_i\) with \(\displaystyle i\in\mathbb Z\) so that \(\displaystyle U_i\cap U_j\neq\emptyset\) only when \(\displaystyle |i-j|\le 1\) and the enumeration goes from left to right on the real line. For an assumed \(\displaystyle n\), let \(\displaystyle x_i\) be the center of \(\displaystyle U_i\) and \(\displaystyle \delta_i = \delta_n(x_i)\), so that \(\displaystyle U_i=(x_i-\delta_i, x_i+\delta_i)\) and \(\displaystyle \delta_i<1/n\) for all \(\displaystyle i\). Now define a continuous \(\displaystyle f_n:\mathbb R\to \mathbb R\) so that it equals \(\displaystyle f(x_i)\) in \(\displaystyle U_i\setminus (U_{i-1}\cup U_{i+1})\), and so that \(\displaystyle f_n\) changes continuously between \(\displaystyle f(x_{i-1})\) and \(\displaystyle f(x_i)\) in the intersection \(\displaystyle U_{i-1}\cap U_i\). Now we show that \(\displaystyle f_n\to f\) pointwise. Fix a point \(\displaystyle x\) and \(\displaystyle \varepsilon = 1/m > 0\), and choose \(\displaystyle n > \max\left\{1/g(x,\varepsilon), m\right\}. \) Examine the construction of \(\displaystyle f_n\) for any such \(\displaystyle n\). Observe that \(\displaystyle g(x,\varepsilon) > 1/n > \delta_i\) and \(\displaystyle 1/n < 1/m\). There are two cases: - \(\displaystyle x\) belongs to the unique \(\displaystyle U_i\). Then using the monotonicity of \(\displaystyle g(x,\varepsilon)\) in \(\displaystyle \varepsilon\) we have
- \(\displaystyle x\) belongs to \(\displaystyle U_{i-1}\cap U_i\). Similar to the previous case,
\(\displaystyle |x_i-x| < \delta_i \le \min\left\{g\left(x_i, \frac1n\right), g\left(x, \varepsilon\right)\right\} \le \min\left\{g(x_i, \varepsilon), g(x, \varepsilon)\right\}. \) Hence \(\displaystyle |f(x) - f_n(x)| = |f(x) - f(x_i)| < \varepsilon. \) \(\displaystyle |f(x) - f(x_{i-1})|, |f(x) - f(x_i)| < \varepsilon. \) Since \(\displaystyle f_n(x)\) is between \(\displaystyle f_n(x_{i-1}) = f(x_{i-1})\) and \(\displaystyle f_n(x_i) = f(x_i)\) by construction, we have \(\displaystyle |f(x) - f_n(x)| < \varepsilon. \) We have that \(\displaystyle |f(x)-f_n(x)| < \varepsilon\) holds for sufficiently large \(\displaystyle n\), which proves the pointwise convergence.
Assume the contrary in view of the above theorem: \(\displaystyle A\subseteq \mathbb R\) is a non-empty closed set and \(\displaystyle f\) has no point of continuity in \(\displaystyle A\). Let's think that \(\displaystyle f\) is defined only on \(\displaystyle A\). Then for all \(\displaystyle x\in A\) there exist rationals \(\displaystyle p<q\) for which \(\displaystyle \limsup_x f> q\), \(\displaystyle \liminf_x f< p\). Apply the Baire category theorem: Choose \(\displaystyle \varepsilon=(q-p)/10\) and find \(\displaystyle k\) for which the set \(\displaystyle S=\{x:g(x)>1/k\}\) is also dense on a certain ball \(\displaystyle A_2\subset A_1\). Partition \(\displaystyle S\) into subsets where \(\displaystyle f(x)> (p+q)/2\) and \(\displaystyle f(x)\leqslant (p+q)/2\), one of them is again dense somewhere in \(\displaystyle A_3\), say the latter. Now take any point \(\displaystyle y\in A_3\cap Q\) and a very close (within distance \(\displaystyle \min(1/k,g(y))\)) to \(\displaystyle y\) point \(\displaystyle x\) with \(\displaystyle g(x)>1/k\) but \(\displaystyle f(x)\leqslant (p+q)/2\). This pair \(\displaystyle x,y\) contradicts the property of \(\displaystyle f\) from the problem statement.
Now the solution follows from the formula with a countable intersection of the unions of intervals:
and the similar formula for \(\displaystyle \{x\colon f(x)\leqslant c\}\). It remains to prove \(\displaystyle (*)\). The left hand side is obviously contained in the right hand side, just put \(\displaystyle y=x\). To prove the opposite inclusion assume the contrary, that \(\displaystyle f(x)<c\), but \(\displaystyle x\) is contained in the right hand side. Choose a positive integer \(\displaystyle n\) such that \(\displaystyle f(x)<c-1/n\) and \(\displaystyle k\) such that \(\displaystyle g(x,1/n)>1/k\). Then, since \(\displaystyle x\) belongs to the right hand side, we see that there exists \(\displaystyle y\) such that \(\displaystyle f(y)\geqslant c\) and \(\displaystyle |x-y|<\min\left\{g\left(y,\frac1n\right),\frac1k\right\}\leqslant \min\left\{g\left(y,\frac1n\right),g\left(x,\frac1n\right)\right\}, \) which yields \(\displaystyle f(x)\geqslant f(y)-1/n\geqslant c-1/n\), a contradiction. | |||||||||||||

© IMC |