# International Mathematics Competition for University Students 2021

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IMC 2021
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## IMC2021: Day 1, Problem 4

Problem 4. Let $\displaystyle f:\RR\to\RR$ be a function. Suppose that for every $\displaystyle \varepsilon > 0$, there exists a function $\displaystyle g:\RR\to(0,\infty)$such that for every pair $\displaystyle (x,y)$ of real numbers,

$\displaystyle \text{if} \quad |x-y| < \min\big\{g(x),g(y)\big\}, \quad\text{then}\quad \big|f(x) - f(y)\big| < \varepsilon.$

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

Solution 1. Since $\displaystyle g$ depends also on $\displaystyle \varepsilon$, let us use the notation $\displaystyle g(x, \varepsilon)$. Considering only $\displaystyle \varepsilon = 1/n$ for positive integer $\displaystyle n$ will suffice to reach our conclusions, hence we may use $\displaystyle \min\{ g(x, 1/m)\ |\ m\le n\}$ in place of $\displaystyle g(x, 1/n)$ and thus assume $\displaystyle g(x,\varepsilon)$ decreasing in $\displaystyle \varepsilon$.

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
• $\displaystyle |x_i-x| < \delta_i \le \min\left\{g\left(x_i, \frac1n\right), g\left(x, \varepsilon\right)\right\} \le \min\left\{g(x_i, \varepsilon), g(x, \varepsilon)\right\}.$

Hence

$\displaystyle |f(x) - f_n(x)| = |f(x) - f(x_i)| < \varepsilon.$

• $\displaystyle x$ belongs to $\displaystyle U_{i-1}\cap U_i$. Similar to the previous case,
• $\displaystyle |f(x) - f(x_{i-1})|, |f(x) - f(x_i)| < \varepsilon.$

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.

Solution 2. This solution uses the Baire characterization theorem: A function $\displaystyle f: \mathbb R\to\mathbb R$ is a pointwise limit of continuous functions if and only if its restriction to every non-empty closed subset of $\displaystyle \mathbb R$ has a point of continuity.

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: If a complete metric space $\displaystyle A$ is a countable union of sets then some of the sets is dense in a positive radius metric ball of $\displaystyle A$. It follows that there exist $\displaystyle p$ and $\displaystyle q$, which serve for a subset $\displaystyle B\subset A$ which is dense on a certain ball (in the induced metric of the real line) $\displaystyle A_1\subset A$. It yields that both sets $\displaystyle Q=f^{-1}(q,\infty)$ and $\displaystyle P=f^{-1}(-\infty,p)$ are dense in $\displaystyle A_1$.

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.

Solution 3. This solution uses the Lebesgue characterization theorem: If $\displaystyle f:\mathbb R\to\mathbb R$ is a function and, for all real $\displaystyle c$, the sublevel and superlevel sets $\displaystyle \{x\ |\ f(x)\geqslant c\}$, $\displaystyle \{x\ |\ f(x)\leqslant c\}$ are countable intersections of open sets then $\displaystyle f$ is a pointwise limit of continuous functions.

Now the solution follows from the formula with a countable intersection of the unions of intervals:

 $\displaystyle \{x\ |\ f(x)\geqslant c\}=\bigcap_{n,k=1}^\infty \bigcup_{\substack{y\in\mathbb R \\ f(y)\geqslant c}} \left( y - \min\left\{\frac1k,g\left(y,\frac1n\right)\right\}, y + \min\left\{\frac 1k,g\left(y,\frac1n\right)\right\}\right)$ $\displaystyle {(*)}$

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.

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