IMC 2003 · Day 1 · P5

B. We shall prove in two different ways that $\lim_{n \to \infty} f_n(x) = g(0)$ for every $x \in (0,1]$. (The second one is more lengthy but it tells us how to calculate $f_n$ directly from $g$.)

Proof I. First we prove our claim for non-decreasing $g$. In this case, by induction, one can easily see that

1. each $f_n$ is non-decrasing as well, and

2. $g(x) = f_0(x) \ge f_1(x) \ge f_2(x) \ge \dots \ge g(0)$ $(x \in (0,1])$.

Then (2) implies that there exists $$h(x) = \lim_{n \to \infty} f_n(x) \quad (x \in (0,1]).$$ Clearly $h$ is non-decreasing and $g(0) \le h(x) \le f_n(x)$ for any $x \in (0,1]$, $n = 0, 1, 2, \dots$. Therefore to show that $h(x) = g(0)$ for any $x \in (0,1]$, it is enough to prove that $h(1)$ cannot be greater than $g(0)$.

Suppose that $h(1) > g(0)$. Then there exists a $0 < \delta < 1$ such that $h(1) > g(\delta)$. Using the definition, (2) and (1) we get $$f_{n+1}(1) = \int_0^1 f_n(t)\,dt \le \int_0^{\delta} g(t)\,dt + \int_{\delta}^1 f_n(t)\,dt \le \delta g(\delta) + (1 - \delta) f_n(1).$$ Hence $$f_n(1) - f_{n+1}(1) \ge \delta (f_n(1) - g(\delta)) \ge \delta (h(1) - g(\delta)) > 0,$$ so $f_n(1) \to -\infty$, which is a contradiction.

Similarly, we can prove our claim for non-increasing continuous functions as well.

Now suppose that $g$ is an arbitrary continuous function on $[0,1]$. Let $$M(x) = \sup_{t \in [0,x]} g(t), \qquad m(x) = \inf_{t \in [0,x]} g(t) \qquad (x \in [0,1])$$ Then on $[0,1]$ $m$ is non-increasing, $M$ is non-decreasing, both are continuous, $m(x) \le g(x) \le M(x)$ and $M(0) = m(0) = g(0)$. Define the sequences of functions $M_n(x)$ and $m_n(x)$ in the same way as $f_n$ is defined but starting with $M_0 = M$ and $m_0 = m$.

Then one can easily see by induction that $m_n(x) \le f_n(x) \le M_n(x)$. By the first part of the proof, $\lim_n m_n(x) = m(0) = g(0) = M(0) = \lim_n M_n(x)$ for any $x \in (0,1]$. Therefore we must have $\lim_n f_n(x) = g(0)$.

Proof II. To make the notation clearer we shall denote the variable of $f_j$ by $x_j$. By definition (and Fubini theorem) we get that $$\begin{align*} f_{n+1}(x_{n+1}) &= \frac{1}{x_{n+1}} \int_0^{x_{n+1}} \frac{1}{x_n} \int_0^{x_n} \frac{1}{x_{n-1}} \int_0^{x_{n-1}} \dots \int_0^{x_2} \frac{1}{x_1} \int_0^{x_1} g(x_0)\,dx_0\,dx_1 \dots dx_n \\ &= \frac{1}{x_{n+1}} \iint\limits_{0 \le x_0 \le x_1 \le \dots \le x_n \le x_{n+1}} g(x_0)\, \frac{dx_0\,dx_1 \dots dx_n}{x_1 \dots x_n} \\ &= \frac{1}{x_{n+1}} \int_0^{x_{n+1}} g(x_0) \left( \iint\limits_{x_0 \le x_1 \le \dots \le x_n \le x_{n+1}} \frac{dx_1 \dots dx_n}{x_1 \dots x_n} \right) dx_0. \end{align*}$$ Therefore with the notation $$h_n(a, b) = \iint\limits_{a \le x_1 \le \dots \le x_n \le b} \frac{dx_1 \dots dx_n}{x_1 \dots x_n}$$ and $x = x_{n+1}$, $t = x_0$ we have $$f_{n+1}(x) = \frac{1}{x} \int_0^x g(t)\, h_n(t, x)\,dt.$$ Using that $h_n(a,b)$ is the same for any permutation of $x_1, \dots, x_n$ and the fact that the integral is 0 on any hyperplanes ($x_i = x_j$) we get that $$n!\, h_n(a, b) = \iint\limits_{a \le x_1, \dots, x_n \le b} \frac{dx_1 \dots dx_n}{x_1 \dots x_n} = \int_a^b \dots \int_a^b \frac{dx_1 \dots dx_n}{x_1 \dots x_n} = \left( \int_a^b \frac{dx}{x} \right)^n = (\log(b/a))^n.$$ Therefore $$f_{n+1}(x) = \frac{1}{x} \int_0^x g(t)\, \frac{(\log(x/t))^n}{n!}\,dt.$$ Note that if $g$ is constant then the definition gives $f_n = g$. This implies on one hand that we must have $$\frac{1}{x} \int_0^x \frac{(\log(x/t))^n}{n!}\,dt = 1$$ and on the other hand that, by replacing $g$ by $g - g(0)$, we can suppose that $g(0) = 0$.

Let $x \in (0,1]$ and $\varepsilon > 0$ be fixed. By continuity there exists a $0 < \delta < x$ and an $M$ such that $|g(t)| < \varepsilon$ on $[0,\delta]$ and $|g(t)| \le M$ on $[0,1]$. Since $$\lim_{n \to \infty} \frac{(\log(x/\delta))^n}{n!} = 0$$ there exists an $n_0$ sucht that $(\log(x/\delta))^n / n! < \varepsilon$ whenever $n \ge n_0$. Then, for any $n \ge n_0$, we have $$\begin{align*} |f_{n+1}(x)| &\le \frac{1}{x} \int_0^x |g(t)|\, \frac{(\log(x/t))^n}{n!}\,dt \\ &\le \frac{1}{x} \int_0^{\delta} \varepsilon\, \frac{(\log(x/t))^n}{n!}\,dt + \frac{1}{x} \int_{\delta}^x |g(t)|\, \frac{(\log(x/\delta))^n}{n!}\,dt \\ &\le \varepsilon\, \frac{1}{x} \int_0^x \frac{(\log(x/t))^n}{n!}\,dt + \frac{1}{x} \int_{\delta}^x M \varepsilon\,dt \\ &\le \varepsilon + M \varepsilon. \end{align*}$$ Therefore $\lim_n f_n(x) = 0 = g(0)$.