IMC 2017 · Day 2 · P9

First of all, the sequence $f_n$ is well defined and it holds that $$\tag{2} f_{n+1}(x) = e^{\int_0^x f_n(t)\,dt}.$$ The mapping $\Phi : C([0,1)) \to C([0,1))$ given by $$\Phi(g)(x) = e^{\int_0^x g(t)\,dt}$$ is monotone, i.e. if $f < g$ on $(0,1)$ then $$\Phi(f)(x) = e^{\int_0^x f(t)\,dt} < e^{\int_0^x g(t)\,dt} = \Phi(g)(x)$$ on $(0,1)$. Since $f_2(x) = e^{\int_0^x 1\,\mathrm{d}t} = e^x > 1 = f_1(x)$ on $(0,1)$, we have by induction $f_{n+1}(x) > f_n(x)$ for all $x \in (0,1)$, $n \in \mathbb{N}$. Moreover, function $f(x) = \frac{1}{1-x}$ is the unique solution to $f' = f^2$, $f(0) = 1$, i.e. it is the unique fixed point of $\Phi$ in $\{ \varphi \in C([0,1)) : \varphi(0) = 1 \}$. Since $f_1 < f$ on $(0,1)$, by induction we have $f_{n+1} = \Phi(f_n) < \Phi(f) = f$ for all $n \in \mathbb{N}$. Hence, for every $x \in (0,1)$ the sequence $f_n(x)$ is increasing and bounded, so a finite limit exists.

Let us denote the limit $g(x)$. We show that $g(x) = f(x) = \frac{1}{1-x}$. Obviously, $g(0) = \lim f_n(0) = 1$. By $f_1 \equiv 1$ and (2), we have $f_n > 0$ on $[0,1)$ for each $n \in \mathbb{N}$, and therefore (by (2) again) the function $f_{n+1}$ is increasing. Since $f_n$, $f_{n+1}$ are positive and increasing also $f'_{n+1}$ is increasing (due to $f'_{n+1} = f_n f_{n+1}$), hence $f_{n+1}$ is convex. A pointwise limit of a sequence of convex functions is convex, since we pass to a limit $n \to \infty$ in $$f_n(\lambda x + (1 - \lambda) y) \le \lambda f_n(x) + (1 - \lambda) f_n(y)$$ and obtain $$g(\lambda x + (1 - \lambda) y) \le \lambda g(x) + (1 - \lambda) g(y)$$ for any fixed $x, y \in [0,1)$ and $\lambda \in (0,1)$. Hence, $g$ is convex, and therefore continuous on $(0,1)$. Moreover, $g$ is continuous in 0, since $1 \equiv f_1 \le g \le f$ and $\lim_{x \to 0+} f(x) = 1$. By Dini's Theorem, convergence $f_n \to g$ is uniform on $[0, 1-\varepsilon]$ for each $\varepsilon \in (0,1)$ (a monotone sequence converging to a continuous function on a compact interval). We show that $\Phi$ is continuous and therefore $f_n$ have to converge to a fixed point of $\Phi$.

In fact, let us work on the space $C([0, 1-\varepsilon])$ with any fixed $\varepsilon \in (0,1)$, $\| \cdot \|$ being the supremum norm on $[0, 1-\varepsilon]$. Then for a fixed function $h$ and $\| \varphi - h \| < \delta$ we have $$\sup_{x \in [0, 1-\varepsilon]} |\Phi(h)(x) - \Phi(\varphi)(x)| = \sup_{x \in [0, 1-\varepsilon]} e^{\int_0^x h(t)\,dt} \Bigl| 1 - e^{\int_0^x \varphi(t) - h(t)\,dt} \Bigr| \le C (e^\delta - 1) < 2 C \delta$$ for $\delta > 0$ small enough. Hence, $\Phi$ is continuous on $C([0, 1-\varepsilon])$. Let us assume for contradiction that $\Phi(g) \ne g$. Hence, there exists $\eta > 0$ and $x_0 \in [0, 1-\varepsilon]$ such that $|\Phi(g)(x_0) - g(x_0)| > \eta$. There exists $\delta > 0$ such that $\| \Phi(\varphi) - \Phi(g) \| < \frac13 \eta$ whenever $\| \varphi - g \| < \delta$. Take $n_0$ so large that $\| f_n - g \| < \min\{ \delta, \frac13 \eta \}$ for all $n \ge n_0$. Hence, $\| f_{n+1} - \Phi(g) \| = \| \Phi(f_n) - \Phi(g) \| < \frac13 \eta$. On the other hand, we have $|f_{n+1}(x_0) - \Phi(g)(x_0)| > |\Phi(g)(x_0) - g(x_0)| - |g(x_0) - f_{n+1}(x_0)| > \eta - \frac13 \eta = \frac23 \eta$, contradiction. So, $\Phi(g) = g$.

Since $f$ is the only fixed point of $\Phi$ in $\{ \varphi \in C([0, 1-\varepsilon]) : \varphi(0) = 1 \}$, we have $g = f$ on $[0, 1-\varepsilon]$. Since $\varepsilon \in (0,1)$ was arbitrary, we have $\lim\limits_{n \to \infty} f_n(x) = \frac{1}{1-x}$ for all $x \in [0, 1)$.