IMC 1996 · Day 1 · P5

(i) With a linear change of the variable (i) is equivalent to:

(i$'$) Let $a$, $b$, $A$ be real numbers such that $b \le 0$, $A > 0$ and $1 + ax + bx^2 > 0$ for every $x$ in $[0,A]$. Denote $I_n = n \int\limits_0^A (1 + ax + bx^2)^n dx$. Prove that $\lim\limits_{n \to +\infty} I_n = -\dfrac{1}{a}$ when $a < 0$ and $\lim\limits_{n \to +\infty} I_n = +\infty$ when $a \ge 0$.

Let $a < 0$. Set $f(x) = e^{ax} - (1 + ax + bx^2)$. Using that $f(0) = f'(0) = 0$ and $f''(x) = a^2 e^{ax} - 2b$ we get for $x > 0$ that $$0 < e^{ax} - (1 + ax + bx^2) < cx^2$$ where $c = \dfrac{a^2}{2} - b$. Using the mean value theorem we get $$0 < e^{anx} - (1 + ax + bx^2)^n < cx^2 n e^{a(n-1)x}.$$ Therefore $$0 < n \int_0^A e^{anx} dx - n \int_0^A (1 + ax + bx^2)^n dx < cn^2 \int_0^A x^2 e^{a(n-1)x} dx.$$ Using that $$n \int_0^A e^{anx} dx = \frac{e^{anA} - 1}{a} \xrightarrow[n \to \infty]{} -\frac{1}{a}$$ and $$\int_0^A x^2 e^{a(n-1)x} dx < \frac{1}{|a|^3 (n-1)^3} \int_0^{\infty} t^2 e^{-t} dt$$ we get (i$'$) in the case $a < 0$.

Let $a \ge 0$. Then for $n > \max\{ A^{-2}, -b \} - 1$ we have $$\begin{align*} n \int_0^A (1 + ax + bx^2)^n dx &> n \int_0^{\frac{1}{\sqrt{n+1}}} (1 + bx^2)^n dx \\ &> n \cdot \frac{1}{\sqrt{n+1}} \cdot \left( 1 + \frac{b}{n+1} \right)^n \\ &> \frac{n}{\sqrt{n+1}}\, e^b \xrightarrow[n \to \infty]{} \infty. \end{align*}$$ (i) is proved.

(ii) Denote $I_n = n \int\limits_0^1 (f(x))^n dx$ and $M = \max\limits_{x \in [0,1]} f(x)$.

For $M < 1$ we have $I_n \le n M^n \xrightarrow[n \to \infty]{} 0$, a contradiction.

If $M > 1$ since $f$ is continuous there exists an interval $I \subset [0,1]$ with $|I| > 0$ such that $f(x) > 1$ for every $x \in I$. Then $I_n \ge n |I| \xrightarrow[n \to \infty]{} +\infty$, a contradiction. Hence $M = 1$. Now we prove that $f'$ has a constant sign. Assume the opposite. Then $f'(x_0) = 0$ for some $x_0 \in (0,1)$. Then $f(x_0) = M = 1$ because $f'' \le 0$. For $x_0 + h$ in $[0,1]$, $f(x_0 + h) = 1 + \dfrac{h^2}{2} f''(\xi)$, $\xi \in (x_0, x_0 + h)$. Let $m = \min\limits_{x \in [0,1]} f''(x)$. So, $f(x_0 + h) \ge 1 + \dfrac{h^2}{2} m$.

Let $\delta > 0$ be such that $1 + \dfrac{\delta^2}{2} m > 0$ and $x_0 + \delta < 1$. Then $$I_n \ge n \int_{x_0}^{x_0 + \delta} (f(x))^n dx \ge n \int_0^{\delta} \left( 1 + \frac{m}{2} h^2 \right)^n dh \xrightarrow[n \to \infty]{} \infty$$ in view of (i) – a contradiction. Hence $f'$ is monotone and $M = f(0)$ or $M = f(1)$.

Let $M = f(0) = 1$. For $h$ in $[0,1]$ $$1 + h f'(0) \ge f(h) \ge 1 + h f'(0) + \frac{m}{2} h^2,$$ where $f'(0) \ne 0$, because otherwise we get a contradiction as above. Since $f(0) = M$ the function $f$ is decreasing and hence $f'(0) < 0$. Let $0 < A < 1$ be such that $1 + A f'(0) + \dfrac{m}{2} A^2 > 0$. Then $$n \int_0^A (1 + h f'(0))^n dh \ge n \int_0^A (f(x))^n dx \ge n \int_0^A \left( 1 + h f'(0) + \frac{m}{2} h^2 \right)^n dh.$$ From (i) the first and the third integral tend to $-\dfrac{1}{f'(0)}$ as $n \to \infty$, hence so does the second.

Also $n \int\limits_A^1 (f(x))^n dx \le n (f(A))^n \xrightarrow[n \to \infty]{} 0$ ($f(A) < 1$). We get $L = -\dfrac{1}{f'(0)}$ in this case.

If $M = f(1)$ we get in a similar way $L = \dfrac{1}{f'(1)}$.