IMC 2004 · Day 1 · P3

(a) Equivalently, we consider the set $$Y = \{ y = (y_1, y_2, \dots, y_n) \mid 0 \le y_1, y_2, \dots, y_n \le 1,\ y_1 + y_2 + \dots + y_n = 1 \} \subset \mathbb{R}^n$$ and the image $f(Y)$ of $Y$ under $$f(y) = \arcsin y_1 + \arcsin y_2 + \dots + \arcsin y_n.$$ Note that $f(Y) = S_n$. Since $Y$ is a connected subspace of $\mathbb{R}^n$ and $f$ is a continuous function, the image $f(Y)$ is also connected, and we know that the only connected subspaces of $\mathbb{R}$ are intervals. Thus $S_n$ is an interval.

(b) We prove that $$n \arcsin \frac{1}{n} \le x_1 + x_2 + \dots + x_n \le \frac{\pi}{2}.$$ Since the graph of $\sin x$ is concave down for $x \in \left[ 0, \frac{\pi}{2} \right]$, the chord joining the points $(0, 0)$ and $\left( \frac{\pi}{2}, 1 \right)$ lies below the graph. Hence $$\frac{2x}{\pi} \le \sin x \quad \text{for all } x \in \left[ 0, \frac{\pi}{2} \right]$$ and we can deduce the right-hand side of the claim: $$\frac{2}{\pi} (x_1 + x_2 + \dots + x_n) \le \sin x_1 + \sin x_2 + \dots + \sin x_n = 1.$$ The value 1 can be reached choosing $x_1 = \frac{\pi}{2}$ and $x_2 = \dots = x_n = 0$.

The left-hand side follows immediately from Jensen's inequality, since $\sin x$ is concave down for $x \in \left[ 0, \frac{\pi}{2} \right]$ and $0 \le \frac{x_1 + x_2 + \dots + x_n}{n} < \frac{\pi}{2}$ $$\frac{1}{n} = \frac{\sin x_1 + \sin x_2 + \dots + \sin x_n}{n} \le \sin \frac{x_1 + x_2 + \dots + x_n}{n}.$$ Equality holds if $x_1 = \dots = x_n = \arcsin\frac{1}{n}$.

Now we have computed the minimum and maximum of interval $S_n$; we can conclude that $S_n = \left[ n \arcsin\frac{1}{n}, \frac{\pi}{2} \right]$. Thus $l_n = \frac{\pi}{2} - n \arcsin\frac{1}{n}$ and $$\lim_{n \to \infty} l_n = \frac{\pi}{2} - \lim_{n \to \infty} \frac{\arcsin(1/n)}{1/n} = \frac{\pi}{2} - 1.$$