IMC 1997 · Day 1 · P5

a) For $x = 0$ the statement is trivial. Let $x \ne 0$. Then $\max\limits_i x_i > 0$ and $\min\limits_i x_i < 0$. Hence $\|x\|_\infty < 1$. From the hypothesis on $x$ it follows that:

i) If $x_j \le 0$ then $\max\limits_i x_i \le x_j + 1$.

ii) If $x_j \ge 0$ then $\min\limits_i x_i \ge x_j - 1$.

Consider $y \in \mathbb{Z}^n_0$, $y \ne 0$. We split the indices $\{1, 2, \dots, n\}$ into five sets: $$I(0) = \{ i : y_i = 0 \},$$ $$I(+,+) = \{ i : y_i > 0, x_i \ge 0 \}, \quad I(+,-) = \{ i : y_i > 0, x_i < 0 \},$$ $$I(-,+) = \{ i : y_i < 0, x_i > 0 \}, \quad I(-,-) = \{ i : y_i < 0, x_i \le 0 \}.$$ As least one of the last four index sets is not empty. If $I(+,+) \ne \emptyset$ or $I(-,-) \ne \emptyset$ then $\|x + y\|_\infty \ge 1 > \|x\|_\infty$. If $I(+,+) = I(-,-) = \emptyset$ then $\sum y_i = 0$ implies $I(+,-) \ne \emptyset$ and $I(-,+) \ne \emptyset$. Therefore i) and ii) give $\|x + y\|_\infty \ge \|x\|_\infty$ which completes the case $p = \infty$.

Now let $1 \le p < \infty$. Then using i) for every $j \in I(+,-)$ we get $|x_j + y_j| = y_j - 1 + x_j + 1 \ge |y_j| - 1 + \max\limits_i x_i$. Hence $$|x_j + y_j|^p \ge |y_j| - 1 + |x_k|^p \quad \text{for every } k \in I(-,+) \text{ and } j \in I(+,-).$$ Similarly $$|x_j + y_j|^p \ge |y_j| - 1 + |x_k|^p \quad \text{for every } k \in I(+,-) \text{ and } j \in I(-,+);$$ $$|x_j + y_j|^p \ge |y_j| + |x_j|^p \quad \text{for every } j \in I(+,+) \cup I(-,-).$$ Assume that $\sum\limits_{j \in I(+,-)} 1 \ge \sum\limits_{j \in I(-,+)} 1$. Then $$\begin{align*} \|x + y\|_p^p - \|x\|_p^p &= \sum_{j \in I(+,+) \cup I(-,-)} (|x_j + y_j|^p - |x_j|^p) + \left( \sum_{j \in I(+,-)} |x_j + y_j|^p - \sum_{k \in I(-,+)} |x_k|^p \right) \\ &\quad + \left( \sum_{j \in I(-,+)} |x_j + y_j|^p - \sum_{k \in I(+,-)} |x_k|^p \right) \\ &\ge \sum_{j \in I(+,+) \cup I(-,-)} |y_j| + \sum_{j \in I(+,-)} (|y_j| - 1) + \sum_{j \in I(-,+)} (|y_j| - 1) - \sum_{j \in I(+,-)} 1 + \sum_{j \in I(-,+)} 1 \\ &= \sum_{i=1}^{n} |y_i| - 2 \sum_{j \in I(+,-)} 1 = 2 \sum_{j \in I(+,-)} (y_j - 1) + 2 \sum_{j \in I(+,+)} y_j \ge 0. \end{align*}$$ The case $\sum\limits_{j \in I(+,-)} 1 \le \sum\limits_{j \in I(-,+)} 1$ is similar. This proves the statement.

b) Fix $p \in (0,1)$ and a rational $t \in (\frac{1}{2}, 1)$. Choose a pair of positive integers $m$ and $l$ such that $mt = l(1-t)$ and set $n = m + l$. Let $$x_i = t, \quad i = 1, 2, \dots, m; \qquad x_i = t - 1, \quad i = m+1, m+2, \dots, n;$$ $$y_i = -1, \quad i = 1, 2, \dots, m; \qquad y_{m+1} = m; \qquad y_i = 0, \quad i = m+2, \dots, n.$$ Then $x \in \mathbb{R}^n_0$, $\max\limits_i x_i - \min\limits_i x_i = 1$, $y \in \mathbb{Z}^n_0$ and $$\|x\|_p^p - \|x + y\|_p^p = m (t^p - (1-t)^p) + (1-t)^p - (m - 1 + t)^p,$$ which is possitive for $m$ big enough.