IMC 1995 · Day 1 · P6

Let $0 < \delta < 1$. First we show that there exists $K_{p,\delta} > 0$ such that $$f(x,y) = \frac{(x-y)^2}{4 - (x+y)^2} \le K_{p,\delta}$$ for every $(x,y) \in D_\delta = \{ (x,y) : |x - y| \ge \delta,\ |x|^p + |y|^p = 2 \}$.

Since $D_\delta$ is compact it is enough to show that $f$ is continuous on $D_\delta$. For this we show that the denominator of $f$ is different from zero. Assume the contrary. Then $|x + y| = 2$, and $\left| \frac{x+y}{2} \right|^p = 1$. Since $p > 1$, the function $g(t) = |t|^p$ is strictly convex, in other words $\left| \frac{x+y}{2} \right|^p < \frac{|x|^p + |y|^p}{2}$ whenever $x \ne y$. So for some $(x,y) \in D_\delta$ we have $\left| \frac{x+y}{2} \right|^p < \frac{|x|^p + |y|^p}{2} = 1 = \left| \frac{x+y}{2} \right|^p$. We get a contradiction.

If $x$ and $y$ have different signs then $(x,y) \in D_\delta$ for all $0 < \delta < 1$ because then $|x - y| \ge \max\{|x|, |y|\} \ge 1 > \delta$. So we may further assume without loss of generality that $x > 0$, $y > 0$ and $x^p + y^p = 2$. Set $x = 1 + t$. Then $$\begin{align*} y &= (2 - x^p)^{1/p} = (2 - (1+t)^p)^{1/p} = \left( 2 - \left( 1 + pt + \frac{p(p-1)}{2} t^2 + o(t^2) \right) \right)^{1/p} \\ &= \left( 1 - pt - \frac{p(p-1)}{2} t^2 + o(t^2) \right)^{1/p} \\ &= 1 + \frac{1}{p} \left( -pt - \frac{p(p-1)}{2} t^2 + o(t^2) \right) + \frac{1}{2p} \left( \frac{1}{p} - 1 \right) (-pt + o(t))^2 + o(t^2) \\ &= 1 - t - \frac{p-1}{2} t^2 + o(t^2) - \frac{p-1}{2} t^2 + o(t^2) = 1 - t - (p-1) t^2 + o(t^2). \end{align*}$$ We have $$(x - y)^2 = (2t + o(t))^2 = 4t^2 + o(t^2)$$ and $$4 - (x+y)^2 = 4 - (2 - (p-1) t^2 + o(t^2))^2 = 4 - 4 + 4(p-1) t^2 + o(t^2) = 4(p-1) t^2 + o(t^2).$$ So there exists $\delta_p > 0$ such that if $|t| < \delta_p$ we have $(x-y)^2 < 5t^2$, $4 - (x+y)^2 > 3(p-1) t^2$. Then $$\tag{$*$} (x-y)^2 < 5t^2 = \frac{5}{3(p-1)} \cdot 3(p-1) t^2 < \frac{5}{3(p-1)} \left( 4 - (x+y)^2 \right)$$ if $|x - 1| < \delta_p$. From the symmetry we have that ($*$) also holds when $|y - 1| < \delta_p$.

To finish the proof it is enough to show that $|x - y| \ge 2\delta_p$ whenever $|x - 1| \ge \delta_p$, $|y - 1| \ge \delta_p$ and $x^p + y^p = 2$. Indeed, since $x^p + y^p = 2$ we have that $\max\{x, y\} \ge 1$. So let $x - 1 \ge \delta_p$. Since $\left( \frac{x+y}{2} \right)^p \le \frac{x^p + y^p}{2} = 1$ we get $x + y \le 2$. Then $x - y \ge 2(x - 1) \ge 2\delta_p$.