IMC 1999 · Day 1 · P6

(a) Let $g(x) = \max(0, f'(x))$. Then $0 < \int_{-1}^{1} f'(x)\,dx = \int_{-1}^{1} g(x)\,dx + \int_{-1}^{1} (f'(x) - g(x))\,dx$, so we get $\int_{-1}^{1} |f'(x)|\,dx = \int_{-1}^{1} g(x)\,dx + \int_{-1}^{1} (g(x) - f'(x))\,dx < 2 \int_{-1}^{1} g(x)\,dx$. Fix $p$ and $c$ (to be determined at the end). Given any $t > 0$, choose for every $x$ such that $g(x) > t$ an interval $I_x = [x, y]$ such that $|f(y) - f(x)| > c g(x)^{1/p} |y - x| > c t^{1/p} |I_x|$ and choose disjoint $I_{x_i}$ that cover at least one third of the measure of the set $\{g > t\}$. For $I = \bigcup_i I_i$ we thus have $$c t^{1/p} |I| \le \int_I f'(x)\,dx \le \int_{-1}^{1} |f'(x)|\,dx < 2 \int_{-1}^{1} g(x)\,dx;$$ so $|\{g > t\}| \le 3 |I| < (6/c)\, t^{-1/p} \int_{-1}^{1} g(x)\,dx$. Integrating the inequality, we get $\int_{-1}^{1} g(x)\,dx = \int_0^1 |\{g > t\}|\,dt < (6/c)\, p/(p-1) \int_{-1}^{1} g(x)\,dx$; this is a contradiction e.g. for $c_p = (6p)/(p-1)$.

(b) No. Given $c > 1$, denote $\alpha = 1/c$ and choose $0 < \varepsilon < 1$ such that $\bigl( (1+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha} < 1/4$. Let $g : [-1,1] \to [-1,1]$ be continuous, even, $g(x) = -1$ for $|x| \le \varepsilon$ and $0 \le g(x) < \alpha \bigl( (|x|+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha-1}$ for $\varepsilon < |x| \le 1$ is chosen such that $$\int_{\varepsilon}^{1} g(t)\,dt > -\varepsilon/2 + \int_{\varepsilon}^{1} \alpha \bigl( (|x|+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha-1} dt = -\varepsilon/2 + 2\varepsilon \left( 1 - \bigl( (1+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha} \right) > \varepsilon.$$ Let $f = \int g(t)\,dt$. Then $f(1) - f(-1) \ge -2\varepsilon + 2 \int_{\varepsilon}^{1} g(t)\,dt > 0$. If $\varepsilon < x < 1$ and $y = -\varepsilon$, then $$|f(x) - f(y)| \ge 2\varepsilon - \int_{\varepsilon}^{x} g(t)\,dt \ge 2\varepsilon - \int_{\varepsilon}^{x} \alpha \bigl( (t+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha-1} dt = 2\varepsilon \bigl( (x+\varepsilon)/(2\varepsilon) \bigr)^{-\alpha} > g(x) |x - y| / \alpha = f'(x) |x - y| / \alpha;$$ symmetrically for $-1 < x < -\varepsilon$ and $y = \varepsilon$.