IMC 1997 · Day 2 · P7

Let $c = \dfrac{1}{2} f''(0)$. We have $$g = \frac{(f')^2 - 2 f f''}{2 (f')^2 \sqrt{f}},$$ where $$f(x) = cx^2 + O(x^3), \quad f'(x) = 2cx + O(x^2), \quad f''(x) = 2c + O(x).$$ Therefore $(f'(x))^2 = 4c^2 x^2 + O(x^3)$, $$2 f(x) f''(x) = 4c^2 x^2 + O(x^3)$$ and $$2 (f'(x))^2 \sqrt{f(x)} = 2 (4c^2 x^2 + O(x^3)) |x| \sqrt{c + O(x)}.$$ $g$ is bounded because $$\frac{2 (f'(x))^2 \sqrt{f(x)}}{|x|^3} \xrightarrow[x \to 0]{} 8 c^{5/2} \ne 0$$ and $f'(x)^2 - 2 f(x) f''(x) = O(x^3)$.

The theorem does not hold for some $C^2$-functions.

Let $f(x) = (x + |x|^{3/2})^2 = x^2 + 2 x^2 \sqrt{|x|} + |x|^3$, so $f$ is $C^2$. For $x > 0$, $$g(x) = \frac{1}{2} \left( \frac{1}{1 + \frac{3}{2} \sqrt{x}} \right)' = -\frac{1}{2} \cdot \frac{1}{(1 + \frac{3}{2} \sqrt{x})^2} \cdot \frac{3}{4} \cdot \frac{1}{\sqrt{x}} \xrightarrow[x \to 0]{} -\infty.$$