IMC 2025 · Day 1 · P2

If $g$ is an arbitrary twice continously differentable function on $[-1, 1]$, then by applying the Caucy–Schwarz inequality for $f''$ and $g$ and integrate by parts twice we get $$\begin{align*} \sqrt{\int_{-1}^{1} (f'')^2 \cdot \int_{-1}^{1} g^2} \ge \int_{-1}^{1} f'' g &= \bigl( f'(1) g(1) - f'(-1) g(-1) \bigr) - \int_{-1}^{1} f' g' = \\ &= \bigl( f'(1) g(1) - f'(-1) g(-1) \bigr) - \bigl( f(1) g'(1) - f(-1) g'(-1) \bigr) + \int_{-1}^{1} f g'' \\ &= \bigl( f'(1) g(1) - f'(-1) g(-1) - g'(1) + g'(-1) \bigr) + \int_{-1}^{1} f g''. \tag{1} \end{align*}$$ In order to get rid of the terms $f'(1) g(1)$ and $f'(-1) g(-1)$ we will chose $g$ such that $g(1) = g(-1) = 0$. Moreover, if $g''$ is constant, so $g$ is an at most quadratic polynomial, then $\int_{-1}^{1} f g'' = g'' \int_{-1}^{1} f = 0$. Hence, it is reasonable to apply (1) with $g(x) = (1 + x)(1 - x) = 1 - x^2$.

With this choice, $$g(1) = g(-1) = 0, \quad g'(1) = -2, \quad g'(-1) = 2, \quad g'' \equiv -2$$ and $$\int_{-1}^{1} g^2 = \int_{-1}^{1} (1 - 2x^2 + x^4)\,dx = \frac{16}{15},$$ so we get $$\begin{gather*} \sqrt{\int_{-1}^{1} (f'')^2 \cdot \frac{16}{15}} \ge 0 - 0 - g'(1) + g'(-1) + 0 = 4, \\ \int_{-1}^{1} (f'')^2 \ge 15. \end{gather*}$$ Equality in Cauchy–Schwarz in (1) holds only if there exists a real $\lambda$ such that $f''(x) = \lambda g(x)$ almost everywhere in $[-1, 1]$; by continuity of $f''$ and $g$ we have $f'' = \lambda g$ everwhere on the interval $[-1, 1]$. Hence $f(x)$ must have the form $$f(x) = \lambda \left( \frac{x^2}{2} - \frac{x^4}{12} \right) + ax + b, \quad \text{with } \lambda, a, b \in \mathbb{R}.$$ From $\int_{-1}^{1} f(x)\,dx = 0$, we get $0 = \frac{3\lambda}{10} + 2b$, hence $b = -\frac{3}{20} \lambda$. Moreover, the condition $f(1) = f(-1) = 1$ implies $$\lambda \left( \frac12 - \frac{1}{12} \right) + a - \frac{3}{20} \lambda = f(1) = 1 = f(-1) = \lambda \left( \frac12 - \frac{1}{12} \right) - a - \frac{3}{20} \lambda$$ hence $a = 0$, and $\lambda = \frac{15}{4}$.

In conclusion, the equality holds if and only if $$f(x) = \frac{15}{4} \cdot \left( \frac{x^2}{2} - \frac{x^4}{12} - \frac{3}{20} \right) = \frac{-5 x^4 + 30 x^2 - 9}{16} \quad \text{for all } x \in [-1, 1].$$