International Mathematics Competition
for University Students
2025

IMC2025: Day 1, Problem 2

Problem 2. Let \(\displaystyle f\colon\mathbb{R} \to \mathbb{R}\) be a twice continuously differentiable function, and suppose that \(\displaystyle \int_{-1}^{1}f(x)\,\mathrm{d}x=0\) and \(\displaystyle f(1)=f(-1)=1\). Prove that

\(\displaystyle \int_{-1}^{1} \left(f''(x)\right)^2\,\mathrm{d}x\geq 15 , \)

and find all such functions for which equality holds.

Alberto Cagnetta, Università degli Studi di Udine, Italy

Solution. If \(\displaystyle g\) is an arbitrary twice continously differentable function on \(\displaystyle [-1,1]\), then by applying the Caucy–Schwarz inequality for \(\displaystyle f''\) and \(\displaystyle g\) and integrate by parts twice we get

$$\begin{align*} \sqrt{\int_{-1}^1(f'')^2\cdot\int_{-1}^1g^2} & \ge \int_{-1}^1f''g = \bigl(f'(1)g(1)-f'(-1)g(-1)\bigr) - \int_{-1}^1f'g' = \\ & = \bigl(f'(1)g(1)-f'(-1)g(-1)\bigr) - \bigl(f(1)g'(1)-f(-1)g'(-1)\bigr) + \int_{-1}^1fg'' \\ &= \bigl(f'(1)g(1) - f'(-1)g(-1) - g'(1)+g'(-1)\bigr) + \int_{-1}^1fg''. \tag{1} \end{align*}$$

In order to get rid of the terms \(\displaystyle f'(1)g(1)\) and \(\displaystyle f'(-1)g(-1)\) we will chose \(\displaystyle g\) such that \(\displaystyle g(1)=g(-1)=0\). Moreover, if \(\displaystyle g''\) is constant, so \(\displaystyle g\) is an at most quadratic polynomial, then \(\displaystyle \int_{-1}^1fg''=g''\int_{-1}^1=f0\). Hence, it is reasonable to apply (1) with \(\displaystyle g(x)=(1+x)(1-x)=1-x^2\).

With this choice,

\(\displaystyle g(1) = g(-1) = 0, \quad g'(1) = -2, \quad g'(-1) = 2, \quad g'' \equiv -2 \)

and

\(\displaystyle \int_{-1}^1g^2=\int_{-1}^1(1-2x^2+x^4)\,\mathrm{d}x=\frac{16}{15}, \)

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 \(\displaystyle \lambda\) such that \(\displaystyle f''(x)=\lambda g(x)\) almost everywhere in \(\displaystyle [-1,1]\); by continuity of \(\displaystyle f''\) and \(\displaystyle g\) we have \(\displaystyle f''=\lambda g\) everwhere on the interval \(\displaystyle [-1,1]\). Hence \(\displaystyle f(x)\) must have the form

\(\displaystyle f(x) = \lambda\left(\frac{x^{2}}{2}-\frac{x^{4}}{12}\right)+ax+b, \quad\text{with }\lambda, a,b\in\mathbb{R}. \)

From \(\displaystyle \int_{-1}^{1}f(x)\mathrm{d}x=0\), we get \(\displaystyle 0=\frac{3\lambda}{10}+2b\), hence \(\displaystyle b=-\frac{3}{20}\lambda\). Moreover, the condition \(\displaystyle f(1)=f(-1)=1\) implies

\(\displaystyle \lambda\left(\frac{1}{2}-\frac{1}{12}\right)+a-\frac{3}{20}\lambda = f(1) = 1 = f(-1) = \lambda\left(\frac{1}{2}-\frac{1}{12}\right)-a-\frac{3}{20}\lambda \)

hence \(\displaystyle a=0\), and \(\displaystyle \lambda=\frac{15}{4}\).

In conclusion, the equality holds if and only if

\(\displaystyle f(x) = \frac{15}{4} \cdot \left(\frac{x^{2}}{2} - \frac{x^{4}}{12} - \frac{3}{20}\right) = \frac{-5 x^4 + 30 x^2 - 9}{16} \quad\text{for all }x\in[-1,1] . \)