Problem 3. Let $\mathcal{C}$ be the set of functions $f:[0,1] \rightarrow \mathbb{R}$, twice differentiable on $[0,1]$, which have at least two zeros, not necessarily distinct, in $[0,1]$, and $\left|f^{\prime \prime}(x)\right| \leq 1$ for all $x$ in $[0,1]$. Determine the maximum value that the integral

$$
\int_{0}^{1}|f(x)| \mathrm{d} x
$$

can take as $f$ ranges over the set $\mathcal{C}$, and the functions that achieve this maximum.

(A function $f$ has two zeros at the same point $a$ if $f(a)=f^{\prime}(a)=0$.)
Solution. If there exists $a \in [0,1]$ such that $f(a) = f'(a) = 0$, consider $x \in [0,1], x \neq a$. Then $f(x) = \frac{1}{2}(x-a)^2 f''(\theta)$ for some $\theta$ between $a$ and $x$. Since $|f''(t)| \leq 1$ for any $t$ in $[0,1]$, it follows that $|f(x)| \leq \frac{1}{2}(x-a)^2$, a relation clearly true also at $x=a$, so

$$
\int_{0}^{1} |f(x)| \, \mathrm{d} x \leq \frac{1}{2} \int_{0}^{1} (x-a)^2 \, \mathrm{d} x = \frac{1}{6}(1 - 3a(1-a)) \leq \frac{1}{6}
$$

## 2 points

If there exist $a < b$ such that $f(a) = f(b) = 0$, consider $x \in [0,1] \setminus \{a, b\}$, and the function $g: [0,1] \rightarrow \mathbb{R}, g(t) = (t-a)(t-b) f(x) - (x-a)(x-b) f(t)$. This function is twice differentiable on $[0,1]$ and has three distinct zeros, $a, b$, and $x$. By applying Rolle's theorem twice, it follows that $g''$ has a zero $\theta$ in the open interval $(0,1)$. It follows that $f(x) = \frac{1}{2}(x-a)(x-b) f''(\theta)$, so $|f(x)| \leq \frac{1}{2}|x-a| \cdot |x-b|$, a relation clearly true also for $x=a$ and $x=b$. Then

$$
\begin{aligned}
\int_{0}^{1} |f(x)| \, \mathrm{d} x & \leq \frac{1}{2} \int_{0}^{1} |x-a| \cdot |x-b| \, \mathrm{d} x = \frac{1}{6} - \frac{a+b}{4} + \frac{ab}{2} + \frac{(b-a)^3}{6} \\
& = \frac{1}{6} - \frac{1}{2} a(1-b) - \frac{1}{12}(b-a)\left(3 - 2(b-a)^2\right) < \frac{1}{6}
\end{aligned}
$$

Equality holds only in the first case, for $a=0$ or $a=1$. From the continuity of the functions in $\mathcal{C}$, it follows that the functions $f$ that achieve equality are those for which $|f(x)| = x^2 / 2, 0 \leq x \leq 1$, or $|f(x)| = (1-x)^2 / 2, 0 \leq x \leq 1$. Therefore, the required functions are $f_1(x) = x^2 / 2, f_2(x) = (1-x)^2 / 2$ and their opposites.

Remark. The function $g$ comes from the following Lagrange interpolation. Let $n$ be a positive integer, let $a_0 < a_1 < \cdots < a_n$ be points in an interval $I \subseteq \mathbb{R}$, and let $f$ be a real function differentiable $n$ times on $I$. Then

$$
f^{(n)}(\theta) = n! \sum_{i=0}^{n} f(a_i) \prod_{j \neq i} \frac{1}{a_i - a_j}
$$

where $\theta$ is an interior point of $I$. In particular, if $f$ vanishes at $n$ of the $a_i$'s, e.g., at $a_0, \ldots, a_{k-1}, a_{k+1}, \ldots, a_n$, then

$$
f(a_k) = \frac{1}{n!} f^{(n)}(\theta) \prod_{j \neq k} (a_k - a_j)
$$

To prove $(**)$, consider the real function $g$ defined on $I$,

$$
g(x) = f(x) - \sum_{i=0}^{n} f(a_i) \prod_{j \neq i} \frac{x - a_j}{a_i - a_j} = f(x) - \left( \sum_{i=0}^{n} f(a_i) \prod_{j \neq i} \frac{1}{a_i - a_j} \right) x^n + \cdots
$$

This function is differentiable $n$ times on $I$ and has $n+1$ distinct zeros in $I$, namely, $a_0, a_1, \ldots, a_n$. According to Rolle's theorem, $g'$ has $n$ distinct zeros in the interior of $I$, one in each of the open intervals $(a_i, a_{i+1}), i=0, \ldots, n-1$. Applying Rolle's theorem to the derivative $g'$, it follows that $g''$ has $n-1$ distinct zeros in the interior of $I$, and so on. Finally, $g^{(n)}$ has a zero $\theta$ in the interior of $I$, from which the conclusion follows.