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Spring 2019 - Problem 1

calculus

Let $f \in C^2\p{\R}$ be a real-valued function that is uniformly bounded on $\R$ . Prove that there exists a point $c \in \R$ such that $f''\p{c} = 0$ .

Solution.

Suppose otherwise, and that $f''\p{x} > 0$ for all $x \in \R$ . In the other case, we may replace $f$ with $-f$ and run the same argument. In this case, $f'$ is strictly increasing, so in particular, there exists $x_0 \in \R$ such that $f'\p{x_0} \neq 0$ . If $f'\p{x_0} > 0$ , then for $x \geq x_0$ , we have

\begin{aligned} f\p{x} &= f\p{x_0} + \int_{x_0}^x f'\p{t} \,\diff{t} \\ &\geq f\p{x_0} + \p{x - x_0}f'\p{x_0} \xrightarrow{x\to\infty} \infty, \end{aligned}

which is impossible. In the other case, if $f'\p{x_0} < 0$ , we get for $x \leq x_0$ that

\begin{aligned} f\p{x} &= f\p{x_0} + \int_{x_0}^x f'\p{t} \,\diff{t} \\ &\leq f\p{x_0} + \p{x - x_0}f'\p{x_0} \xrightarrow{x\to-\infty} -\infty, \end{aligned}

which also contradicts the boundedness of $f$ . Thus, $f''$ must vanish for some $c \in \R$ , which was what we wanted to show.