This is a problem from Berkeley prelim exams, Spring '99
Suppose that $ f $ is a twice differentiable real-valued function on
$\mathbb{R}$ such that $ f(0) = 0 $, $ f'(0) > 0 $, and $ f''(x) \geq f(x) $ for all $ x \geq 0 $. Prove that
$ f(x) > 0 $ for all $ x > 0 $.
Are there general techniques to solve problems like this?
Best Answer
The basic intuition is that in a neighbourhood of $0$, $f$ is positive. This leads to $f'$ being increasing (as $f'' > f$), and thus $f$ is increasing. If you picked the "leftmost" point where $f$ is 0, then in that interval, you need $f'$ to become $0$ at some point.
To formalize this, an elementary proof:
Since $f(0) = 0$, and $f'(0) \gt 0$, there is a $\delta \gt 0$ such $f(x) \gt 0$ for all $x \in (0, \delta)$.
This we can see by using the $\epsilon-\delta$ definition of derivative and choosing $\epsilon = \frac{f'(0)}{2}$.
Now assume there is some point $y \gt 0$ where $f(y) \le 0$. This implies there is some point $y' \gt 0$ such that $f(y') = 0$
Now let $S$ be the set defined by $S = \{ y: y \ge \delta, f(y) = 0\}$.
Since S is bounded below, $c = \inf S$ (greatest lower bound) is well defined. By continuity of $f$ we have that $f(c) = 0$ (we can pick a sequence $c_n \to c$ and $c_n \in S$). Note that $c \ge \delta \gt 0$.
Now for any $0 \lt x \lt c$, we have that $f(x) \gt 0$. This is because, for $x \lt c$, we cannot have $f(x) = 0$ (as $c = \inf S$), and if $f(d) \lt 0$ for some $d$, then by continuity, there is a point $e \lt d \lt c$ such that $f(e) = 0$.
Thus for $x \in (0,c)$, we have $f(x) \gt 0$ and $f(0) = f(c) = 0$.
In this interval $f''(x) \ge f(x) \gt 0$. Thus $f'(x)$ is increasing, and since $f'(0) \gt 0$, we have that $f'(x) \gt 0$ for all $x \in (0,c)$.
But since $f(0) = f(c)$, we must have that $f'(\eta) = 0$ for some $\eta \in (0,c)$, by Rolle's theorem.
A contradiction.
Thus there is no $y \gt 0$ for which $f(y) \le 0$.