Putnam 2023, Problem A-3

Determine the smallest positive real number \( r \) such that there exist differentiable functions \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R} \to \mathbb{R} \) satisfying
(a) \( f(0) > 0 \),
(b) \( g(0) = 0 \),
(c) \( |f'(x)| \leq |g(x)| \) for all \( x \),
(d) \( |g'(x)| \leq |f(x)| \) for all \( x \), and
(e) \( f(r) = 0 \).

Solution:
The answer is $r = \frac{\pi}{2}$, which manifestly is achieved by setting $f(x) = \cos x$ and $g(x) = \sin x$.

**First solution.** Suppose by way of contradiction that there exist some $f, g$ satisfying the stated conditions for some $0 < r < \frac{\pi}{2}$. We first note that we can assume that $f(x) \neq 0$ for $x \in [0, r)$. Indeed, by continuity, $\{x \mid x \geq 0$ and $f(x) = 0\}$ is a closed subset of $[0, \infty)$ and thus has a minimum element $r'$ with $0 < r' \leq r$. After replacing $r$ by $r'$, we now have $f(x) \neq 0$ for $x \in [0, r)$.

Next we note that $f(r) = 0$ implies $g(r) \neq 0$. Indeed, define the function $k: \mathbb{R} \to \mathbb{R}$ by $k(x) = f(x)^2 + g(x)^2$. Then $|k'(x)| = |2f(x)f'(x) + g(x)g'(x)| \leq 4|f(x)g(x)| \leq 2k(x)$, where the last inequality follows from the AM-GM inequality. It follows that $\left|\frac{d}{dx}(\log k(x))\right| \leq 2$ for $x \in [0, r)$; since $k(x)$ is continuous at $x = r$, we conclude that $k(r) \neq 0$.

Now define the function $h: [0, r) \to (-\pi/2, \pi/2)$ by $h(x) = \tan^{-1}(g(x)/f(x))$. We compute that
$$h'(x) = \frac{f(x)g'(x) - g(x)f'(x)}{f(x)^2 + g(x)^2}$$
and thus
$$|h'(x)| \leq \frac{|f(x)||g'(x)| + |g(x)||f'(x)|}{f(x)^2 + g(x)^2} \leq \frac{|f(x)|^2 + |g(x)|^2}{f(x)^2 + g(x)^2} = 1.$$
Since $h(0) = 0$, we have $|h(x)| \leq x < r$ for all $x \in [0, r)$. Since $r < \pi/2$ and $\tan^{-1}$ is increasing on $(-r, r)$, we conclude that $|g(x)/f(x)|$ is uniformly bounded above by $\tan r$ for all $x \in [0, r)$. But this contradicts the fact that $f(r) = 0$ and $g(r) \neq 0$, since $\lim_{x \to r^-} g(x)/f(x) = \infty$. This contradiction shows that $r < \pi/2$ cannot be achieved.

**Second solution.** (by Victor Lie) As in the first solution, we may assume $f(x) > 0$ for $x \in [0, r)$. Combining our hypothesis with the fundamental theorem of calculus, for $x > 0$ we obtain
$$|f'(x)| \leq |g(x)| \leq \left| \int_0^x g'(t) dt \right|$$
$$\leq \int_0^x |g'(t)| dt \leq \int_0^x |f(t)| dt.$$
Define $F(x) = \int_0^x f(t) dt$; we then have
$$f'(x) + F(x) \geq 0 \quad (x \in [0, r]).$$
Now suppose by way of contradiction that $r < \frac{\pi}{2}$. Then $\cos x > 0$ for $x \in [0, r]$, so
$$f'(x)\cos x + F(x)\cos x \geq 0 \quad (x \in [0, r]).$$
The left-hand side is the derivative of $f(x)\cos x + F(x)\sin x$. Integrating from $x = y$ to $x = r$, we obtain
$$F(r)\sin r \geq f(y)\cos y + F(y)\sin y \quad (y \in [0, r]).$$
We may rearrange to obtain
$$F(r)\sin r \sec^2 y \geq f(y)\sec y + F(y)\sin y \sec^2 y \quad (y \in [0, r]).$$
The two sides are the derivatives of $F(r)\sin r \tan y$ and $F(y)\sec y$, respectively. Integrating from $y = 0$ to $y = r$ and multiplying by $\cos^2 r$, we obtain
$$F(r)\sin^2 r \geq F(r)$$
which is impossible because $F(r) > 0$ and $0 < \sin r < 1$.