Putnam 2019, Problem A-6

Let \(g\) be a real-valued function that is continuous on the closed interval \([0,1]\) and twice differentiable on the open interval \((0,1)\). Suppose that for some real number \(r > 1\),
\[\lim_{x \to 0^+} \frac{g(x)}{x^r} = 0.\]
Prove that either
\[\lim_{x \to 0^+} g'(x) = 0 \quad \text{or} \quad \limsup_{x \to 0^+} x^r |g''(x)| = \infty.\]

Solution:
**Solution 1.** (by Harm Derksen) We assume that $\limsup_{x \to 0^+} x^r |g''(x)| < \infty$ and deduce that $\lim_{x \to 0^+} g'(x) = 0$. Note that
$$\limsup_{x \to 0^+} x^r \sup\{|g''(\xi)| : \xi \in [x/2,x]\} < \infty.$$

Suppose for the moment that there exists a function $h$ on $(0,1)$ which is positive, nondecreasing, and satisfies
$$\lim_{x \to 0^+} \frac{g(x)}{h(x)} = \lim_{x \to 0^+} \frac{h(x)}{x^r} = 0.$$

For some $c > 0$, $h(x) < x^r < x$ for $x \in (0,c)$. By Taylor's theorem with remainder, we can find a function $\xi$ on $(0,c)$ such that $\xi(x) \in [x-h(x),x]$ and
$$g(x-h(x)) = g(x) - g'(x)h(x) + \frac{1}{2}g''(\xi(x))h(x)^2.$$

We can thus express $g'(x)$ as
$$\frac{g(x)}{h(x)} + \frac{1}{2}x^r g''(\xi(x)) \frac{h(x)}{x^r} - \frac{g(x-h(x))}{h(x-x)} \frac{h(x-h(x))}{h(x)}.$$

As $x \to 0^+$, $g(x)/h(x)$, $g(x-h(x))/h(x-h(x))$, and $h(x)/x^r$ tend to 0, while $x^r g''(\xi(x))$ remains bounded (because $\xi(x) \geq x-h(x) \geq x-x^r \geq x/2$ for $x$ small) and $h(x-h(x))/h(x)$ is bounded in $(0,1]$. Hence $\lim_{x \to 0^+} g'(x) = 0$ as desired.

It thus only remains to produce a function $h$ with the desired properties; this amounts to "inserting" a function between $g(x)$ and $x^r$ while taking care to ensure the positive and nondecreasing properties. One of many options is $h(x) = x^r \sqrt{f(x)}$ where
$$f(x) = \sup\{|z^{-r}g(z)| : z \in (0,x)\},$$
so that
$$\frac{h(x)}{x^r} = \sqrt{f(x)}, \quad \frac{g(x)}{h(x)} = \sqrt{f(x)}x^{-r}g(x).$$

**Solution 2.** We argue by contradiction. Assume that $\limsup_{x \to 0^+} x^r |g''(x)| < \infty$, so that there is an $M$ such that $|g''(x)| < Mx^{-r}$ for all $x$; and that $\lim_{x \to 0^+} g'(x) \neq 0$, so that there is an $\epsilon_0 > 0$ and a sequence $x_n \to 0$ with $|g'(x_n)| > \epsilon_0$ for all $n$.

Now let $\epsilon > 0$ be arbitrary. Since $\lim_{x \to 0^+} g(x)x^{-r} = 0$, there is a $\delta > 0$ for which $|g(x)| < \epsilon x^r$ for all $x < \delta$. Choose $n$ sufficiently large that $\frac{\epsilon_0 x_n^r}{2M} < x_n$ and $x_n < \delta/2$; then $x_n + \frac{\epsilon_0 x_n^r}{2M} < 2x_n < \delta$. In addition, we have $|g'(x_n)| > \epsilon_0$ and $|g''(x)| < Mx^{-r} \leq Mx_n^{-r}$ in this range. It follows that
$$\frac{\epsilon_0^2}{2} \frac{x_n^r}{2M} < |g(x_n + \frac{\epsilon_0 x_n^r}{2M}) - g(x_n)|$$
$$\leq |g(x_n + \frac{\epsilon_0 x_n^r}{2M})| + |g(x_n)|$$
$$< \epsilon \left((x_n + \frac{\epsilon_0 x_n^r}{2M})^r + x_n^r\right)$$
$$< \epsilon(1 + 2^r)x_n^r,$$
whence $4M(1 + 2^r)\epsilon > \epsilon_0^2$. Since $\epsilon > 0$ is arbitrary and $M, r, \epsilon_0$ are fixed, this gives the desired contradiction.

**Remark.** Harm Derksen points out that the "or" in the problem need not be exclusive. For example, take
$$g(x) = \begin{cases} x^5 \sin(x^{-3}) & x \in (0,1] \\ 0 & x = 0. \end{cases}$$

Then for $x \in (0,1)$,
$$g'(x) = 5x^4 \sin(x^{-3}) - 3x \cos(x^{-3})$$
$$g''(x) = (20x^3 - 9x^{-1}) \sin(x^{-3}) - 18 \cos(x^{-3}).$$

For $r = 2$, $\lim_{x \to 0^+} x^{-r}g(x) = \lim_{x \to 0^+} x^3 \sin(x^{-3}) = 0$, $\lim_{x \to 0^+} g'(x) = 0$ and $x^r g''(x) = (20x^5 - 9x) \sin(x^{-3}) - 18x^2 \cos(x^{-3})$ is unbounded as $x \to 0^+$. (Note that $g'(x)$ is not differentiable at $x = 0$.)