# Ngô Quốc Anh

## December 19, 2007

### Quen quen :-?

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:34

Let $f \in C^1$ and ${\lim_{x \to \infty} {f'^2}(x) + {f^3}(x) = 0$. Prove that ${\lim_{x \to \infty} f(x) = 0$.

Proof. Let $h(x) = \{ f'(x) \}^2 + \{ f(x) \}^3$. For any $\epsilon > 0$, there is $x_0$ such that for $x>x_0$, $|h(x)| < \epsilon$. Consider the set $U = \{ x>x_0 \, : \, |f(x)| > \sqrt[3]{2\epsilon} \}$. If $U$ is empty, then the result immediately follows. So assume $U$ is nonempty and choose any $a \in U$. If $f(a) > 0$, then $\{ f(a) \}^3 > 2\epsilon$, so $\{ f'(a) \}^2 + 2\epsilon < h(a) < \epsilon,$ which is a contradiction. Thus $f(a) < 0$ and $\{ f(a) \}^3 < -2\epsilon$. Then

$-\epsilon < \{ f'(a) \}^2 + \{ f(a) \}^3 < \{ f'(a) \}^2 - 2\epsilon,$

or $\epsilon < \{ f'(a) \}^2$. Since $f'$ is continuous, this inequality implies $f'(a)$ does not vanish at any point in $U$.

Now assume there is a point $b \in U$ and $x_0 < a < b$ such that $f(b) < f(a)$, Applying intermediate value theorem, we may assume $a$ is in the neighborhood of $b$ in $U$. By mean value theorem, there is $c \in U$ such that $f'(c) < 0$. Since $f'$ cannot change its sign on $U$, $f$ is strictly decreasing on $U$. Especially, $I = [a, \infty) \subset U$.

Now let $f(x) = - u^{-2}$, where $u = u(x) > 0$ on $I$. Then

$h(x) = \{ f'(x) \}^2 + \{ f(x) \}^3 = 4 u'^2 u^{-6} - u^{-6}$,

or

$2u' = - \sqrt{1 + u^6 h(x)}$.

Since $\epsilon < \{ f'(x) \}^2$ and $f'(x) < 0$, we have $-\sqrt{\epsilon} > f'(x) = 2u'u^{-3}$ and $u$ is strictly decreasing. So $u$ must converge and we have

$\lim_{x\to\infty} 2u' = - \lim_{x\to\infty} \sqrt{1 + u^6 h(x)} = -1,$

which contradicts that $u$ converges.

So, for all $b \in U$ and $x_0 < a < b$, $f(b) \geq f(a)$. This means $f$ is monotone increasing on $U$. If $f(x) < -\sqrt[3]{2\epsilon}$ for all $x > x_0$, then $f(x)$ must converge to some $-L$ where $L \geq \sqrt[3]{2\epsilon}$. But this implies $\lim_{x\to\infty} f'(x) = \sqrt{L} > 0$, leading to a contradiction. Hence $|f(x_1)| \leq \sqrt[3]{2\epsilon}$ for some sufficient large $x_1$, and by our preceeding argument, $|f(x)| \leq \sqrt[3]{2\epsilon}$ for all $x \geq x_1$. So $f(x) \to 0$ as $x \to \infty$.