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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: