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If $f : \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and
$\lim\limits_{x \rightarrow \infty} f'(x)^2 + f^3(x) = 0$ , show that $\lim\limits_{x\rightarrow \infty} f(x) = 0$.

I really have no clue how to start, I tried things like MVT and using definition of derivatives but I really can't figure this out.

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marked as duplicate by Paramanand Singh, Community May 3 '15 at 9:36

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    $\begingroup$ Is $f^3(x)$ , the third derivative or power? And does $f'(x)^2$ means here $[f'(x)]^2$ or $f'(x^2)$ , I just want to be sure. $\endgroup$ – Mann May 2 '15 at 9:42
  • $\begingroup$ @Mann they are $(f(x))^3$ and $(f'(x))^2$ $\endgroup$ – Soham May 2 '15 at 9:43
  • $\begingroup$ This is a duplicate. See math.stackexchange.com/q/881706/72031 $\endgroup$ – Paramanand Singh May 3 '15 at 9:21
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Assume the claim is wrong, ie., there exists $a>0$ and a sequence $\xi_n\to \infty$ with $|f(\xi_n)|>a$. There exists $x_0$ with $|f'(x)^2+f(x)^3|<\frac{a^3}8$ for $x\ge x_0$. Assume $x_0<x_1<x_2$ and $f'(x_{1,2})=0$. Then $|f(x_{1,2})|<\frac a2$. If some $\xi_n$ is in $[x_1,x_2]$ then there is a local extremum at some $\eta\in(x_1,x_2)$ where $f'(\eta)=0$ and $|f(\eta)|>|f(\xi_n)|>a$. Since this contradicts $|f'(\eta)^2+f(\eta)^3|<\frac{a^3}8$, we conclude that no $\xi_n$ is between two critical points, hence there is at most one critical point $>x_0$. Hence we may assume (by increasing $x_0$ if necessary) that $f$ is strictly monotonic on $[x_0,\infty)$.

Assume $f$ is strictly increasing. Since $f(x)<\frac a2$ for $x>x_0$, $f$ is bounded from above, hence converges. Then $f'(x)^2$ converges as well and as it is never positive, $f'(x)$ converges, and must converge to $0$. Then also $f(x)\to 0$ and we are done.

Hence we may assume that $f$ is strictly decreasing. If $f$ is bounded from below, the same aregument as above applies and we are done again.

Hence $f$ is strictly decreasing and not bounded from below. As $f(x)\to-\infty$, we may assume wlog. that $f(x)^3<-\frac a{16}$ for $x\ge x_0$. Hence $$\tag1 f'(x)^2+\frac12f(x)^3>0$$

Let $b\in\mathbb R $ and $g(x)=-8(x-b)^{-2}$ for $x<b$. Then $g'(x)=16(x-b)^{-3}$ so that $$\tag2g'(x)^2+\frac12g(x)^3=0.$$ Now adjust $b$ so that $x_0<b$ and $g(x_0)=f(x_0)$. That is, we let $$ b=x_0+\sqrt{-\frac8{f(x_0)}}$$ (Note that the radicand is positive). At any point $x\in[x_0,b)$ where $f(x)=g(x)$, we have $f'(x)<g'(x)<0$ by $(1)$ and $(2)$ so that in $f(\xi)<g(\xi)$ in an interval $(x,x+\epsilon)$. We conclude that $f(x)\le g(x)$ for all $x\in[x_0,b)$. But $g(x)\to -\infty$ as $x\to b^-$ whereas $f(x)$ is bounded on $[x_0,b]$. - Contradiction!

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I'll try to solve it by studying the points at which $f$ is zero.

1) First, suppose that $f$ has infinitely many changes of signs, then note $x_n$ all its zero. Then there is $y_n \in ]x_n, x_{n+1}[$ such that $f'(y_n) = 0$ and $f$ is extremum at $y_n$. But then $$f'(y_n)^2 + f(y_n)^3 = f(y_n)^3 \to 0 $$ and $|f(x)|\leqslant |f(y_n)|$ for every $x \in ]x_n, x_{n+1}[$, so we clearly have $f(x) \to 0$ in this case.

Suppose that $f(x)$ only vanishes for a finite number of $x$. Then, for $x$ far enough, $f$ is positive of negative.

2) If $f$ is positive, as $f'(x)^2 + f(x)^3 \to 0$, then it should be clear that $f(x) \to 0$.

3) and if $f$ is negative, I'm working on it, but this seems to be the difficult part of the problem.

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