# Math Help - second-differentiable function proving help

1. ## second-differentiable function proving help

f is a second-differentiable function at $(0,\infty)$ so $f''(x)>0$ to every $x\in(0,\infty)$
i need to prove that if: $lim{}_{x\rightarrow\infty}f(x)=\ell$ $(\ell$ is finite), so -
(1) $f'(x)<0$ to every $x\in(0,\infty)$
(2) $sup\: f'((0,\infty))=0$
(3) $lim{}_{x\rightarrow\infty}f'(x)=0$

2. ## Re: second-differentiable function proving help

Here's some ideas on how to prove them:

1. If $f'(x_0) = r > 0$ for some $x_0\in(0,\infty)$, then f'(x) would have to be greater than r for all $x > x_0$, which would mean it is unbounded as $x \to \infty$.

2. If the sup is not zero, then it must be less than zero. So the derivative is less than some fixed negative number, which again means f(x) is unbounded.

3. This follows immediately from #1 and #2.

- Hollywood

3. ## Re: second-differentiable function proving help

it's greating getting ideas and clues.. that's what's the best. thx!
but i didn't quite understand you well.. can you repeat your ideas in other way?

4. ## Re: second-differentiable function proving help

Let's start with #1. I should have said $f'(x_0)=r\ge{0}$, by the way, and "it" at the end is f(x):
1. If $f'(x_0) = r \ge 0$ for some $x_0\in(0,\infty)$, then f'(x) would have to be greater than r for all $x > x_0$, which would mean f(x) is unbounded as $x \to \infty$.

The proof is by contradiction, so you assume that $f'(x_0)\ge{0}$ for some real number $x_0$.

If you look at the graph of f'(x) vs. x, you know that it always goes up as it goes to the right (that's what f''(x)>0 means). So if $(x_0,r)$ is a point on the graph, every point to the right of $x=x_0$ is above $y=r$. To prove this rigorously, use the Mean Value Theorem (for f'(x) between $x_0$ and x, where x is some real number greater than $x_0$). So in this first part, you should prove that if $f'(x_0)\ge{0}$ for some real number $x_0$, then $f'(x)>0$ for all $x>x_0$.

The second half is essentially the same argument using f(x) instead of f'(x). You choose $x_1>x_0$, so $f'(x_1)>0$ and prove that $f(x)>f(x_1)+f'(x_1)(x-x_1)$ for all $x>x_1$. Since $f'(x_1)$ is positive, the right-hand side is unbounded, so the left-hand side is unbounded. This contradicts the hypothesis that $\lim_{x\to\infty}f(x)=\ell$. Therefore the assumption that $f'(x_0)\ge{0}$ for some real number $x_0$ is false.

- Hollywood

5. ## Re: second-differentiable function proving help

great! i got all, thank you.. except -
Originally Posted by hollywood
You choose $x_1>x_0$, so $f'(x_1)>0$ and prove that $f(x)>f(x_1)+f'(x_1)(x-x_1)$ for all $x>x_1$. Since $f'(x_1)$ is positive, the right-hand side is unbounded, so the left-hand side is unbounded.

6. ## Re: second-differentiable function proving help

You need to move from $x_0$ to $x_1$ because $f'(x_0)$ could be zero, and we need the derivative to be greater than zero.

You can use the Mean Value Theorem from $x_0$ to $x_1$ to prove that $f'(x_1)>0$. Then use the Mean Value Theorem (again!) to prove my inequality. Remember $f''(x)$ is always positive, so if $c>x_1$, $f'(c)>f'(x_1)$.

As x goes to infinity, $f'(x_1)$ and $f(x_1)$ are fixed, and $f'(x_1)$ is positive. So for any M, you can determine an x for which $f(x)>M$. This is inconsistent with $\lim_{x\to\infty}f(x)=\ell$.

- Hollywood

7. ## Re: second-differentiable function proving help

ok.. got it!
and what about number 2 - sup f'((0,infty))=0

8. ## Re: second-differentiable function proving help

By (1) you have $\sup{f'(x)}\le{0}$. So you need to show that $f'(x)>-\epsilon$ for all $\epsilon>0$. Suppose this were not true - that there is some $\epsilon>0$ for which $f'(x)\le-\epsilon$ for all $x \in (0,\infty)$. You can use the same argument as #1 to show that $f(x)$ goes to $-\infty$, which contradicts $\lim_{x\to\infty}f(x)=\ell$.

- Hollywood

9. ## Re: second-differentiable function proving help

great! and now, how #3 follows immediately from #1 and #2?
sorry i can't figure it out by my own...

10. ## Re: second-differentiable function proving help

Well, maybe not so immediately. Let $\epsilon>0$. From #2, you can conclude that there is an N such that $f'(N)>-\epsilon$. If x>N, then you can use the mean value argument to prove that $f'(x)>-\epsilon$ (since f''(x)>0). Of course, you have (from #1) that f'(x)<0. So you have what you need to prove the limit - for all $\epsilon$, there is an N such that if x>N, $|f(x)|<\epsilon$.

- Hollywood