11
$\begingroup$

Some time ago, I asked this here. A restricted form of the second question could be this:

If $f$ is a function with continuous first derivative in $\mathbb{R}$ and such that $$\lim_{x\to \infty} f'(x) =a,$$ with $a\gt 0$, then $$\lim_{x\to\infty}f(x)=\infty.$$

To prove it, I tried this:

There exist $x_0\in \mathbb{R}$ such that for $x\geq x_0$, $$f'(x)\gt \frac{a}{2}.$$ There exist $\delta_0\gt 0$ such that for $x_0\lt x\leq x_0+ \delta_0$ $$\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&\gt -\frac{a}{4}\\ \frac{f(x)-f(x_0)}{x-x_0}&\gt f'(x_0)-\frac{a}{4}\\ &\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\ f(x)-f(x_0)&\gt \frac{a}{4}(x-x_0)\end{align*}.$$ We can assume that $\delta_0\geq 1$. If $\delta_0 \lt 1$, then $x_0+2-\delta_0\gt x_0$ and then $$f'(x_0+2-\delta_0)\gt \frac{a}{2}.$$ Now, there exist $\delta\gt 0$ such that for $x_0+2-\delta_0\lt x\leq x_0+2-\delta_0+\delta$ $$f(x)-f(x_0+2-\delta_0)\gt \frac{a}{4}(x-(x_0+2-\delta_0))= \frac{a}{4}(x-x_0-(2-\delta_0))\gt \frac{a}{4}(x-x_0).$$ It is clear that $x\in (x_0,x_0+2-\delta_0+\delta]$ and $2-\delta_0+\delta\geq 1$.

Therefore, we can take $x_1=x_0+1$. Then $f'(x_1)\gt a/2$ and then there exist $\delta_1\geq 1$ such that for $x_1\lt x\leq x_1+\delta_1$ $$f(x)-f(x_1)\gt \frac{a}{4}(x-x_1).$$ Take $x_2=x_1+1$ and so on. If $f$ is bounded, $(f(x_n))_{n\in \mathbb{N}}$ is a increasing bounded sequence and therefore it has a convergent subsequence. Thus, this implies that the sequence $(x_n)$: $$x_{n+1}=x_n+1,$$ have a Cauchy's subsequence and that is a contradiction. Therefore $\lim_{x\to \infty} f(x)=\infty$.

I want to know if this is correct, and if there is a simpler way to prove this. Thanks.

  • 0
    Intuitively (and I am being *very* inaccurate here) if you think about $f=\int f'$ then the fact that approaching infinity you have a positive value means the area under the curve of $f'$ approaches infinity, which in turns means $f$ approaches infinity.2011-07-15
  • 0
    What about my intent? I'm just curious because in my proof, I don't use the mean value theorem. I know, by the answers, that is simpler with t.m.t., but, since I wrote, I want to know.2011-07-15
  • 0
    I would accept your proof, but it looks like you use the mean value theorem in disguise in your sentence $f(x)-f(x_1)\gt \frac{a}{2}(x-x_1).$2011-07-15
  • 0
    @leo: I have read the proof. Have a problem digesting the part of the proof that asserts that we can take $\delta_0 \ge 1$. You tried the jump from local (there exists $\delta_0$) to global (we can take $\delta_0 \ge 1$) and I think it didn't work. In the long displayed line just before "it is clear that" the $x$ have the wrong range, they are not near $x_0$. Can be fixed, by using MVT or something a little weaker. The rest of the argument is fine.2011-07-15
  • 0
    @Ross Millikan: You're right, but it's possible obtain something like $f(x)-f(x_1)\gt \frac{a}{4}(x-x_1)$. I'll fix it.2011-07-15
  • 0
    @user6312: I put "is clear" because $x\in (x_0+2-\delta_0,x_0+2-\delta_0+\delta]\subset (x_0,x_0+2-\delta_0+\delta]$, and $1\gt\delta_0$, then $-\delta_0\gt -1$, sum $2$ both sides: $2-\delta_0\gt 1$, therefore $2-\delta_0+\delta\gt 1$2011-07-15
  • 0
    What do you think now.2011-07-15
  • 0
    @everybody: I couldn't parse the title, so I tried changing it to something which I think is more descriptive. Feel free to keep working on it (or change it back if I have done something unwanted).2011-07-16
  • 0
    In first place, choose a title is difficult. Must be clear and descriptive, and yes, this is more descriptive. Thank you @Pete.2011-07-16

5 Answers 5