If $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function with the property that there is some positive number $c$ such that for every $x\in\mathbb{R}$ $$f'(x)\geq c$$ then $f$ is both one one and onto. Strictly increasing condition gives that $f$ is one one function. How to prove onto? Please help. Thanks a lot.

  • $\begingroup$ Hint: Look at the limits as $|x|\to\infty$ and use the intermediate value theorem. $\endgroup$ – Tim B. Oct 18 '16 at 18:37
  • $\begingroup$ $\lim_{x\to\infty}f(x)=+\infty$ but what about negative limit? $\endgroup$ – neelkanth Oct 18 '16 at 18:38
  • $\begingroup$ Even though I dont' understand how you could get the positive limit and not the negative one: If it's easier for you, look at $\lim_{x\to\infty}-f(-x)$, this function has the same property and tells you something about the negative limit. $\endgroup$ – Tim B. Oct 18 '16 at 18:40
  • $\begingroup$ $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{f(x)}{x}x$ this way... $\endgroup$ – neelkanth Oct 18 '16 at 18:42

You have $f(y) = \int_x^y f'(t) dt \ge c(y-x)$ hence $\lim_{y \to \infty} f(y) = \infty$. If $y < x$ then $f(y) = - \int_y^x f'(t) dt \le -c(x-y)$ and hence $\lim_{y \to -\infty} f(y) = -\infty$.

  • $\begingroup$ +1 yours solution is ok....i am waiting others one..thank you... $\endgroup$ – neelkanth Oct 18 '16 at 19:01
  • $\begingroup$ Glad you approve. $\endgroup$ – copper.hat Oct 18 '16 at 19:03

By mean-value theorem, $a<b$

$$\displaystyle f(b)-f(a)=(b-a)f'(d)$$

$$\displaystyle \geq c(b-a) >0$$

Therefore f(x) is increasing and continuous, one one and onto.


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