# Proving an unknown function with some properties is bijective

Let $$f : \mathbb{R} \rightarrow \mathbb{R}$$ be a continuously differentiable function with the property that $$\exists c > 0$$ such that $$f'(x) > c$$ for all $$x\in\mathbb{R}$$.

I want to show $$f$$ is bijective

Injectivity follows easily because $$f$$ is strictly increasing. How can I show $$f$$ is onto? Usually I just compute the inverse function, but this isn't possible here.

• Well, you need to prove that there is no upper or lower bound on the values of $f(x)$. – Don Thousand Apr 1 at 13:30
• Suppose the lower bound was $f(x)$. Then by the fact that $f$ is strictly increasing, we can take $f(x - a) < f(x)$ for $a > 0$. Does that work? – wutv1922 Apr 1 at 13:31
• You have a stronger hypothesis than strictly increasing, and you will have to use it. For example, $\exp(x)$ is strictly increasing but not bijective (but it also does not satisfy the stronger hypothesis.) – hunter Apr 1 at 13:34
• How can you assume that lower bound is attained by $f$? – Dbchatto67 Apr 1 at 13:35
• @hunter I'm guessing I need to use the fact that it's continuously differentiable. I still can't get anything. – wutv1922 Apr 1 at 13:48

Notice that $$f(x)=\int_0^{x}f'(t)dt \implies \lim_{x\to\pm\infty}|f(x)|>\lim_{x\to\pm\infty}|x|c=\infty$$.

Claim one: for any $$x \in \mathbb{R}$$, $$f(x + 1) > f(x) + c$$.

Proof: suppose that $$f(x + 1) \leq f(x) + c$$. By the mean value theorem, there is a point $$y \in [x, x+1]$$ with $$f'(y) = \frac{f(x + 1) - f(x)}{(x+1) - 1} = f(x+1) - f(x) \leq c,$$ a contradiction.

Claim two: $$f(x)$$ is surjective.

Proof: Say we want to show that $$b$$ is in the range of $$f(x)$$. We'll break into cases, although the argument in both cases is really the same. First, if $$b = f(0)$$, we're obviously done.

If $$b > f(0)$$, then for any positive integer $$N$$, we have $$f(N) > f(0) + Nc$$ by iteratively applying claim one, and we can pick $$N$$ large enough to make the right hand side greater than $$b$$. Then applying the intermediate value theorem we see that $$b$$ is in the range of $$f$$.

If $$b < f(0)$$, then for any positive integer $$N$$, we have $$f(-N) < f(0) - Nc$$ by applying claim one iteratively again. Now pick $$N$$ large enough to make the left hand side less than $$b$$, and apply the intermediate value theorem.

In the two non-trivial cases, we are using that $$c \neq 0$$.

Claim three: $$f(x)$$ is injective. As you say, this is immediate, since $$f$$ is increasing.