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Prove: $$x \leq f(x) \leq 2x, \forall x\geq0$$


  1. $f$ is differentiable

  2. $f(0) = 0$

  3. $1 \leq f'(x) \le 2, \forall x\ge0$

I've tried to do it by limit defn but couldn't seem to get to the right solution:

$$ 1 \le \lim_{x \to c} \frac{f(x)-f(c)}{x-c} \le 2$$

how do i manipulate them in such a way that I get $$x \leq f(x) \leq 2x, \forall x\geq0 $$

I've also noticed that $f(x)$ is an increasing function as $f'(x) > 0$. Is this information of any use?

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Hint: consider $g(x) = f(x) - x$ and observe that $g(0) = 0$ and $0 \leq g'(x) \leq 1$. Use this and the mean value theorem to prove that $0 \leq f(x) - x \leq 1$. – t.b. Apr 11 '12 at 8:36
+1 for posting your work! – Bruno Joyal Apr 11 '12 at 8:38
There have been several answers now that use the definite integral from $0$ to $t$ to $1 \leq f'(x) \leq 2$ conclude that $$ \begin{gather*} \int_{0}^t 1\,dx \leq \int_{0}^t f'(x)\,dx \leq \int_{0}^t 2\,dt \qquad\implies \qquad t \leq f(t) \leq 2t \end{gather*} $$ which certainly looks more elegant than the given answers applying the mean value theorem for differentiation. Note however, that Riemann integration is not good enough to justify this conclusion because $f$ is not assumed to be continuously differentiable or has Riemann integrable derivative, while ... – t.b. Apr 11 '12 at 20:18
... applying Lebesgue theory smells circular: In order to prove that $f$ is the definite integral of $f'$ one needs to establish that $f$ is absolutely continuous. However, this seems to need what one wants to prove in the first place. – t.b. Apr 11 '12 at 20:21
You can use the code \to to obtain $\to$ and replace -> – Pedro Tamaroff Apr 11 '12 at 20:29
up vote 5 down vote accepted

The key here is to use the Mean Value Theorem. Suppose $f(x)=kx>2x$ for some $x$. Then by the MVT we have some $c\in [0,x]$ such that $f'(c)=\frac{f(x)-f(0)}{x-0}=k>2$, contradicting the fact that $f'(c)\leq 2$ for all $c\geq 0$. Similarly, if $f(x)=kx<x$ for some $x$ then by the MVT we have some $c\in [0,x]$ such that $f'(c)=\frac{f(x)-f(0)}{x-0}=k<1$, contradicting the fact that $f'(c)\geq 1$ for all $c\geq 0$.

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i managed to obtain a direct proof using t.b.'s hint, with help from some of your workings, thanks! – adsisco Apr 11 '12 at 13:33

Hint: You can study $g(x)=f(x)-x$ and $h(x)=2x-f(x)$, computing their derivatives.

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I would apply monotony of the defined integral in the close interval [0,x] to the three functions in the inequality number 3 : the constant function 1 , f' , and the constant function 2.

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This argument doesn't seem to work. See my comment to the question. – t.b. Apr 11 '12 at 20:23
Yes, you are right...only we know that the derivate is bounded. and if it had discontinuties they would be from second type because it is the derivate I don't think it can be solved in this way.Thank you. – alpha.Debi Apr 11 '12 at 20:31

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