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Let $f$ be a continuous function, with the initial condition: f''>0.

I need to prove that $f(2x)-f(x) < f(3x)-f(2x)$.

By f''>0 I can learn that f' is monotonous increasing, and so I tried to use Lagrange or Taylor.

Thank you.

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    @Myself: sorry for the misunderstanding ;) I checked your profile and since there was nothing on it I made an assumption, no harm intended :D. Maybe it's for "romance language"-speaking countries, I don't know. ;)2011-03-01

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You are given that f''(x)>0 everywhere. If this condition holds, then $f(x)$ is convex on its domain.

Any convex function will satisfy Jensen's Inequality. Using Jensen's inequality with two terms, you have

\begin{equation} f(2x)<\frac{f(x)+f(3x)}{2} \end{equation}

i.e.,

\begin{equation} f(2x) - f(x) < f(3x) -f(2x) \end{equation}

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    @Myself - I didn't think I was cheating here because the OP didn't say anything about using only "Calculus I" methods.2011-03-01
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Let $g(y) = f(2x+y)$. Then g(y) = g(0) + y g'(0) + \frac{y^2}{2} g''(\xi) g(-y) = g(0) - y g'(0) + \frac{y^2}{2} g''(\eta) Note that g'' > 0. Hence, adding the above two we get g(y)+g(-y) = 2g(0) + \frac{y^2}{2} g''(\xi) + \frac{y^2}{2} g''(\eta) > 2g(0)

Hence, we get $f(2x+y) + f(2x-y) > 2f(2x)$ Plug in $y=x$ and rearrange to get $f(2x)-f(x) < f(3x)-f(2x)$

(PS: This is nothing but the derivation of Jensen's in the case when $\lambda = \frac{1}{2}$)

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I'll try to expand my comment a bit:

First assume $x>0$.

Since $f$ is continuous everywhere, it is continuous on $[x,2x]$. By the mean value theorem, there is some $y_1\in [x,2x]$ such that f(2x)-f(x) = xf'(y_1). Similarly, there is some $y_2\in [2x,3x]$ suc that f(3x)-f(2x) = xf'(y_2).

Therefore it is sufficient to show that xf'(y_2) > xf'(y_1), or (because $x>0)$ that f'(y_2) > f'(y_1). We know that $y_2 \geq y_1$ and that f' is strictly increasing so we'll just have to exclude the possibility that $y_1 = y_2$. But in that case $y_2 = y_1 = 2x$ which leads to a contradiction. ( f(2x) = f(x) + \int_x^{2x} f'(t)dt < f(x) + \int_x^{2x} f'(2x) dt = f(x) + xf'(2x) = f(2x). Although there are probably neater ways to find a contradiction.)

Now assume $x<0$. Construct $g(x) = f(-x)$ and reason with $g(x)$; I haven't done it but it shouldn't be too hard.