# How to prove this sign of the derivative?

Suppose that $u:[0,\delta]\rightarrow\mathbf{R}$, $u\in C^2((0,\delta))\cap C([0,\delta])$ such that $$u(0)=0,$$ $$u>0 \ \ in \ (0,\delta],$$$$u''>0$$ Then $u'>0$ in $(0,\delta)$.

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Firstly, $u''>0$ in $(0,\delta)$ implies that $u'$ is strictly increasing in $(0,\delta)$. Hence we need only show that for any given $x \in (0,\delta)$, $u'(x)\geq 0$.

So let $x \in (0,\delta)$. By MVT, $\exists c\in(0,x)$ such that $u'(c)=\frac{u(x)-u(0)}{x-0}=\frac{u(x)}{x}>0$ as $u(0)=0$ and $u(x)>0$. As $u'$ is strictly increasing this implies $u'(x)>0$ as needed.

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Perfect. Thank you very much E. Lim. –  José Carlos Dec 18 '12 at 12:02
..this shows that $u'(c)>0$ for some $c\in(0,\delta)$: it does not show that $u'(x)>0$ for all $x\in(0,\epsilon)$ for some $\epsilon>0$. With $u'$ increasing, we can conclude that $u'>0$ on $[c,\epsilon)$, but there is still $(0,c)$ left to deal with. –  user12477 Dec 18 '12 at 12:10
Well, this is because $c$ is chosen depending on $x$ (for any $x$) and it is always be chosen to be less than $x$. –  E.Lim Dec 18 '12 at 14:32
I think the correct answer is:\\ CLAIM: $u'(x)>0$, $\forall x\in(0,\delta)$.\\ Suppose that exists $x_0\in(0,\delta)$ such that $u'(x_0)\leq0$. Since $u''>0$, the function $u'$ is increasing. Then, $u'(x)\leq0$ for all $x\in(0,x_0)$. By MVT, exists $c\in(0,x_0)$ such that $$u'(c)=\frac{u(x_0)-u(0)}{x_0-0}=\frac{u(x_0)}{x_0},$$ and this is impossible, since that $u'(c)\leq0$ and $u(x_0)/x_0>0$. What you think?! –  José Carlos Dec 18 '12 at 20:05

The condition $u''>0$ tells that the function is concave, that is the curve is facing upwards in the coordinate axes. Just like $y=x^2$. For this to happen while $u>0$ be positive, while $u(0)=0$, $u$ either has a local minima at $0$ or is merely increasing from $0$ towards the positive side of $x$-axis. So it's 1st derivative is positive when you are approaching the curve at $0$ from the positive side.

Just imagine the curve with these conditions.

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Your answer is correct but intuitive. I need to formalize this. –  José Carlos Dec 18 '12 at 10:47

Suppose that there is no such $\epsilon$. Then on every interval of the form $(0,\epsilon)$, and in particular on the interval $(0,\delta)$, there is a number $x_0$ such that $u'(x_0)\leq0$. As noted, we can take $x_0\in(0,\delta)$. Since $u''>0$ on this interval, we find $u'(x)<0$ for all $x\in(0,x_0)$. Now apply the mean value theorem to $u$ at two points $x_1,x_2\in(0,x_0)$ to obtain a contradiction.

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What is the contradiction @user12477??? Because when you apply the mean value theorem, you get $$u(x_2)<u(x_1).$$ Why this is not possible? –  José Carlos Dec 18 '12 at 18:52
Your comment to E.Lim's answer finishes my attempt correctly - I should have said to apply the MVT to $x_1=0$ and $x_2=x_0$ as you did. –  user12477 Dec 18 '12 at 21:09
Thank you @user12477! –  José Carlos Dec 18 '12 at 21:34