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Let $I \subseteq \Bbb{R}$ be an open interval. If $f:I\to \mathbb{R}$ is differentiable and $f'$ is monotone nondecreasing then $f$ is convex.

I would like a hint to solve this exercise. I applied the mean value theorem but I failed to solve it.

Thanks.

roger
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    Hint: What does growing monotone means ? – Nizar Apr 23 '15 at 11:39
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    Assume convexity fails, i.e. there is a tangent that is above some points of the graph. Let $P=(a,f(a))$ be the point of contact of the tangent and $Q=(b,f(b))$ a point of the graph that is below the tangent. Apply the mean value theorem in the interval between $a$ and $b$ to get a point $c$ where the slope is the slope of the segment $PQ$. The derivatives at $x=a$ and at $x=c$ decreased. – Alamos Apr 23 '15 at 11:50

1 Answers1

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$(1)$ If $f'$ is nondecreasing monotone, then $f(x)\leq f(a)+f'(a)(x-a)$ for any $x,a\in I$.

Let $x,y \in I$ with $x<y$ and $\theta \in [0,1]$, take $a=(1-\theta)x+\theta y$ and show that $f(a)\leq (1-\theta)f(x)-\theta f(y)$. Now, by $(1) $ :

$f(x)\leq f(a)+f'(a)(x-a)$ $(2) $

$f(y)\leq f(a)+f'(a)(y-a)$$(3) $

Of $(2)$ and $ (3) $ :

$\frac{f(x)-f(a)}{x-a}\leq f'(a)\leq \frac{f(y)-f(a)}{y-a}$.

Then $(1- \theta)f(x)+\theta f(y)$.

Angel
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