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Let $f:[a,b]\rightarrow \mathbb R$ be a continuous function such that

$$\frac{f(x) - f(a)}{x-a}$$

is an increasing function of $x\in [a,b]$. Is $f$ necessarily convex? What if we also assume that

$$\frac{f(b) - f(x)}{b-x}$$

is increasing in $x$?

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    I am adding the condition that $f$ be continuous.2012-07-18

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A counterexample (to both) can be constructed from the function $f(x)=x^2$ on the interval $[0,1]$ by adding a little bump to the graph, say, near the point $(1/2,1/4)$.

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    Or, more crudely, try $f(x) = x^2$ when $x \in [0,\frac{1}{2}]$, and $f(x) = x^2+1$, when $x \in (\frac{1}{2},1]$. $f$ is not continuous at $x=\frac{1}{2}$, hence it is not convex.2012-07-18
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    I don't think one can make a small enough bump for non-convexity to hold, while still satisfying both given assumptions. Perhaps I'm not interpreting "bump" correctly. Could you elaborate, perhaps?2012-07-18
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    Or for a continuous counterexample, $f(x) = x^2$ on $[0,\frac{1}{2}]$, $f(x) = (x-\frac{1}{2})^2+\frac{3}{4}(x-\frac{1}{2})+\frac{1}{4}$ on $[\frac{1}{2},1]$. A little harder to verify, but it is non-convex, and $\frac{f(x)}{x}$ is increasing.2012-07-18
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    @Cameron What I had in mind: if you replaced a part of parabola with its chord, both quotients remain **strictly increasing**, but the function is no longer strictly convex. Now replace the chord with a slightly concave arc.2012-07-18
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    Ah! I thought you meant a vertical shift to give a jump discontinuity. That makes more sense.2012-07-18
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Yes $f$ is necessarily convex (with the added assumption continuous), if $ \displaystyle x\mapsto \frac{f(x)-f(a)}{x-a}$ is increasing over $(a,b]$. Let $a