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For any real-valued function, $f$, if for all, $x, y, z \in \mathbb{R}$ and $x \leq z$

$$ f(x+y) - f(x) \leq f(z+y) - f(z)$$

then $f$ is convex. Is this argument generic?

Edit: Sorry for the obvious typographical error. It is easy to note that the convexity stems from

$$ \frac{f(x+y) - f(x)}{y} \leq \frac{f(z+y) - f(z)}{y}$$

or, $ f'(x) \leq f'(z)$ in general.

Rodrigo de Azevedo
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Eval
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1 Answers1

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Let $f(w) = w^2,\;\; w \in \mathbb R$. This is a convex function. Consider values $x=1, y=2, z=3$. Then the candidate condition for convexity offered by the OP $f(x+y) - f(x) \geq f(z+y) - f(z)$, becomes

$$(1+2)^2 - 1^2 \geq (3+2)^2 - 3^2 \implies 9-1 \geq 25 - 9 \implies 8 \geq 16$$

which does not hold.

Alecos Papadopoulos
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