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How do you prove a function like $ax - x\log(x)$ is convex? The definition doesn't seem to work easily due to the non-linearity of the log function.

Any ideas?

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3 Answers 3

A function $f(x) \in C^2(\Omega)$ is convex if its second derivative is non-negative.

$$f(x) = x \log(x) \implies f'(x) = x \cdot \dfrac1x + \log(x) \implies f''(x) = \dfrac1x > 0$$

EDIT If $f(x) = ax - x\log(x)$, then $f''(x) = - \dfrac1x$ and hence the function is concave.

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Differentiating twice, we get $$\frac{d^2}{dx^2}x\log(x) = \frac{1}{x}$$ The second derivative is strictly positive everywhere the function is defined, so $x\log(x)$ is strictly convex.

This implies that $ax - x\log(x)$ is strictly concave.

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The answer is quite easy if you use derivatives, so here's one which does not use them. Note that for a continuous function $f:P\to\mathbb{R}$, it is sufficient that it satisfies the inequality:

$$f\left(\frac{x}{2}+\frac{y}{2}\right)\geqslant \frac{f(x)}{2}+\frac{f(y)}{2}$$

for all $x,y\in{P}$ to be concave.

By the AM-GM inequality, we have $\frac{x+y}{2}>\sqrt{xy}$ for $x\ne{y}$. Thus, for your function, we have:

$$\frac{ax+ay}{2}-\ln\frac{x+y}{2}>\frac{ax+ay}{2}-\ln\sqrt{xy}=\frac{ax+ay}{2}-\frac{1}{2}\ln{xy}=\left(\frac{ax-\ln{x}}{2}\right)+\left(\frac{ay-\ln{y}}{2}\right)$$

EDIT: In fact, since in our case, the inequality is strict, your function is strictly concave.

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