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I have a log-concave function $f(\cdot)$ that is defined over $\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$. If I know that $f(0)=0$, $\lim_{x \rightarrow \infty}f(x)=0$, and in between it has strictly positive values, then can I conclude that it has a unique maximizer $x^{*}$? and what other properties can I know about this function?

My basic understanding of a log-concave function is that the log of it is concave, so if it starts from zero and ends at zero, then I can only imagine that it will have a unique maximizer, but I can't prove it, and maybe my imagination is flawed.

Alammouri
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  • You would need to have another property, e.g. $f$ is twice differentiable and non-zero, in order to guarantee a unique maximum. You can, however, guarantee that $f$ attains a maximum, and that $f$ attains this maximum over a unique interval $[a,b]$. – Ben Grossmann Dec 20 '16 at 02:14
  • Let's assume that it is twice differentiable and non-zero. How can I prove that it has a unique maximizer? – Alammouri Dec 20 '16 at 02:32
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    Maximizing $f$ is the same as maximizing $g(x) = \log(f(x))$. Your assumptions tell you $g(x) \to -\infty$ for $x \to 0$ and $x \to \infty$ and that $g$ is concave. Even if $g$ is arbitrarily smooth, this is clearly not enough to guarantee the uniqueness of the maximizer. – Dominik Dec 20 '16 at 09:13

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The function $f(x) = \begin{cases} 0, & x=0 \\ e^{\min(0, 3-({1 \over x}+x)}), & \text{otherwise} \end{cases}$ is log concave and has a maximum value of $1$, but $[1,2] \subset f^{-1} (\{0\})$, so it is not unique.

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copper.hat
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