Can a smooth probability density function on $\mathbb{R}^n$ have no local maxima?

Question

I am thinking about probability measures on unbounded spaces. Is it possible to find a smooth, bounded probability measure on $\mathbb{R}^n$ that has no local maxima? I have seen some related questions, but not enough to help me understand how to answer this one. I haven't been able to think of a reason why such a measure can't exist, but am having trouble coming up with anything.

Answer 1

This is not possible in dimension $n=1$. Let $f$ be your probability distribution having no extrema. Then $f$ needs be monotone, which contradicts the requirement that it is a probability distribution.

For higher dimensions this is possible by decaying in sufficiently many dimensions. For example in dimension $n=2$ pick a smooth, positive function $f:\mathbb{R} \rightarrow \mathbb{R}$ which is bounded and strictly monotone. Then consider $F(x,y) = f(x) e^{-y^2 (1+x^2)^2}$. Then we get $$ \int_{\mathbb{R}^2} F(x,y) dx dy = \int_{\mathbb{R}^2} \frac{f(x)}{1+x^2} e^{-z^2}dx dz<\infty. $$ Note that $F$ does not admit a local extrema as the gradient never vanishes, indeed $$ \partial_2 F(x,y) = -2y (1+x^2)^2 f(x) e^{-y^2 (1+x^2)^2}. $$ Thus, $\partial_2 F(x,y)=0$ implies that $y=0$. However, $F(x,0)=f(x)$ which is strictly monotone and does not admit a local extremum. Now we can normalize $F$ such that it becomes a probability distribution and we get our counterexample.

Similarly we can argue in dimension $n \geq 3$. Just consider $$G: \mathbb{R} \times \mathbb{R}^{n-1} \rightarrow \mathbb{R}, G(x,y) = f(x) e^{-\Vert y \Vert^2 (1+x^2)^2}.$$

Note that we could make our probability distribution even real-analytic by choosing $f(x) = \arctan(x) + 42.$