Let $f$ be a probability density function that is continuous and positive everywhere on $\mathbb{R}^n$, $n>1$. Is it possible that $f$ has no local maximum?
For $n=1$ it is easy to show that this can't happen. But the bivariate case seems different. For example, could there be ridge that rises forever in some direction, but narrows sufficiently fast as it does so to ensure integration to 1?
An example or reference would be especially appreciated.
Consider the following example: $$ f(x,y)=e^xe^{-2(1+e^x)^2|y|}. $$ For every fixed $x\in\mathbb R$ the maximum of $f(x,y)$ in $y$ is $e^x$, and since it's reached at the same value $y=0$ this means that if there were a local maximum for $f(x,y)$ it would have $y=0$ as well. But $e^x$ doesn't have any maximum. So, there's no local maxima for $f(x,y)$.
Integrating the function we get $$ \int_{\mathbb R^2}f(x,y)dxdy=\int_{-\infty}^\infty e^x\left(2\int_0^\infty e^{-2(1+e^x)^2y}dy\right)dx=2\int_{-\infty}^\infty e^x\frac{dx}{2(1+e^x)^2}=\left.\frac{e^x}{1+e^x}\right|_{-\infty}^\infty=1. $$ So $f(x,y)$ is indeed a PDF.
Your idea is correct. An example is the density $$ f(x,y) = \frac{cy^2}{1+y^8+x^2} $$ with $c = \frac{1}{2 \sqrt{\pi} \Gamma(3/8) \Gamma(9/8)}$.
