For any probability distribution $P$ on the sphere $\Bbb S^n\subset \Bbb R^{n+1}$ let $\lambda_{max}(P)$ denote the largest eigenvalue of its covariance matrix $$ Cov(P)=\left(\int_{\Bbb S^n} x_ix_j P(dx)-\int_{\Bbb S^n} x_i P(dx)\int_{\Bbb S^n} x_j P(dx)\right)_{i,j=1,\ldots,n+1}. $$
Question: Assume that $P$ has a density $p$ with respect to the Lebesgue surface measure and that $p$ is unimodal, i.e. it has exactly one local maximum. Is it true that then $$ \lambda_{max}(P)<\lambda_{max}(\mathcal U), $$ where $\mathcal U$ denotes the uniform distribution on $\Bbb S^n$?
This seems like a relatively obvious question, and I'd expect that either a counter example or a proof is known - and I am looking for a reference. I'd be surprised if the answer to the question was no. Note that in any case the unimodality assumption cannot be dropped.
PS: I am aware that $Cov(\mathcal U)=(n+1)^{-1}\cdot 1_{(n+1)\times(n+1)}$ and hence $\lambda_{max}(\mathcal U)=(n+1)^{-1}$. That is not part of my question.
"Exactly one local maximum" is too weak a condition to imply anything. Indeed, take any reasonable smooth density with finitely many local maxima, draw a region of very small area encompassing all these maxima and, say, homeomorphic to a ball, and pull the density up there keeping the boundary values, so that there will be only one maximum after that but there would be not much change in the $L^1$ norm (you'll need to renormalize a bit, but it is not an issue either). So, if you have something like what you are interested in in the class of unimodal probability densities, you'll actually have it for all densities, and you have indicated that you know yourself that the latter is false.