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Consider a function $f(S): \mathcal{S} \to \mathbb{R}$, where $\mathcal{S}$ is the convex cone of all positive semidefinite complex $M\times M$ Hermitian matrices. The function $f(\cdot )$ is concave over $\mathcal{S}$.

I now define another function $g(L): \mathcal{RSM}(M,N) \to \mathbb{R}$ where $\mathcal{RSM}(M,N)$ is the relaxed Stiefel manifold in $\mathbb{C}^M$, i.e., if $L \in \mathcal{RSM}(M,N)$, then $L^\mathrm{H} L \preceq I_N$. The notation $A \preceq B$ means that $A-B$ is negative semi-definite. The function $g(L)$ is defined as $g(L)=f(LL^{\mathrm{H}})$.

My question is concerning what convexity-properties of $g(\cdot )$ that can be $in \; general$ inherited from $f(\cdot)$. For several different $f(\cdot)$ I have observed that the function $g(\cdot )$ is quasi-convex over $\mathcal{RSM}(M,N)$. Is this true in general?

If no general conclusions can be made, I am in particular interested in the function $$f(S) = \log (\det (I+HSH^\mathrm{H})),$$ where $H$ is an $P \times M$ matrix where $P>N$.

Moreover, I am even more interested in the function $$f(S_1,S_2,\ldots,S_K) = \log (\det (I+\sum_{k=1}^K H_kS_kH_k^\mathrm{H}))$$ in which case, the notation above needs slight revision, i.e., $f(\cdot,\ldots,\cdot): \mathcal{S}^K \to \mathbb{R}$ etc.

Finally, $A^\mathrm{H}$ denotes the conjugate transpose operator.

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  • $\begingroup$ The function of interest to you turns out to be geodesically convex on the set of (strictly) positive hermitian matrices. For a proof, please see Corollary 2.11 of our paper: epubs.siam.org/doi/pdf/10.1137/140978168 -- on a different note, this function is not quasi-convex in the Euclidean sense $\endgroup$
    – Suvrit
    Commented Jun 3, 2019 at 17:45
  • $\begingroup$ Thanks for your comment. I have read your paper but failed to see how the reduced rank is entering the problem. I have the feeling we are not studying the same problem. Could you please expand your answer? We have also tried numerically over a long simulation to came up with a counterexample when the quasiconvexity in the euclidean sense fails. It would be of a great help for me if you expand also on this. Thank you for your time. $\endgroup$
    – Jesús Rodríguez
    Commented Jun 6, 2019 at 11:44
  • $\begingroup$ The paper I linked indeed works only for strictly +ve matrices, not for semidefinite ones (because in that case, we no longer have a clean geometry). There seems to be some mismatch btw in your question above, because you say $f$ is convex, however the determinant is not convex on $S$, so I am not sure what you are after. $\endgroup$
    – Suvrit
    Commented Jun 6, 2019 at 14:07
  • $\begingroup$ Sorry, the function f(S) is concave and equal to log det(). $\endgroup$
    – Jesús Rodríguez
    Commented Jun 6, 2019 at 16:19
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    $\begingroup$ I see; I misread it then. IMHO the question could be improved by just stating it shortly as: "Is the map $S \mapsto \log\det(I+SS^*)$ quasi-convex?". It turns out (as a simple numerical counterexample will show) that this map is also not quasi-convex (I tried, however, without the assumption $SS^* \le I$, with that assumption it seems q-c holds for the map mentioned in this comment). $\endgroup$
    – Suvrit
    Commented Jun 10, 2019 at 13:46

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