Log-convexity of determinant

Question

Let $f(z):=\langle g(z),g(z)\rangle,$ where $z \mapsto g(z)$ is holomorphic and $\langle \bullet,\bullet\rangle$ is an inner-product on some function space, such as $L^2$, such that $\langle g(z),g(z)\rangle = \int_X \vert g(z,x)\vert^2 d\mu(x).$

Then one has by a fairly explicit computation

$$ \partial_z \partial_{\bar z} \log(f(z)) = f(z)^{-2} ( f(z) \Vert \partial_z g(z) \Vert^2 - \vert \langle \partial_z g(z),g(z) \rangle \vert^2) \ge 0,$$ where positivity just follows from the Cauchy-Schwarz inequality.

I wonder if one can generalize this formula to $f(z):= \operatorname{det}(\langle g_i(z),g_j(z) \rangle)_{i,j}$, where $g_i$ are holomorphic and if the log derivative still remains non-negative?

By a direct computation one has of course also in this case $$ \partial_z \partial_{\bar z} \log(f(z)) = f(z)^{-2} ( f(z) \partial_z \partial_{\bar z} f(z) - \vert \partial_z f(z) \vert^2),$$

but I do not quite see how to say anything more about this last expression, i.e. is $$f(z) \partial_z \partial_{\bar z} f(z) \ge \vert \partial_z f(z) \vert^2?$$

Numerical experiments suggest that this holds, but it is hard for me to verify it.

Answer 1

Your claim can be deduced from the case $n=1$. Let $H$ be the Hilbert space you start with. Let $H^{\otimes n}=H\otimes\dots\otimes H$ be the $n$-th tensor power. Equip it with the usual inner product and complete to get a Hilbert space $V$ again. Let $g=g_1\wedge\dots\wedge g_n$ be the function $$ g=\sum_\sigma\mathrm{sign}(\sigma)\ g_{\sigma(1)}\cdots g_{\sigma(n)}, $$ where the sum runs over all permutations. Then, because of $\mathrm{sign}(\sigma)=\mathrm{sign}(\sigma^{-1})$ we get $$ <g,g> =\sum_\sigma\sum_\tau\mathrm{sign}(\sigma^{-1}\tau)<g_{\sigma(1)},g_{\tau(1)}>\cdots <g_{\sigma(n)},g_{\tau(n)}> $$ Replacing $\tau$ with $\sigma\tau$ and reordering the product we get $$ <g,g> =\sum_\sigma\sum_\tau\mathrm{sign}(\tau)<g_{1},g_{\tau(1)}>\cdots <g_{n},g_{\tau(n)}> $$ this equals $n!\ \det(<g_i,g_j>)$. You apply the $n=1$ case to this $g$ and get the claim.