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.



Log-convexity of determinant
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.


