I feel as if I ought to know this, but it's escaping me. Let $dA$ denote the usual Lebesgue area measure in $\Bbb C$.
Let's say I have a function $\phi$ which is nonnegative and in $L^{\infty}(\Bbb D)$, where $\Bbb D$ is the unit disc (or some finite Borel measure space).
Let $g_a(z)$ be a function which is square integrable on the disc for all $a \in \Bbb D$ and holomorphic for all $z \in \Bbb D$ for all $a \in \Bbb D$, and suppose we know there exists a $\delta > 0$ such that
$$\int_{\Bbb D} \sqrt{\phi(z)} |g_a(z)|^2 dA(z) > \delta$$ for all $a \in \Bbb D$. Does there exist some other $\delta'>0$ also independent of $a$ such that
$$\int_{\Bbb D} \phi(z) |g_a(z)|^2 dA(z) > \delta'$$ for all $a \in \Bbb D$?
Notice that at its heart, this is really an elementary measure theory question about the Lebesgue integral.
If $\|g_a\|_{L^2(D)}\le M$ for all $a$, then the claim can be proven by Holder inequality: $$ \delta \le \int_D \sqrt\phi |g_a|^2 \le \|g_a\|_{L^2(D)} \|\sqrt\phi |g_a|\|_{L^2(D)} \le M \cdot\left( \int_D \phi |g_a|^2 \right)^{1/2} $$