= Solution
Put $r_t=\sqrt{p_t}-\sqrt p-(t/2)g\sqrt p$. By <differentiability in quadratic mean>, $\|r_t\|_{L^2(\mu)}=o(|t|)$ and
$$
\frac{\sqrt{p_t}-\sqrt p}{t}\longrightarrow\frac12g\sqrt p,
\qquad
\sqrt{p_t}+\sqrt p\longrightarrow2\sqrt p
$$
in <L2 space>. The second limit follows because the first gives $\|\sqrt{p_t}-\sqrt p\|_2=O(|t|)$.
Both <probability density functions> integrate to one. Consequently the <L2 inner product> satisfies
$$
0=\frac1t\int(p_t-p)\,d\mu
=\left\langle\frac{\sqrt{p_t}-\sqrt p}{t},\sqrt{p_t}+\sqrt p\right\rangle_{L^2(\mu)}.
$$
The <Cauchy-Schwarz inequality> permits passage to the limit in this <inner product>, giving
$$
\boxed{Pg=\int g\,dP=0.}
$$
Thus every <score function> is a <mean-zero function>. This proof uses square-root <differentiability in quadratic mean>, without requiring differentiation under the density <integral>.
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