 Solution

Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.