Differentiability in quadratic mean ($\Vert \sqrt{p_t}-\sqrt{p}-\frac{t}{2}g\sqrt{p}{\Vert }_2=o(|t|)$)

A dominated statistical path is differentiable in quadratic mean at $P$ if its square-root probability density function has an L2 space derivative $(1/2)g\sqrt{p}$, with $g\in L^2(P)$. The score function is $g$. This formulation controls mass near zeros of $p$ as well as the derivative on the support of $P$; a pointwise log-density derivative is insufficient by itself.

Let ${\delta }_t=\sqrt{p_t}-\sqrt{p}=(t/2)g\sqrt{p}+r_t$, with $\Vert r_t{\Vert }_2=o(|t|)$. Since $p_t-p=2\sqrt{p}{\delta }_t+{\delta }_t^2$, the Cauchy-Schwarz inequality gives $\Vert p_t-p-tpg{\Vert }_1\le 2\Vert r_t{\Vert }_2+\Vert {\delta }_t{\Vert }_2^2=o(|t|)$. Thus differentiability in quadratic mean suffices to differentiate bounded density integrals, even when the density itself has no useful pointwise derivative.

For a differentiable-in-quadratic-mean path, normalization gives $⟨(\sqrt{p_t}-\sqrt{p})/t,\sqrt{p_t}+\sqrt{p}⟩=0$. The two factors converge in L2 space to $g\sqrt{p}/2$ and $2\sqrt{p}$. Continuity of the inner product therefore gives $Pg=0$. This proves score function centering without differentiating the density integral.