= Solution
Use the same $M_n,v_i,S,\Delta$ as in part (a). The error of the <weighted realized variance> splits into its centered fluctuation and its deterministic <quadrature error>:
$$
\widehat\Lambda_n(g)-\Lambda(g)=M_n+B_n,\qquad
B_n=\sum_i\int_{t_{i-1}}^{t_i}\bigl(g(t_{i-1})-g(s)\bigr)\sigma(s)^2\,ds.
$$
The <Hölder continuity> of $g$ controls this <bias>:
$$
|B_n|\leq R\|\sigma^2\|_\infty\sum_i\frac{\delta_i^{1+\alpha}}{1+\alpha}
\leq\frac{R\|\sigma^2\|_\infty}{1+\alpha}\Delta^\alpha.
$$
Since $\mathbb E M_n=0$, the <bias-variance decomposition of mean squared error> has no cross term. Part (a) therefore gives
$$
\mathbb E\bigl(\widehat\Lambda_n(g)-\Lambda(g)\bigr)^2
\leq2R^2S\Delta+\frac{R^2S}{(1+\alpha)^2}\Delta^{2\alpha}
\leq3R^2S\max\{\Delta,\Delta^{2\alpha}\}.
$$
Hence \b[$\widetilde D=3$ is a universal choice]. This argument separately controls the statistical fluctuation and the <Riemann sum> approximation.
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