Consider the following Ito integral $$I_\Delta = \int_{t-\Delta}^t\sigma_s\int_{t-\Delta}^s\sigma_u dB_u dB_s,$$ where $\sigma_s$ is a bounded stochastic process. I want to show that $$I_\Delta = o_p(\Delta).$$
By Ito isometry, we have $$\mathbb{E}\left[\int_{t-\Delta}^t\sigma_s\int_{t-\Delta}^t\sigma_u dB_u dB_s \right] = \mathbb{E}\int_{t-\Delta}^t\sigma^2_sds = O(\Delta)$$ if $\sigma_s$ is bounded. But this is not quite what I need.
The claim is false in general. Choose $\sigma_t\equiv 1$ so that $I_\Delta=\int_{t-\Delta}^tB_sdB_s-B_{t-\Delta}(B_t-B_{t-\Delta})$. Now recall that $d(B_t)^2=2B_tdB_t+dt$ and thus we can write $$\frac{I_\Delta}{\Delta}=\frac{(B_t-B_{t-\Delta})^2}{2\Delta}-\frac{1}{2}$$ However, $\frac{B_t-B_{t-\Delta}}{\sqrt{\Delta}}\sim \mathcal{N}(0,1)$ for all $\Delta>0$, so $\frac{I_\Delta}{\Delta}$ does not converge in probability to $0$ (if it did, it would converge in distribution to $0$, but it does not).
One approach is to try the Kolmogorov Hölder continuity theorem. Using 4th moment of a Wiener stochastic integral with Ito isometry property?, we have that
$$E[I_{\Delta}^{4}]\leq c_1 \left(\int_{t-\Delta}^{t}E\sigma_{s}^{2}ds \right)\leq c_2 |\Delta|^{2}.$$
Therefore, we get Hölder continuity for $\gamma<\frac{1}{4}$.
