Evaluating $\mathbb E [\lvert \frac{1}{B_{t}}\int\limits_{0}^{t} K_{s}dB_{s}\rvert ^{1/4}]$ as $t \to 0$ Let $B$ be a standard Brownian motion starting in $0$ and $K$ be a continuous, adapted process starting in $0$ and such that $\sup\limits_{0\leq s \leq t}\lvert K_{s}\rvert < \infty$
I want to show that  $\mathbb E [\lvert \frac{1}{B_{t}}\int\limits_{0}^{t} K_{s}dB_{s}\rvert ^{1/4}]\to 0, \; t \to 0$
Any ideas?
I thought of using the Cauchy-Schwarz inequality so that eventually I get a term $$ \mathbb E [\lvert \frac{1}{B_{t}}\int\limits_{0}^{t} K_{s}dB_{s}\rvert ^{2}]$$
and perhaps then to use Ito's isometry but the problem is the $B_{t}$ in the denominator.
 A: You had a good idea. Let me present the solution under the assumption that $t \mapsto \mathbb E[K_t^2]$ is continuous at zero (which is the case if $\sup_{s \le t}K_s$ is a bounded random variable) Let $X_t = \frac{1}{B_t}$ and $Y_t = \int_0^t K_sdB_s$. Then by Holder inequality with exponents $p=8$ and $q = \frac{8}{7}$ we get $$ \mathbb E|X_tY_t|^{\frac{1}{4}}  \le \big( \mathbb E|X_t|^{\frac{2}{7}}\big)^{\frac{7}{8}} \cdot \big( \mathbb E|Y_t|^2\big)^{\frac{1}{8}}$$
Now, note that $X_t \sim \frac{1}{\sqrt{t}} \frac{1}{G}$ where $G \sim \mathcal N(0,1)$ hence $$\big(\mathbb E|X_t|^{\frac{2}{7}}\big)^{\frac{7}{8}} = \big(\frac{1}{t}\big)^{\frac{1}{8}} \big( \mathbb E|\frac{1}{G}|^{\frac{2}{7}}\big)^{\frac{7}{8}} = \frac{C}{\sqrt[8]{t}}$$ where $C$ is a finite constant (because function $x \mapsto \big(\frac{1}{|x|}\big)^{\frac{2}{7}}$ is integrable near $0$). As for the second expectation, simply apply Ito's-Isometry getting $$ \big( \mathbb E|Y_t|^2\big)^\frac{1}{8} = \Big( \int_0^t \mathbb E|K_s|^2 ds\Big)^{\frac{1}{8}} $$ Hence $$ \mathbb E|X_tY_t|^{\frac{1}{4}} \le C \Big( \frac{1}{t} \int_0^t \mathbb E[K_s^2]ds\Big)^{\frac{1}{8}} $$
Now, since $s \mapsto \mathbb E[K_s^2]$ is continuous due to our assumption, we get $\frac{1}{t}\int_0^t \mathbb E[K_s^2]ds \to \mathbb E[K_0^2] = 0$ which proves our result.
