Maximal inequality for the Itō integral Let $W$ be a Brownian motion and $X$ be a predictable process with $$\operatorname E\left[\int_0^t|X_s|^2\:{\rm d}s\right]<\infty\;\;\;\text{for all }t\ge0.$$ Now, let $p\ge2$.

How can we show that $$\operatorname E\left[\sup_{s\in[0,\:t]}\left|\int_0^sX_r\:{\rm d}W_r\right|^p\right]\le C\operatorname E\left[\left|\int_0^t\left|X_s\right|^2\:{\rm d}s\right|^{\frac p2}\right]\tag1$$ for some $C\ge0$.

Let $$q:=\frac p{p-1}.$$ Clearly, we somehow need to apply Doob's inequality and the Itō formula. Doob's inequality yields $$\operatorname E\left[\sup_{s\in[0,\:t]}\left|\int_0^sX_r\:{\rm d}W_r\right|^p\right]\le q^p\operatorname E\left[\left(\left|\int_0^tX_s\:{\rm d}W_s\right|^2\right)^{\frac p2}\right]\tag2.$$ How do we need to proceed from here?
 A: If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := \int_0^t X_s \, dW_s$$ and its quadratic variation $$\langle M \rangle_t = \int_0^t X_s^2 \, ds$$ are bounded processes. As $p > 2$ the mapping $x \mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives
$$|M_t|^p = p \int_0^t |M_s|^{p-1} dM_s + \frac{p(p-1)}{2} \int_0^t |M_s|^{p-2} \, d\langle M \rangle_s.$$
Since $M$ is bounded, the first term on the right-hand side is a martingale; thus
$$\mathbb{E}\left( \sup_{s \leq t} |M_s|^p \right) \leq q^p \mathbb{E}(|M_t|^p) = q^p \frac{p (p-1)}{2} \mathbb{E} \int_0^t |M_s|^{p-2} \, d\langle M \rangle_s$$
and so
$$\mathbb{E}\left( \sup_{s \leq t} |M_s|^p \right) \leq q^p \frac{p(p-1)}{2} \mathbb{E} \left( \langle M \rangle_t \sup_{s \leq t} |M_s|^{p-2} \right).$$
Now an application of Hölder's inequality yields
$$\mathbb{E}\left( \sup_{s \leq t} |M_s|^p \right) \leq q^p \frac{p(p-1)}{2}\left( \mathbb{E} \left[ \sup_{s \leq t} |M_s|^p \right] \right)^{1-2/p} (\mathbb{E}[\langle M \rangle_t^{p/2}]^{2/p}.$$
Hence,
$$\left(\mathbb{E}\left[ \sup_{s \leq t} |M_s|^p \right] \right)^{2/p}\leq q^p \frac{p(p-1)}{2}  (\mathbb{E}[\langle M \rangle_t^{p/2}]^{2/p}$$
and this proves the assertion.
Remark: The Burkholder-Davis-Gundy inequality states that $$\mathbb{E} \left( \sup_{s \leq t} \left| \int_0^s X_s \, dW_s \right|^p \right)$$ is comparable with $$\mathbb{E} \left( \left| \int_0^t X_s^2 \, ds \right|^{p/2} \right),$$
see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.
