Exercise 3.3.25 of Karatzas and Shreve This is the Exercise 3.25 of Karatzas and Shreve on page 163

Whith $W=\{W_t, \mathcal F_t; 0\leq t<\infty\}$ a standard, one-dimensional Brownian motion and $X$ a measurable, adapted process satisfying
  $$E\int_0^T|X_t|^{2m}dt<\infty$$
  for some real numbers $T>0$ and $m\geq1$, show that
  $$E\left|\int_0^TX_tdW_t\right|^{2m}\leq(m(2m-1))^mT^{m-1}E\int_0^T|X_t|^{2m}dt$$
  (Hint: Consider the martingale $\{M_t=\int_0^tX_sdW_s, \mathcal F_t; 0\leq t\leq T\}$, and apply Ito's rule to the submartingale $|M_t|^{2m}$.)

By the hint, using Ito's rule I get
$$E|M_T|^{2m}=m(2m-1)E\int_0^T|M_t|^{2m-2}dt$$
I don't know how to continue. I tried to use Holder's inequality, but failed.
Thanks very much!

First of all, your calculation is not correct. Itô's formula gives
$$\begin{align*} M_T^{2m} &= 2m \cdot \int_0^T M_t^{2m-1} \, dM_t + m \cdot (2m-1) \cdot \int_0^t M_t^{2m-2} \, d \langle M \rangle_t \\ \Rightarrow \mathbb{E}(M_T^{2m}) &= m \cdot (2m-1) \cdot \mathbb{E} \left( \int_0^T M_t^{2m-2} \cdot X_t^2 \, dt \right)\tag{1} \end{align*}$$
where we used in the second step that the stochastic integral is a martingale and $d\langle M \rangle_t = X_t^2 \, dt$. Note that $(M_t^{2m-2})_{t \geq 0}$ is a submartingale; therefore we find by the tower property
$$\mathbb{E}(M_t^{2m-2} \cdot X_t^2) \leq \mathbb{E}((M_T^{2m-2} \mid \mathcal{F}_t) \cdot X_t^2) = \mathbb{E}(M_T^{2m-2} \cdot X_t^2). \tag{2}$$
Hence, by Fubini's theorem,
$$\mathbb{E} \left( \int_0^T M_t^{2m-2} \cdot X_t^2 \, dt \right) \leq \mathbb{E} \left( M_T^{2m-2} \cdot \int_0^T X_t^2 \, dt \right).$$
By applying Hölder's inequality (for the conjugate coefficients $p=\frac{2m}{2m-2}$, $q=m$), we find
$$\mathbb{E} \left( \int_0^T M_t^{2m-2} \cdot X_t^2 \, dt \right) \leq \left[\mathbb{E} \left( M_T^{2m} \right) \right]^{1-\frac{1}{m}} \cdot \left[ \mathbb{E} \left( \int_0^T X_t^2 \, dt \right)^m \right]^{\frac{1}{m}}. \tag{3}$$
Combining $(1)$ and $(3)$ yields,
$$\bigg(\mathbb{E}(M_T^{2m}) \bigg)^{\frac{1}{m}} \leq m \cdot (2m-1) \cdot \left[ \mathbb{E} \left( \int_0^T X_t^2 \, dt \right)^m \right]^{\frac{1}{m}}.$$
Finally, by Jensen's inequality,
$$\left(\int_0^T X_t^2 \, dt \right)^m \leq T^{m-1} \cdot \int_0^T X_t^{2m} \, dt.$$
This finishes the proof.
