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I want to prove the following inequality: $$E\left[\left|\int_{t0}^t f(s,\omega)dW_s \right|^{2n}\right]\le (t-t_0)^{n-1}[n(2n-1)]^n\int_{t0}^tE[|f(s,\omega)|^{2n}]ds$$

I have been trying with Ito-Isometry and Cauchy Schwarz Inequality to get the Expectation under the integral. Where does the factor $[n(2n-1)]^n$ come from? Applying Ito Formula:

$$E[\lvert\int_{t_0}^t f(s,\omega)dW_s\rvert^{2n}]=E[\int f^2(t,\omega)*n(2n-1)|\int f(s,\omega)dW_s|^{2n-2}dt]$$ Repeating the recursion n times we will get $$E[\lvert\int_{t0}^t f(s,\omega)dW_s \rvert^{2n}]= E[\int f^{2n}(t,\omega)*n(2n-1)*(n-1)*(2n-3)*.....*2 dt]\le E[\int f^{2n}(t,\omega)*n(2n-1)^n dt]\le n(2n-1)^nE[\int |f|^{2n}(t,\omega) dt]$$

Now how does the $(t-t_0)^{n-1}$ comes in?

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    Hint: Use Ito^'s formula for the Itô process $X_t := \int_{t_0}^t f(s) \, dW_s$ and $f(x) := x^{2n}$. – 2017-02-27
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    @saz but I will have 2 Integral, one of $X_t$ and one from dt arised from Itos formula – 2017-02-27
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    Yes, that's correct. The expectation value of the stochastic integral (with respect to $X_t$) vanishes. Just start doing the computations and add them to your question. – 2017-02-27
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    @saz Thanks! But now how can I get $(t-t0)^{n-1}$ in? – 2017-02-27
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    It's not that simple because you cannot iterate Itô's formula (the problem is that you get expressions of the form $f^2(t) \int_0^t X_s^{2n-4} \, dX_s$ and the expectation of this term is not necessarily $0$). Are you sure that there is no constant additional constant on the right-hand side of your inequality? – 2017-02-27
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    @saz I dont understand, every Iteration will leave me a expression with square of the diffusion coefficient and two time differentiation, i.e $f^2$ and a dt term. The dW term vanishes. I can iterate n times then the derivative will be a constant.( ($f^2)''->2$). Or did I miss something? – 2017-02-27
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    What I'm saying is that the dW term does **not** vanish. Next time you apply Itô's formula you get something like $\mathbb{E}(f^2 \int \dots dW)$ and this term does not vanish. – 2017-02-27
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    @saz ok I got your point. But I thought it would vanish becasue if I multiplying it out by the factors before then the $dW_t$ term becomes a mix term $dW_t*dt$ and to the Ito Table it would vanish. If it doesn't, how am I suppose to solve such $dW_t*dt$ integral? – 2017-02-27

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If we set $$X_t := \int_{t_0}^t f(s) \, dW_s$$ then Itô's formula shows

$$\mathbb{E} \left( \left| \int_{t_0}^t f(s) \, dW_s \right|^{2n} \right) = \mathbb{E}(|X_t|^{2n}) = n (2n-1) \mathbb{E} \left( \int_{t_0}^t |X_s|^{2n-2} f(s)^2 \, ds. \right) \tag{1}$$

Because of the additional factor $f(s)^2$ on the right-hand side we cannot simply iterate this argumentation (because then the stochastic integral doesn't vanish). Instead, we have to apply Hölder's inequality:

$$\mathbb{E}(f(s)^2 |X_s|^{2n-2}) \leq \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \bigg[ \mathbb{E}(|X_s|^{2n}) \bigg]^{1-1/n}.$$

Since $|X_s|^{2n}$ is a submartingale, we have $\mathbb{E}(|X_s|^{2n}) \leq \mathbb{E}(|X_t|^{2n})$ for all $s \in [t_0,t]$. Combining both inequalities with $(1)$, we get

$$\mathbb{E}(|X_t|^{2n}) \leq n(2n-1) \bigg[\mathbb{E}(|X_t|^{2n}) \bigg]^{1-1/n} \int_{t_0}^t \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \, ds.$$

Hence,

$$\mathbb{E}(|X_t|^{2n}) \leq (n(2n-1))^n \left( \int_{t_0}^t \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \, ds \right)^n.$$

Applying Jensen's inequality proves the assertion.

Remark: There is a similar inequality for the running maximum: $$\mathbb{E} \left( \sup_{t_0 \leq s \leq t} |X_s|^{2n} \right) \leq \left( \frac{2n}{2n-1} \right)^{2n} (t-t_0)^{n-1} (n(2n-1))^n \int_{t_0}^t \mathbb{E}(|f(s)|^{2n}) \, ds;$$ this is a direct consequence of the Burkholder-Davis-Gundy inequality.

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    This is a great answer, and I don't want to speak for @saz, but as I understand it the $t-t_0$ term will show up when you rescale the final Lebesgue integral to apply Jensen's? – 2017-02-27
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    @saz thanks! but each Iteration gives me an additional (1-1/n) power, i.e. $$[E(|X_t|^{2n})]^{1-1/n}^2$$ for second iteration and so on. How do I get it down to zero? – 2017-02-27
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    @Nadiels Thanks! I missed that – 2017-02-27
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    @quallenjäger What do you mean? There is no iteration... – 2017-02-28
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    @saz In the last step, how did you get rid of $E[(|X_t|^{2n}]^{1-1/n}$? Did you divide both side by this factor and then take the power n on both side? – 2017-02-28
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    @quallenjäger Yes, exactly. – 2017-02-28
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    @saz can I be sure that this number won't be zero? – 2017-02-28
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    @quallenjäger If $\mathbb{E}(|X_t|^{2n})=0$, then the inequality which you want to prove is trivially satisfied; therefore we can assume WLOG that $\mathbb{E}(|X_t|^{2n})>0$. – 2017-02-28
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    @saz Thank you for your patient and your time! – 2017-02-28
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    @quallenjäger You are welcome. – 2017-02-28