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Let $(X,\mathcal{F},\mu)$ be a probability space. Let $p \geq 2$.

Let $F$ be a finite set of functions of the form $f: X \rightarrow \mathbb{C}$, $\| f \|_{L^p(\mu)} \leq 1$. Let $(\Omega,\mathcal{G},\nu)$ be another probability space. Let $Y_1,\ldots,Y_n : \Omega \rightarrow F$ be i.i.d. random variables. Let $$ Y = \mathbb{E}Y_j = \int_{\Omega} Y_j d\nu = \sum_{f \in F}\mathbb{P}(Y_j=f) f . $$

In a paper I am reading, it is asserted that $$ \mathbb{E} \left( \int_{X} \left( \frac{1}{n}\sum_{j=1}^{n} |Y_j(x) - Y(x)|^2 \right)^{p/2} d\mu(x) \right) \leq C^p. $$

I can't figure out why this is true. Can someone help me out?

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We have $\| Y_j \|_{L^p(\mu)} \leq 1$ and $\| Y \|_{L^p(\mu)} \leq 1$. Then \begin{align*} \mathbb{E} \left( \int_{X} \left( \frac{1}{n}\sum_{j=1}^{n} |Y_j(x) - Y(x)|^2 \right)^{p/2} d\mu(x) \right) &= \mathbb{E} \left( \int_{X} \left( \frac{1}{n}\sum_{j=1}^{n} |Y_j(x) - Y(x)|^2 \right)^{p/2} d\mu(x) \right)^{\frac{2}{p} \cdot \frac{p}{2}} \\ &= \mathbb{E} \left( \left\| \frac{1}{n}\sum_{j=1}^{n} |Y_j(x) - Y(x)|^2 \right\|_{L^{p/2}(d\mu(x))} \right)^{\frac{p}{2}} \\ &\leq \mathbb{E} \left( \frac{1}{n}\sum_{j=1}^{n} \left\| |Y_j(x) - Y(x)|^2 \right\|_{L^{p/2}(d\mu(x))} \right)^{\frac{p}{2}} \\ &= \mathbb{E} \left( \frac{1}{n}\sum_{j=1}^{n} \left\| Y_j(x) - Y(x) \right\|_{L^{p}(d\mu(x))}^2 \right)^{\frac{p}{2}} \\ &\leq \mathbb{E} \left( \frac{1}{n}\sum_{j=1}^{n} \left(\left\| Y_j(x) \right\|_{L^{p}(d\mu(x))} + \left\| Y(x) \right\|_{L^{p}(d\mu(x))}\right)^2 \right)^{\frac{p}{2}} \\ &\leq \mathbb{E} \left( \frac{1}{n}\sum_{j=1}^{n} (1+1)^2 \right)^{\frac{p}{2}} \\ &\leq 2^p \end{align*}

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