An inequality for random variables taking values in a function space. 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?
 A: 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*}
