Let $X$ be a random variable such that takes values on a functional space of type $L^{2}$ ($L^{2}(I)$, that is, $L^{2}$ functions $f: I\rightarrow\mathbb{R}$, with $I$ a compact of $R^{p}$). Let $f,g\in L^{2}(I)$. My question is if \begin{equation}\nonumber \text{E}[\|X - f\|_{L^{2}}]\geq\|f - g\|_{L_{2}} \end{equation} or \begin{equation}\nonumber \text{E}[\|X - g\|_{L^{2}}]\geq\|f - g\|_{L_{2}} \end{equation} is true. It can be expressed also as \begin{equation}\nonumber \max\{\text{E}[\|X - f\|_{L^{2}}, \text{E}[\|X - g\|_{L^{2}}\}\geq\|f - g\|_{L^{2}}. \end{equation}
Thans in advance.
It is true with a constant $2$. $$ \begin{align*} \|f-g\|_{L^2} = \mathbb{E}_X(\|f-g\|_{L^2}) &≤ \mathbb{E}_X(\|f-X\|_{L^2} + \|X-g\|_{L^2}) \\ &\leq \mathbb{E}_X(\|f-X\|_{L^2}) + \mathbb{E}_X(\|X-g\|_{L^2}) \\ &\leq 2\max\left(\mathbb{E}_X(\|f-X\|_{L^2}),\mathbb{E}_X(\|X-g\|_{L^2})\right). \end{align*} $$ Is this sufficient for you?