$L^{2}$ inner product inequality.

Question

Let $X$ be a random variable which takes values on $L^{2}$ space, $L^{2}(K)$ (the set of $L^{2}$ functions $f:K\rightarrow\mathbb{R}$ with $K$ a compact subset of $\mathbb{R}^{p}$). Let $f_{0}\in L^{2}(K)$ be a function such that \begin{equation}\nonumber \text{E}[\|X - f_{0}\|_{L^{2}}^{2}] = \inf_{g\in L^{2}(K)}\text{E}[\|X - g\|_{L^{2}}^{2}]. \end{equation} This is the same to say that \begin{equation}\nonumber \text{E}[\langle X - f_{0},X - f_{0}\rangle_{L^{2}}] = \inf_{g\in L^{2}(K)}\text{E}[\langle X - g,X - g\rangle_{L^{2}}]. \end{equation} We also have that \begin{equation}\nonumber \text{E}[|X(k) - f_{0}(k)|] = \inf_{g\in L^{2}(K)}\text{E}[|X(k) - g(k)|], \end{equation} for all $k\in K$. My question is whether or not it is satisfied the following inequality \begin{equation}\nonumber \text{E}[\langle X - f_{0}, X - g\rangle_{L^{2}}]\leq 2\cdot\text{E}[\langle X - g, X - g\rangle_{L^{2}}], \end{equation} for all $g\in L^{2}(K)$.

Answer 1

By Cauchy-Schwarz in $L^2(K)$, we have $$ |\langle X-f_0, X-g\rangle| \leq \sqrt{\langle X- f_0, X-f_0\rangle}\sqrt{\langle X- g, X-g\rangle}, $$ which is a product of two real-valued random variables on our background space (call it $\Omega$). Thus, applying Cauchy-Schwarz to $L^2(\Omega)$, we get $$ |\mathbb{E}\langle X-f_0, X-g\rangle|\leq \sqrt{\mathbb{E}\langle X- f_0, X-f_0\rangle}\sqrt{\mathbb{E}\langle X- g, X-g\rangle}\leq \mathbb{E} \langle X-g,X-g\rangle, $$ by assumption, so it doesn't seem to me that you even need the factor $2$?