tl;dr: Is it possible that the best approximation to a nonnegative function of three variables with a bivariate function is no better than the best univariate function?
Let $w$ be a density on $\mathbb{R}^n$ and denote the corresponding weighted-$L^2$ space by $L^2_w(\mathbb{R}^n)$. Define $M=M(\mathbb{R}^n)\subset L^2_w(\mathbb{R}^n)$ to be the subset of all Lipschitz continuous and nonnegative functions $f:\mathbb{R}^n\to[0,\infty)$. Suppose we have $f\in M(\mathbb{R}^3)$ that is not independent of $x_1$ or $x_2$ (we assume nothing about the dependence on $x_3$). Now consider the $L^2_w$ projection of $f$ onto the following subsets of $M$: $M_1=$ functions that are independent of $(x_2,x_3)$, and $M_2=$ functions that are independent of $x_3$. Obviously, $\inf_{g\in M_2}\Vert f-g\Vert_{2,w}^2\le\inf_{g\in M_1}\Vert f-g\Vert_{2,w}^2$.
Since $f$ genuinely depends on both $x_1$ and $x_2$ it should be that this is in fact a strict inequality: $\inf_{g\in M_2}\Vert f-g\Vert_{2,w}^2 < \inf_{g\in M_2}\Vert f-g\Vert_{2,w}^2$. If $f$ was independent of $x_3$, we can prove this by taking $f\in M_2$ and appealing to continuity, but when $f$ depends on $x_3$, this argument breaks.
Is there a counterexample, or a simple proof of this fact? Does something similar work for general $L^p_w$ spaces with $p\ne 2$?
