No. Vector space structure is not enough, we actually need a compatible lattice structure to make things work. To apply the conditional expectation operator $E(\bullet\mid Y)$ onto the Hilbert space consisting of $L^2(\mathcal{X})$ random variables defined on the probability space $\mathcal{X}$, it requires us to consider following construction. A probability Hilbert space $H$(of $L^2(\mathcal{X})$ with $<x,y>=E(x-Ex)(y-Ey)$) if it satisfies following conditions
(1) it has unitary element $\boldsymbol{1}_{\mathcal{X}}$.
(2) it adopts a lattice structure compatible with the vector space
structure $(H,\ge)$ such that $x\vee\boldsymbol{1}=\boldsymbol{0}$ iff
$x=\boldsymbol{0}$
(3) For $x\ge \boldsymbol{0}$ and $y\ge \boldsymbol{0}$ $<x,y>\geq 0$.
equality holds iff $x\wedge y=\boldsymbol{0}$
A closed subspace $G$ of a probability Hilbert subspace $H$ is a closed probability space such that it contains $\boldsymbol{1}_G$ and closed under $\vee \wedge$ operations of the lattice. The probability span of $y\in H$ is defined to be the intersection of all probability Hilbert subspaces that contains ${y}$(which is like the way we define ideals/spectrum). Then $E(\bullet\mid y)$ is a projection operator onto the probability span of $y$.
[Small&McLeish] contains more details of construction of probability spans.
[Small&McLeish]Small, Christopher G., and Don L. McLeish. Hilbert space methods in probability and statistical inference. Vol. 920. John Wiley & Sons, 2011.