Consider some random variables $Y_1,\ldots, Y_n\in L^2(\Omega,\mathcal{F},P)$ and the inner product $$\langle X,Y\rangle_{L^2}=E[XY].$$ Moreover, let $\mathcal{L}^Y=\overline{\operatorname{span}}(Y_1,\ldots,Y_n)\subset L^2$ be the closure of the all linear combinations of $Y_1,\ldots,Y_n$. By the Hilbert projection theorem, for any $X\in L^2$ there exists a unique $P_Y^\bot X\in\mathcal{L}^Y$, that minimizes the distance to $\mathcal{L}^Y$, i.e. for any $Z\in\mathcal{L}^Y$ it holds $$E[(X-P_Y^\bot X)^2]=\lVert X-P_Y^\bot X\rVert_{L^2}^2\le \lVert X-Z\rVert_{L^2}^2=E[(X-Z)^2]$$
Something similiar is true for the conditional expectation $E[X|Y_1,\ldots, Y_n]$. It is $\sigma(Y_1,\ldots, Y_n)$-measurable and for any other $\sigma(Y_1,\ldots, Y_n)$-measurable random variable $Z$, it holds $$E[(X-E[X|Y_1,\ldots,Y_n])^2]\le E[(X-Z)^2].$$
with equality if and only if $Z=E[X|Y_1,\ldots,Y_n]$.
Now I am wondering, if these two random variables coincide?
I am tempted to say, that $P_Y^\bot X$ is a linear combination of $Y_1,\ldots, Y_n$, such that there is a Borel function $f$ with $P_Y^\bot X=f(Y_1,\ldots, Y_n)$ yielding that $P_Y^\bot$ is $\sigma(Y_1,\ldots, Y_n)$-measurable. This would imply
$$E[(X-E[X|Y_1,\ldots,Y_n])^2]\le E[(X-P_Y^\bot X)^2].$$
However, in reality $P_Y^\bot X$ only needs to sit in the closure of the $\operatorname{span}(Y_1,\ldots, Y_n)$. So I am not convinced yet.
The other way around, I am tempted to say, that any Borel function $f$ can be somehow represented or approximated by linear combinations of its arguments, which should then imply $E[X|Y_1,\ldots,Y_n]\in\mathcal{L}^Y$ and
$$ E[(X-P_Y^\bot X)^2] \le E[(X-E[X|Y_1,\ldots,Y_n])^2],$$
such that $ E[X|Y_1,\ldots,Y_n]=P^\bot_Y X$ with the inequality above.
But this argumentation is even less convincing to me.
Is my claim true? How should I prove it? Or do there exist any counter examples?
No, they are not equal. The span of $Y_1, \ldots , Y_n$ is a much smaller subspace than $L^2(\Omega, \sigma(Y_1, \ldots , Y_n), \mathbb{P})$; for instance, the former often doesn't contain $Y_1^2$.
For instance, take normalised Lebesgue measure on $[-1,1]$ and $Y(\omega) = \omega$ (so that $Y$ is a uniform random variable on $[-1,1]$) and $X = Y^2$. Then, $E(X|Y) = Y^2$, but $$\mathrm{proj}_{\mathrm{span}(Y)}(X) = Y\arg\min_a E[(Y^2 -aY)^2] = Y\frac{E(Y^3)}{E(Y^2)} =0\neq Y^2$$ where we minimised a quadratic polynomial in $a$ to perform the optimisation.