[proofplan] We minimize the $L^2$ distance from $Y$ to the closed subspace $H$. A minimizing sequence is shown to be Cauchy by applying the parallelogram identity to two near-minimizers and observing that their midpoint is still in $H$. Completeness of $L^2$ and closedness of $H$ give the minimizer. Orthogonality follows by varying the minimizer along each direction in $H$, and uniqueness follows because the difference of two minimizers is both in $H$ and orthogonal to itself. [/proofplan] [step:Define the distance from $Y$ to $H$ and choose a minimizing sequence] Define the $L^2$ norm $\|\cdot\|_{L^2}: L^2(\Omega,\mathcal F,\mathbb P)\to [0,\infty)$ by \begin{align*} \|X\|_{L^2} := \mathbb E[|X|^2]^{1/2}. \end{align*} Set \begin{align*} d := \inf_{Z\in H}\|Y-Z\|_{L^2}. \end{align*} Since $0\le d<\infty$, by the definition of the infimum there exists a sequence $(Z_n)_{n\in\mathbb N}$ in $H$ such that \begin{align*} \|Y-Z_n\|_{L^2}^2 \le d^2+\frac{1}{n} \end{align*} for every $n\in\mathbb N$. [/step] [step:Use the parallelogram identity to prove the minimizing sequence is Cauchy] For $m,n\in\mathbb N$, define \begin{align*} X_m := Y-Z_m, \qquad X_n := Y-Z_n. \end{align*} The parallelogram identity in the real inner product space $L^2(\Omega,\mathcal F,\mathbb P)$ gives \begin{align*} \|X_m-X_n\|_{L^2}^2 = 2\|X_m\|_{L^2}^2 + 2\|X_n\|_{L^2}^2 - 4\left\|\frac{X_m+X_n}{2}\right\|_{L^2}^2. \end{align*} Because $H$ is a linear subspace, the midpoint \begin{align*} \frac{Z_m+Z_n}{2} \end{align*} belongs to $H$. Also, \begin{align*} \frac{X_m+X_n}{2} = Y-\frac{Z_m+Z_n}{2}. \end{align*} Therefore, by the definition of $d$, \begin{align*} \left\|\frac{X_m+X_n}{2}\right\|_{L^2} \ge d. \end{align*} Using the minimizing bound for $Z_m$ and $Z_n$, we obtain \begin{align*} \|Z_m-Z_n\|_{L^2}^2 &= \|X_n-X_m\|_{L^2}^2 \\ &\le 2\left(d^2+\frac{1}{m}\right) + 2\left(d^2+\frac{1}{n}\right) - 4d^2 \\ &= \frac{2}{m}+\frac{2}{n}. \end{align*} Hence $(Z_n)_{n\in\mathbb N}$ is Cauchy in $L^2(\Omega,\mathcal F,\mathbb P)$. [guided] The key point is that two near-minimizers cannot be far apart. To make this quantitative, define \begin{align*} X_m := Y-Z_m, \qquad X_n := Y-Z_n. \end{align*} These are the two residuals. The parallelogram identity gives the exact relation \begin{align*} \|X_m-X_n\|_{L^2}^2 = 2\|X_m\|_{L^2}^2 + 2\|X_n\|_{L^2}^2 - 4\left\|\frac{X_m+X_n}{2}\right\|_{L^2}^2. \end{align*} The last term is where convexity of the linear subspace enters. Since $H$ is a real linear subspace, \begin{align*} \frac{Z_m+Z_n}{2}\in H. \end{align*} Moreover, \begin{align*} \frac{X_m+X_n}{2} = \frac{Y-Z_m+Y-Z_n}{2} = Y-\frac{Z_m+Z_n}{2}. \end{align*} Thus the average residual is the residual associated to an admissible element of $H$. By the definition of the distance $d$ from $Y$ to $H$, \begin{align*} \left\|\frac{X_m+X_n}{2}\right\|_{L^2} = \left\|Y-\frac{Z_m+Z_n}{2}\right\|_{L^2} \ge d. \end{align*} Substituting this lower bound into the parallelogram identity and using \begin{align*} \|X_m\|_{L^2}^2 \le d^2+\frac{1}{m}, \qquad \|X_n\|_{L^2}^2 \le d^2+\frac{1}{n}, \end{align*} we get \begin{align*} \|Z_m-Z_n\|_{L^2}^2 &= \|X_n-X_m\|_{L^2}^2 \\ &\le 2\left(d^2+\frac{1}{m}\right) + 2\left(d^2+\frac{1}{n}\right) - 4d^2 \\ &= \frac{2}{m}+\frac{2}{n}. \end{align*} The right-hand side tends to $0$ as $m,n\to\infty$, so $(Z_n)_{n\in\mathbb N}$ is Cauchy in $L^2(\Omega,\mathcal F,\mathbb P)$. [/guided] [/step] [step:Pass to the limit inside the closed subspace] Since $L^2(\Omega,\mathcal F,\mathbb P)$ is complete, there exists $P\in L^2(\Omega,\mathcal F,\mathbb P)$ such that \begin{align*} Z_n \to P \end{align*} in $L^2(\Omega,\mathcal F,\mathbb P)$. Since each $Z_n$ belongs to $H$ and $H$ is closed in the $L^2$ norm, we have $P\in H$. The reverse triangle inequality gives \begin{align*} \left|\|Y-Z_n\|_{L^2}-\|Y-P\|_{L^2}\right| \le \|Z_n-P\|_{L^2}. \end{align*} Letting $n\to\infty$ yields \begin{align*} \|Y-P\|_{L^2}=d. \end{align*} Therefore \begin{align*} \mathbb E[|Y-P|^2] = d^2 = \inf_{Z\in H}\mathbb E[|Y-Z|^2]. \end{align*} [/step] [step:Vary the minimizer along each direction in $H$ to obtain orthogonality] Let $W\in H$ be arbitrary, and define the residual \begin{align*} R := Y-P. \end{align*} For every $a\in\mathbb R$, the element $P+aW$ belongs to $H$. Since $P$ minimizes the squared distance, the function \begin{align*} \varphi:\mathbb R &\to \mathbb R \\ a &\mapsto \|Y-(P+aW)\|_{L^2}^2 \end{align*} has a minimum at $a=0$. Expanding the square gives \begin{align*} \varphi(a) &= \|R-aW\|_{L^2}^2 \\ &= \mathbb E[|R-aW|^2] \\ &= \mathbb E[|R|^2]-2a\,\mathbb E[RW]+a^2\mathbb E[|W|^2]. \end{align*} The product $RW$ is integrable because \begin{align*} |RW|\le \frac{1}{2}|R|^2+\frac{1}{2}|W|^2 \end{align*} and $R,W\in L^2(\Omega,\mathcal F,\mathbb P)$. Since the quadratic polynomial $\varphi$ is minimized at $a=0$, its linear coefficient must vanish. Hence \begin{align*} \mathbb E[RW]=0. \end{align*} Because $W\in H$ was arbitrary, \begin{align*} \mathbb E[(Y-P)W]=0 \end{align*} for every $W\in H$. [guided] Fix an arbitrary direction $W\in H$. We want to show that the residual $R:=Y-P$ has zero inner product with this direction. Since $H$ is a real linear subspace and $P,W\in H$, every perturbation \begin{align*} P+aW \end{align*} belongs to $H$ for every $a\in\mathbb R$. Thus the minimizing property of $P$ says that the real-valued function \begin{align*} \varphi:\mathbb R &\to \mathbb R \\ a &\mapsto \|Y-(P+aW)\|_{L^2}^2 \end{align*} has a minimum at $a=0$. Now compute this function exactly. Since $R=Y-P$, \begin{align*} \varphi(a) &= \|R-aW\|_{L^2}^2 \\ &= \mathbb E[|R-aW|^2] \\ &= \mathbb E[|R|^2]-2a\,\mathbb E[RW]+a^2\mathbb E[|W|^2]. \end{align*} The expectation $\mathbb E[RW]$ is finite because the elementary inequality \begin{align*} |RW|\le \frac{1}{2}|R|^2+\frac{1}{2}|W|^2 \end{align*} and the facts $R,W\in L^2(\Omega,\mathcal F,\mathbb P)$ imply $RW\in L^1(\Omega,\mathcal F,\mathbb P)$. A real quadratic polynomial minimized at $a=0$ must have zero linear coefficient. Equivalently, if $\mathbb E[RW]\neq 0$, choosing $a$ with the same sign as $\mathbb E[RW]$ and sufficiently small would make \begin{align*} -2a\,\mathbb E[RW]+a^2\mathbb E[|W|^2] <0, \end{align*} contradicting the minimality of $a=0$. Hence \begin{align*} \mathbb E[RW]=0. \end{align*} Since the direction $W\in H$ was arbitrary, the residual satisfies \begin{align*} \mathbb E[(Y-P)W]=0 \end{align*} for every $W\in H$. [/guided] [/step] [step:Use orthogonality of residuals to prove uniqueness] Suppose $P_1,P_2\in H$ both minimize the squared distance from $Y$ to $H$. Applying the previous orthogonality argument to each minimizer gives \begin{align*} \mathbb E[(Y-P_1)W]=0 \qquad\text{and}\qquad \mathbb E[(Y-P_2)W]=0 \end{align*} for every $W\in H$. Take \begin{align*} W:=P_1-P_2\in H. \end{align*} Subtracting the two orthogonality identities gives \begin{align*} 0 &= \mathbb E[(Y-P_1)W]-\mathbb E[(Y-P_2)W] \\ &= \mathbb E[(P_2-P_1)(P_1-P_2)] \\ &= -\mathbb E[|P_1-P_2|^2]. \end{align*} Thus \begin{align*} \mathbb E[|P_1-P_2|^2]=0, \end{align*} so $P_1=P_2$ as elements of $L^2(\Omega,\mathcal F,\mathbb P)$. Therefore the minimizer is unique. Denoting it by $P_HY$, the existence, minimizing property, and orthogonality assertion are all proved. [/step]