[proofplan] We prove the projection result directly in the [Hilbert space](/page/Hilbert%20Space) $L^2(\Omega,\mathcal F,\mathbb P)$. First we take a minimizing sequence in the closed linear subspace $\mathcal H$ and use the parallelogram identity to show that the sequence is Cauchy. Closedness of $\mathcal H$ gives a minimizer. We then prove uniqueness by applying the same convexity argument to two hypothetical minimizers, and finally derive the orthogonality condition by minimizing a one-variable quadratic along affine lines inside $\mathcal H$. The converse follows from the Pythagorean identity. [/proofplan] [step:Construct a minimizing sequence and prove that it is Cauchy] Define the distance from $Y$ to $\mathcal H$ by \begin{align*} d:=\inf_{Z\in\mathcal H}\|Y-Z\|_{L^2}. \end{align*} By the definition of the infimum, there exists a sequence $(Z_k)_{k=1}^\infty$ in $\mathcal H$ such that \begin{align*} \|Y-Z_k\|_{L^2}\to d. \end{align*} For $m,n\in\mathbb N$, the midpoint \begin{align*} M_{m,n}:=\frac{Z_m+Z_n}{2} \end{align*} belongs to $\mathcal H$ because $\mathcal H$ is a linear subspace. Hence \begin{align*} \|Y-M_{m,n}\|_{L^2}\ge d. \end{align*} Applying the parallelogram identity in $L^2(\Omega,\mathcal F,\mathbb P)$ to the two vectors $Y-Z_m$ and $Y-Z_n$ gives \begin{align*} \|Z_m-Z_n\|_{L^2}^2 &=2\|Y-Z_m\|_{L^2}^2+2\|Y-Z_n\|_{L^2}^2 -4\left\|Y-\frac{Z_m+Z_n}{2}\right\|_{L^2}^2 \\ &\le 2\|Y-Z_m\|_{L^2}^2+2\|Y-Z_n\|_{L^2}^2-4d^2. \end{align*} Since $\|Y-Z_k\|_{L^2}\to d$, the right-hand side tends to $0$ as $m,n\to\infty$. Therefore $(Z_k)_{k=1}^\infty$ is Cauchy in $L^2(\Omega,\mathcal F,\mathbb P)$. [guided] The quantity we want to minimize is the distance from $Y$ to the subspace $\mathcal H$, so we first name it: \begin{align*} d:=\inf_{Z\in\mathcal H}\|Y-Z\|_{L^2}. \end{align*} Because $d$ is an infimum, we may choose a sequence $(Z_k)_{k=1}^\infty$ in $\mathcal H$ whose distances to $Y$ approach $d$: \begin{align*} \|Y-Z_k\|_{L^2}\to d. \end{align*} The main point is to prove that the approximants $Z_k$ approach one another. For $m,n\in\mathbb N$, define the midpoint \begin{align*} M_{m,n}:=\frac{Z_m+Z_n}{2}. \end{align*} Since $\mathcal H$ is a linear subspace, it is closed under addition and scalar multiplication, so $M_{m,n}\in\mathcal H$. Therefore the definition of $d$ gives \begin{align*} \|Y-M_{m,n}\|_{L^2}\ge d. \end{align*} Now apply the parallelogram identity to the two vectors $Y-Z_m$ and $Y-Z_n$ in the Hilbert space $L^2(\Omega,\mathcal F,\mathbb P)$. Their difference is $Z_n-Z_m$, and their average is \begin{align*} \frac{(Y-Z_m)+(Y-Z_n)}{2}=Y-\frac{Z_m+Z_n}{2}. \end{align*} Thus \begin{align*} \|Z_m-Z_n\|_{L^2}^2 &=2\|Y-Z_m\|_{L^2}^2+2\|Y-Z_n\|_{L^2}^2 -4\left\|Y-\frac{Z_m+Z_n}{2}\right\|_{L^2}^2 \\ &\le 2\|Y-Z_m\|_{L^2}^2+2\|Y-Z_n\|_{L^2}^2-4d^2. \end{align*} The last inequality uses the midpoint estimate $\|Y-M_{m,n}\|_{L^2}\ge d$. Since both $\|Y-Z_m\|_{L^2}$ and $\|Y-Z_n\|_{L^2}$ tend to $d$, the upper bound tends to $0$ as $m,n\to\infty$. Hence $(Z_k)_{k=1}^\infty$ is Cauchy in $L^2(\Omega,\mathcal F,\mathbb P)$. [/guided] [/step] [step:Use completeness and closedness to obtain a minimizer] The space $L^2(\Omega,\mathcal F,\mathbb P)$ is complete, so there exists $P\in L^2(\Omega,\mathcal F,\mathbb P)$ such that \begin{align*} Z_k\to P \end{align*} in $L^2(\Omega,\mathcal F,\mathbb P)$. Since each $Z_k\in\mathcal H$ and $\mathcal H$ is closed in $L^2(\Omega,\mathcal F,\mathbb P)$, we have $P\in\mathcal H$. By the [reverse triangle inequality](/theorems/2300), \begin{align*} \left|\|Y-Z_k\|_{L^2}-\|Y-P\|_{L^2}\right| \le \|Z_k-P\|_{L^2}, \end{align*} so $\|Y-Z_k\|_{L^2}\to\|Y-P\|_{L^2}$. Since also $\|Y-Z_k\|_{L^2}\to d$, it follows that \begin{align*} \|Y-P\|_{L^2}=d. \end{align*} Thus $P$ is a minimizer in $\mathcal H$. [guided] The previous step showed that $(Z_k)_{k=1}^\infty$ is Cauchy in $L^2(\Omega,\mathcal F,\mathbb P)$. The space $L^2(\Omega,\mathcal F,\mathbb P)$ is a Hilbert space, hence complete, so there is an element $P\in L^2(\Omega,\mathcal F,\mathbb P)$ such that \begin{align*} Z_k\to P \end{align*} in the $L^2$ norm. We must still check that the limit belongs to the constraint set. This is exactly where closedness of $\mathcal H$ is used. Since every $Z_k$ lies in $\mathcal H$ and $\mathcal H$ is closed in $L^2(\Omega,\mathcal F,\mathbb P)$, the limit $P$ also lies in $\mathcal H$. It remains to prove that $P$ actually attains the infimum. The reverse triangle inequality gives \begin{align*} \left|\|Y-Z_k\|_{L^2}-\|Y-P\|_{L^2}\right| \le \|(Y-Z_k)-(Y-P)\|_{L^2} =\|Z_k-P\|_{L^2}. \end{align*} Since $\|Z_k-P\|_{L^2}\to 0$, we get \begin{align*} \|Y-Z_k\|_{L^2}\to\|Y-P\|_{L^2}. \end{align*} But the sequence was chosen so that $\|Y-Z_k\|_{L^2}\to d$. Therefore \begin{align*} \|Y-P\|_{L^2}=d, \end{align*} so $P$ is a minimizer in $\mathcal H$. [/guided] [/step] [step:Prove uniqueness of the minimizer by midpoint convexity] Suppose $P,Q\in\mathcal H$ both satisfy \begin{align*} \|Y-P\|_{L^2}=d, \qquad \|Y-Q\|_{L^2}=d. \end{align*} Since $\mathcal H$ is a linear subspace, \begin{align*} R:=\frac{P+Q}{2} \end{align*} belongs to $\mathcal H$, and therefore $\|Y-R\|_{L^2}\ge d$. Applying the parallelogram identity to $Y-P$ and $Y-Q$ gives \begin{align*} \|P-Q\|_{L^2}^2 &=2\|Y-P\|_{L^2}^2+2\|Y-Q\|_{L^2}^2 -4\left\|Y-\frac{P+Q}{2}\right\|_{L^2}^2 \\ &\le 2d^2+2d^2-4d^2 \\ &=0. \end{align*} Hence $\|P-Q\|_{L^2}=0$, so $P=Q$ in $L^2(\Omega,\mathcal F,\mathbb P)$. The minimizer is unique. Denote it by $P_{\mathcal H}Y$. [/step] [step:Derive the orthogonality condition from one-dimensional minimization] Let $P:=P_{\mathcal H}Y$. Fix $Z\in\mathcal H$ and define the function \begin{align*} \varphi:\mathbb R&\to\mathbb R \\ t&\mapsto \|Y-(P+tZ)\|_{L^2}^2. \end{align*} Since $P+tZ\in\mathcal H$ for every $t\in\mathbb R$ and $P$ minimizes the distance from $Y$ to $\mathcal H$, the function $\varphi$ has a minimum at $t=0$. Expanding the square using the $L^2$ inner product, \begin{align*} \varphi(t) &=\|Y-P-tZ\|_{L^2}^2 \\ &=\|Y-P\|_{L^2}^2-2t(Y-P,Z)_{L^2}+t^2\|Z\|_{L^2}^2. \end{align*} This quadratic polynomial has a minimum at $t=0$, so its derivative at $0$ is $0$: \begin{align*} \varphi'(0)=-2(Y-P,Z)_{L^2}=0. \end{align*} Therefore \begin{align*} (Y-P_{\mathcal H}Y,Z)_{L^2}=0 \end{align*} for every $Z\in\mathcal H$. [guided] Fix an arbitrary element $Z\in\mathcal H$, and write $P:=P_{\mathcal H}Y$ for the unique minimizer. To extract an orthogonality condition, we move away from $P$ in the direction $Z$. Define \begin{align*} \varphi:\mathbb R&\to\mathbb R \\ t&\mapsto \|Y-(P+tZ)\|_{L^2}^2. \end{align*} Because $\mathcal H$ is a linear subspace and both $P$ and $Z$ belong to $\mathcal H$, the point $P+tZ$ belongs to $\mathcal H$ for every $t\in\mathbb R$. Since $P$ minimizes the distance from $Y$ to $\mathcal H$, the function $\varphi$ has a minimum at $t=0$. Now compute $\varphi$ explicitly. Using bilinearity and symmetry of the real $L^2$ inner product, \begin{align*} \varphi(t) &=\|Y-P-tZ\|_{L^2}^2 \\ &=(Y-P-tZ,Y-P-tZ)_{L^2} \\ &=\|Y-P\|_{L^2}^2-2t(Y-P,Z)_{L^2}+t^2\|Z\|_{L^2}^2. \end{align*} This is a polynomial in $t$. Since it has a minimum at $t=0$, its derivative at $0$ must vanish: \begin{align*} \varphi'(0)=-2(Y-P,Z)_{L^2}=0. \end{align*} Therefore \begin{align*} (Y-P_{\mathcal H}Y,Z)_{L^2}=0. \end{align*} Because $Z\in\mathcal H$ was arbitrary, the orthogonality condition holds for every element of $\mathcal H$. [/guided] [/step] [step:Recover minimality and uniqueness from the orthogonality condition] Conversely, suppose $P\in\mathcal H$ satisfies \begin{align*} (Y-P,Z)_{L^2}=0 \end{align*} for every $Z\in\mathcal H$. For any $W\in\mathcal H$, the difference $P-W$ belongs to $\mathcal H$, so \begin{align*} (Y-P,P-W)_{L^2}=0. \end{align*} Using the decomposition \begin{align*} Y-W=(Y-P)+(P-W), \end{align*} we obtain \begin{align*} \|Y-W\|_{L^2}^2 &=\|(Y-P)+(P-W)\|_{L^2}^2 \\ &=\|Y-P\|_{L^2}^2+2(Y-P,P-W)_{L^2}+\|P-W\|_{L^2}^2 \\ &=\|Y-P\|_{L^2}^2+\|P-W\|_{L^2}^2 \\ &\ge \|Y-P\|_{L^2}^2. \end{align*} Thus $P$ minimizes the distance from $Y$ to $\mathcal H$. Since the minimizer has already been shown to be unique, $P=P_{\mathcal H}Y$. This proves both the characterization and the theorem. [/step]