Let $H_1,H_2$ be Hilbert spaces. Suppose that $A\subseteq H_1$ y $B\subseteq H_2$ are closed, convex, and not empty. Let $H_1,H_2$ be Hilbert spaces. Suppose that $A\subseteq H_1$ and $B\subseteq H_2$ are closed, convex, and not empty. Show that:
$$P_{A\times B}\langle x,y\rangle= \langle P_A(x),P_B(y)\rangle$$
for every $\langle x,y\rangle \in H_1\times H_2.$ Note: here I am using the $\langle .,. \rangle$ to denote a pair of elements in the space $H_1\times H_2.$, also P is an orthogonal proyection in this hilbert space. $$ $$
P.D. I've already done a similar problem but I was working on the space $\mathbb{R}$ with $A=[a,b]$, $B=[c,d]$ and $$H_1\times H_2 = D =[a,b]\times[c,d]$$ where I had $$P_D\langle x,y\rangle =\langle P_{[a,b]}(x),P_{[c,d]}(y)\rangle.$$ In this case I used the Closest Point Theorem in a Hilbert Space and the fact that if $x\notin A$, then $P_A(x)\in \partial A$ (I proved this fact by the way). What I did was that I defined a function $L_D$ and I listed the cases where $\langle x,y\rangle \in D$ and where $\langle x,y\rangle \notin D$. I finished with 9 different cases and then to each individual case I applied $L^2_D$, after that, I got that $L_D=L^2_D$ which tells me that the operator $L_D$ is idempotent, then that told me that $L_D$ is actually an orthogonal projection, and with that, I finally got the equality I wanted. The problem here is that I do not have any intervals to do the same, and although I have A and B, I'm not sure how to proceed to prove this problem. I know I need to prove that the operator is idempotent, but I don't know how to reach an expression so that I can verify that property.
 A: I think that I got the answer so I will post it, if anyone shall need it in the future: $$ $$
In this proof I'm going to use the the Closest Point Theorem in a Hilbert Space and the next statement (let's call it Theorem 2):
$$ $$Let $ H $ be a Hilbert space, and $A \subset H$ where $A$ is a closed, convex and nonempty set, then the following statements are true:
$$\text{if}\; x \notin A\;  \text{then} \;  P_ {A} (x) \in \partial A\;  ,   d (x, A) = d (x, \partial A)\;\text{and}\;P_{A}(x)=P_{\partial A}(x).$$
Now that we know with what we are working with we can begin:
$$ $$ Let $L_D\langle x,y\rangle=\langle P_{A}(x),P_{B}(y)\rangle$ with $D=A\times B$  such that
$$P_{A}(x)=\left\{\begin{matrix}
        a_x & \text{si} & x \notin A\\
        x   & \text{si} & x \in A\\
        \end{matrix}\right.\;\;\;
 P_{B}(y)=\left\{\begin{matrix}
        b_y & \text{si} & y \notin B\\
        y   & \text{si} & y \in B\\
        \end{matrix}\right.
        $$
Where $ a_x $ and $ b_y $ are elements of the boundary of $ A $ and $ B $ and where $P_{B}$ and $P_{A}$ are orthogonal proyections. Now, using Closest Point Theorem in a Hilbert Space and Theorem 2, let's consider the following cases:
$$\langle x, y\rangle \in A\times B \Rightarrow L_D\langle x,y\rangle = \langle x,y\rangle \; \text{si}\; x\in A \wedge y\in B$$
\begin{align*}
  \langle x, y \rangle \notin A\times B \Rightarrow  L_D\langle x,y\rangle &= \langle a_x,b_y\rangle\; \text{si}\; x\notin A \wedge y\notin B\\
  L_D\langle x,y\rangle &= \langle x,b_y\rangle\; \text{si}\; x\in A \wedge y\notin B\\
  L_D\langle x,y\rangle &= \langle a_x,y\rangle\; \text{si}\; x\notin A \wedge y\in B\\
\end{align*}
Thus we have the following:
$$L_D\langle x,y\rangle=\left\{ \begin{array}{lcc}
         \langle a_x,b_y \rangle &   \text{si}  & x \notin A \wedge  y \notin B. \\
         \langle x,b_y \rangle &   \text{si}  & x \in A \wedge  y \notin B. \\
         \langle a_x,y \rangle &   \text{si}  & x \notin A \wedge  y \in B. \\
         \langle x,y \rangle &   \text{si}  & x \in A \wedge  y \in B. \\
         \end{array}
   \right.$$
Now let's apply the $ L_D $ operator to itself and see what happens:
$$L_D^2\langle x,y\rangle = \langle P_{A}(P_{A}(x)),P_{B}(P_{B}(y))\rangle$$
\begin{align*}
    L_D^2\langle x,y\rangle &= \langle P_{A}(P_{A}(x)),P_{B}(P_{B}(y))\rangle = \langle P_{A}(a_x),P_{B}(b_y)\rangle= \langle a_x,b_y\rangle\; \text{si}\; x\notin A \wedge y\notin B\\ &\text{Note that this is true by Theorem 2 (for the second equality)} \\ & \text {and by the Closest Point Theorem in a Hilbert Space (for the third equality)}\\
   L_D^2\langle x,y\rangle &= \langle P_{A}(P_{A}(x)),P_{B}(P_{B}(y))\rangle = \langle P_{A}(x),P_{B}(b_y)\rangle= \langle x,b_y\rangle\; \text{si}\; x\in A \wedge y\notin B\\ &\text{Note that this is true by Theorem 2} \\ & \text {and by the Closest Point Theorem in a Hilbert Space (for the second and third equality,} \\ & \text {where 2 is used for the input $ "y" $ and the Closest point Theorem for the input $ "x" $)} \\
  L_D^2\langle x,y\rangle &= \langle P_{A}(P_{A}(x)),P_{B}(P_{B}(y))\rangle = \langle P_{A}(a_x),P_{B}(y)\rangle= \langle a_x,y\rangle\; \text{si}\; x\notin A \wedge y\in B\\ &\text{Note that this is true by Theorem 2} \\ & \text {and by the Closest Point Theorem in a Hilbert Space (for the second and third equality,} \\ & \text {where 2 is used for the input $ "x" $ and the Closest point Theorem for the input $ "y" $)} \\
  L_D^2\langle x,y\rangle &= \langle P_{A}(P_{A}(x)),P_{B}(P_{B}(y))\rangle = \langle P_{A}(x),P_{B}(y)\rangle= \langle x,y\rangle\; \text{si}\; x\in A \wedge y\in B\\ &\text{Note that this is fulfilled by the Closest point Theorem} \\ & \text {(for the second and for the third equality)} \\
\end{align*}
With the above we can see that the following is true:
$$L_D^2\langle x,y\rangle = L_D\langle x,y\rangle  $$
In this way we ensure that the operator $ L_D $ is idempotent and by basic properties of orthogonal projections (the fact that an operator is an orthogonal projection if only if it is idempotent) we can ensure that $ L_D $ is an orthogonal projection and it is also unique .
$$\therefore P_{A\times B}\langle x,y \rangle =\langle P_{A}( x ),P_{B}(y)\rangle_\square$$
