Question

Let V be a Hilbert space. Let S1 and S2 be two hyperplanes in V defined by

$S_1 = \left \{ x\in V |\left \langle a_1,x \right \rangle = b_1 \right \},S_2 = \left \{ x\in V |\left \langle a_2,x \right \rangle = b_2 \right \}$

Let $y\in V$ be given. We consider the projection of y onto , i.e., the solution of

(1)

(a) Prove that is a plane, i.e., if $x,z \in S_1 \cap S_2$ , then $(1+t)z - tx \in S_1 \cap S_2$ for any $t\in\mathbb{R}$.

(b) Prove that z is a solution of (1) if and only if $z \in S_1 \cap S_2$ and

$\left \langle z-y,z-x \right \rangle = 0,\forall x \in S_1 \cap S_2$ (2)

(c) Find an explicit solution of (1). (

d) Prove the solution found in part (c) is unique.

Sin S2 min TESInS2

Answer 1

a) We know $x,z\in S_1\cap S_2$ if and only if

$\langle a_1,x\rangle=b_1=\langle a_1,z\rangle,~~~~\langle a_2,x\rangle=b_2=\langle a_2,z\rangle$

Thus, for any $t\in\mathbb R$ we have

a1 .tx + (1 _ t)z)-t(a1·z) + (1 _ t)(a1·z) thị + (1 _ t)b1-b1

and

$\langle a_2,tx+(1-t)z\rangle=t\langle a_2,x\rangle+(1-t)\langle a_2,z\rangle=tb_2+(1-t)b_2=b_2$

This shows that $tx+(1-t)z\in S_1\cap S_2$ . Hence, is a plane.

Sin S2

b) Let be a solution. Then

$||z-y||=\min_{x\in S_1\cap S_2}||x-y||$

shows that it is in
Sin S2
because $\begin{align*}S_1,S_2\end{align*}$ are closed. (Alternatively, otherwise, the "line segment" $\begin{align*}tz+(1-t)y\end{align*}$ will intersect at a point which will make $\begin{align*}||x-y||\end{align*}$ smaller). Also, for any $x\in S_1\cap S_2$ let $\begin{align*}\alpha=\langle z-y,z-x\rangle\end{align*}$ . Suppose that $\begin{align*}\alpha\neq 0\end{align*}$ for some $\begin{align*}x\end{align*}$ in
Sin S2
, and without loss of generality let us assume $\begin{align*}||z-x||=1\end{align*}$ ; then we have

$\begin{align*}||y-(\alpha x+(1-\alpha)z||^2&=||(y-z)-\alpha(x-z)||^2\\ &=||y-z||^2+\alpha^2||x-z||^2-2\alpha\langle z-y,z-x\rangle\\ &=||y-z||^2+\alpha^2||x-z||^2-2\alpha^2\\ &=||y-z||^2-\alpha^2\\ &<||y-z||^2\end{align*}$

a contradiction to

$||z-y||=\min_{x\in S_1\cap S_2}||x-y||$

since $\begin{align*}\alpha=\langle z-y,z-x\rangle\end{align*}$ is zero.

Because this argument is reversible, we get the only if part via backtracking.