Theorem 12.6.12.label Let $H$ be a Hilbert space over $K \in \RC$ and $A \subset H$ be a closed subspace, then:
- (1)
For each $x \in H$ and $y \in A$, then $\norm{x - y}_{H} = d(x, A)$ if and only if $x - y \perp A$.
- (2)
For any $x \in H$ and $y, z \in A$,
\[\norm{y - z}_{H}^{2} \le 2\norm{y - x}_{H}^{2} + 2\norm{z - x}_{H}^{2} - 4d(x, A)^{2}\] - (3)
For each $x \in H$, there exists a unique $x_{A} \in A$ such that $\norm{x - x_A}_{H} = d(x, A)$, and the mapping $P_{A}: H \to A$ with $x \mapsto x_{A}$ is an orthogonal projection.
- (4)
$H = A \oplus A^{\perp}$ as a direct sum of closed subspaces, where for every $x \in H$, $x = P_{A}x + P_{A^\perp}x$.
Proof, [Theorem 5.24, Fol99]. (1): Suppose that $\norm{x - y}_{H} = d(x, A)$. Let $z \in A$ and assume without loss of generality that $\dpn{y, z}{H}> 0$, then
is a quadratic function. Since $y - tz \in A$ for all $t \in \real$, it has a minimum at $t = 0$. Therefore $\dpn{(x - y), z}{H}= 0$, and $x - y \perp A$.
Now suppose that $x - y \perp A$, then for any $z \in A$,
by the Pythagorean Theorem.
(2): By the parallelogram law,
Since $(y + z)/2 \in A$,
(3): Let $\seq{y_n}\subset A$ with $\norm{y_n - x}_{H} \to d(x, A)$ as $n \to \infty$, $\seq{y_n}$ is a Cauchy sequence by (2). Since $H$ is complete and $A$ is a closed subspace, there exists $x_{A} \in A$ such that $y_{n} \to x_{A}$ as $n \to \infty$ by Proposition 6.7.3. In which case, by continuity of the norm, $\norm{x_A - x}_{H} = d(x, A)$. By (1), $x_{A}$ is the unique element of $A$ such that $\norm{x_A - x}_{H} = d(x, A)$.
Let $x, y \in H$ and $\lambda \in K$, then $(\lambda x + y) - (\lambda P_{A}x + P_{A}y) \perp A$, so by (1), $P_{A}(\lambda x + y) = \lambda P_{A}x + P_{A}y$, and $P_{A}$ is linear. In addition,
so $P_{A}$ is self-adjoint.
(4): By (1), for each $x \in H$, $P_{A^\perp}x = x - P_{A}x$. Since $A \cap A^{\perp} = \bracs{0}$, $H = A \oplus A^{\perp}$.$\square$
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