Corollary 2.76 (Orthogonal decomposition). Let H be a Hilbert space,
and let $Y \subset H$ be a closed subspace. Then $Y^\bot $ is a closed subspace with
$$H = Y \oplus Y^\bot$$,
meaning that every element $h \in H$ can be written in the form
$$h = y + z$$
with $y \in Y$ and $z \in Y^\bot$ and $y$ and $z$ are unique with these properties
The uniqueness part is easily shown. But for the existance part it's said that we want to use the unique approximation within a closed convex set (i.e $\forall w\in V \exists ! v_o\in K \text{ s.t. } \|w-v_0\|=\inf_{k\in K}\|k-v_0\|$). The argumentation goes as follows:
Fix $h \in H$, and
apply Theorem 2.73 with $K = Y$ to find a point $y \in Y$ that is closest to $h$.
Let $z := h − y$, so that for any $v \in Y$ and any scalar $t$ we have (noting that the first inequality holds because $(tv+y)\in Y$ and is strict for all $t\ne0$ by the lemma and the second equality holds for the properties of the inner product)
$$\|z\|^2\le\|h-(tv+y)\|^2=\|z-tv\|=\|z\|^2-2\Re(t\langle v,z\rangle)+|t|^2\|v\|^2$$
(And now the problematic part)
However, this shows that
$$\Re(t\langle v,z\rangle)=0$$
for all scalars $t$ and $v \in Y$ , and so
$$\langle v,z\rangle=0$$
How should I interpret that "However"? Where should I use the fact that this particular $y$ is the unique closest in $Y$?
Many thanks in advance
Best Answer
This "however" has no reversal meaning to me. The details goes as follows:
For any $t \neq 0$, any $v \in Y \setminus \{0\}$, $\|z\|^2 \leq \|z\|^2 - 2t\Re(v,z) + |t|^2\|v\|^2$ and therefore
$$2\Re(t(v,z)) \leq |t|^2\|v\|^2$$
In particular take $t = \frac{\bar{(v,z)}}{\|v\|^2}$. We obtain
$$2\frac{|(v,z)|^2}{\|v\|^2} = 2\Re(\frac{|(v,z)|^2}{\|v\|^2}) \leq \frac{|(v,z)|^2}{\|v\|^4}\|v\|^2 = \frac{|(v,z)|^2}{\|v\|^2}$$
which implies $|(v,z)|^2 = 0$ and finally $(v,z) = 0$.