Projection theorem doubt. Why $\operatorname{Re}\langle z,l\rangle=0$?

Question

Theorem: Let $H$ be a Hilbert space, L a closed subspace then for any $x\in H$, there exists a unique projection.

Let $x\in H$, and $x\notin L\:\:\exists y\in L$ such that $\lVert x-y\rVert=d>0$.

Proof: By the definition of projection $\langle x-y,l\rangle=0\:\:\forall l\in L$. Consider $t\in\mathbb{R}$, then $\lVert x-(y+tl)\rVert^2\geqslant\lVert x-y\rVert^2=d^2$.

Defining $x-y=z$, we have:

$\lVert z-tl\rVert^2=\langle z-tl,z-tl\rangle=d^2-t\langle z, l\rangle-t\langle l, z\rangle+t^2\lVert l^2\rVert=d^2+t^2\lVert l\rVert^2-2t \operatorname{Re}\langle z,l\rangle$

$\lVert z-tl\rVert^2\geqslant d^2\implies d^2+t^2\lVert l\rVert^2-2t \operatorname{Re}\langle z,l\rangle\geqslant d^2\implies t^2\lVert l\rVert^2-2t \operatorname{Re}\langle z,l\rangle\geqslant 0$

The last inequality forces $\operatorname{Re}(\langle z,l\rangle)=0$.

Question:

Why $\operatorname{Re}\langle z,l\rangle=0\mkern1mu$? I have been thinking about this and I cannot understand it.

Why use the real part?

Thanks in advance!

Answer 1

Suppose $$ 2t \Re \langle z,l \rangle\le t^2 l^2. $$

• If $t>0$, then $$ 2 \Re \langle z,l \rangle\le t l^2 $$ for every $t$, hence as $t\to 0^{+}$ we have $\Re \langle z,l \rangle\le0$.
• If $t<0$, then $$ 2 \Re \langle z,l \rangle\ge t l^2 $$ so for $t \to 0^{-}$ we obtain that $\Re \langle z,l \rangle\ge0$

From these two inequalities you obtain $\Re \langle z,l \rangle=0$

Answer 2

Let $a=\|l\|^2$ and let $b=\operatorname{Re}\langle z,l\rangle$. You know that$$(\forall t\in\mathbb{R}):at^2-2bt\geqslant 0.\tag1$$If $a=0$, it is trivial that it follows from $(1)$ that $b=0$. Suppose now that $a\neq0$. If $b\neq0$, then the equation $at^2-2bt=0$ has two roots: $0$ and $\frac{2b}a$. So, $at^2-2bt<0$ when $t$ is between the roots. But, from $(1)$, we know that that can't take place. So, $b=0$. But $b=\operatorname{Re}\langle z,l\rangle$.