Let $X$ be an Hilbert Space, $X=Y\bigoplus Z$ where $Y$, $Z$ are both closed subspaces.
Let $P:X \rightarrow X$ $P(y+z)= y$ be the canonical projection, then
$||P|| \leq 1 \implies Y=Z^{\bot}$
This should be quite standard but the proof provided by my lecture does not convince me, they argue in the following way, let $\zeta\in \mathbb{C} $ then $$||y+\zeta z||^2 = ||y||^2 + |\zeta|^2 ||z||^2+ \Re\langle y, \zeta z\rangle = ||P(y+\zeta z)||^2 + |\zeta|^2 ||z||^2+ \Re\langle y, \zeta z\rangle $$
So by the bound on the norm
$$|\zeta|^2 ||z||^2+ \Re\langle y, \zeta z\rangle \geq 0$$
Since this houlds for all $\zeta$ they conclude that $\langle y,z\rangle=0$
How does this follow?
Alternatively, could you indicate a valid proof?
Thanks
The idea is to divide the inequality $$ |\zeta|^2 \|z\|^2 + 2\Re \langle y ,\zeta z\rangle \ge0 $$ by $\zeta$ for suitable values of $\zeta$.
First, take $\zeta\in \mathbb R$, $\zeta> 0$, then divide by $\zeta$, which gives $$ |\zeta| \|z\|^2 + 2\Re \langle y , z\rangle \ge0, $$ then $\zeta\searrow 0$. Do the same for a negative $\zeta$. At the end, $\Re \langle y ,\zeta z\rangle=0$ is proved.
Then do the same for $\zeta = i\cdot t$, $t\ne 0$, $t\in \mathbb R$. This gives $\Im \langle y ,\zeta z\rangle=0$.