In an euclidean vector space $E$ how can I prove that a projection $p$ is orthogonal if and only if $$||p(x)||\leq ||x||\quad \forall x\in E$$
I can prove that if $p$ is an orthogonal projection then we have the inequality but I'm stuck in the other implication. Any suggestion is welcome.
A property of all projections is that $P^2=P$.
Let $v=Pu-u$. Then $$ Pv=P(Pu-u)=P^2u-Pu=0 $$ Next, by assumption $$ \begin{align} \|Pu\|^2 &=\|P(Pu+tv)\|^2\\ &\le\|Pu+tv\|^2\\ &=\|Pu\|^2+2t\langle v,Pu\rangle+t^2\|v\|^2 \end{align} $$ The only way that this can be true for all $t$, is if $\langle v,Pu\rangle=0$. That is, $$ \langle Pu-u,Pu\rangle=0 $$ Therefore, $P$ is orthogonal.