# How can I prove that: $\|P_{1}x\|^2+\ldots+\|P_{n}x\|^2 \leq \|x\|^2$?

Let $X$ be a Hilbert space and $P_{j}\in B(X)$ a projector, for any $j=\overline{1,n}$. If $P=P_{1}+P_{2}+\ldots+P_{n}$ is projector, prove that for any $x \in X$:

$$\|P_{1}x\|^2+\ldots+\|P_{n}x\|^2 \leq \|x\|^2.$$

I tried to write $\|P_{i}x\|^2=\langle P_{i}x,P_{i}x\rangle=\langle x,P_{i}^{2}x\rangle \leq \|P_{i}^{2}\|\|x\|^{2}$, and then to sum but it is not anything concrete.

Thanks :)

-

$$\|P_1x\|^2+\cdots+\|P_nx\|^2=\|P_1x+\cdots+P_nx\|^2=\|(P_1+\cdots+P_n)x\|^2.$$

The result then follows because $\|P_1+\cdots+P_n\|\leq 1$. The first equation is true because $P_iP_j=0$ for $i\neq j$, from which it follows that $P_ix\perp P_jx$ for $i\neq j$. To see that $P_iP_j=0$ for $i\neq j$, you can apply the result at this question after adding an $(n+1)^\text{st}$ projection $P_{n+1}=I-(P_1+\cdots+P_n)$.

-
Thanks. One more question please: you wrote : $\|(P_1+\cdots+P_n)x\|^2$ and from here I can say: $\|(P_1+\cdots+P_n)x\|^2 \leq \|P_{1}+\ldots P_{n}\|^{2} \|x\|^{2}=\|P\|^{2}\|x\|^{2}=\|x\|^{2}$ because $\|P\|=1$? – Iuli Feb 5 '13 at 23:06
Yes, almost; really $\|P\|\leq 1$, with equality unless $P=0$. – Jonas Meyer Feb 5 '13 at 23:08
$P=0$ or $P \neq 0$? Thanks :) – Iuli Feb 5 '13 at 23:27
@Iuli: Equality unless $P=0$, meaning $\|P\|\leq 1$, and $\|P\|=1$ unless (i.e., with the only exception being) $P=0$. You're welcome. – Jonas Meyer Feb 6 '13 at 2:23