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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 :) |
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$$\|P_1x\|^2+\cdots+\|P_nx\|^2=\|P_1x+\cdots+P_nx\|^2=\|(P_1+\cdots+P_n)x\|^2.$$ 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 $P_{n+1}=I-(P_1+\cdots+P_n)$. |
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