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I was assigned the following homework problem:

"Let $P: H \to H$ be bounded and linear. Assume it satisfies $P^2 = P$ and $P^\star = P$. Show $\|P\| \le 1$."

This isn't too hard to show: for any $v\in H$, $$\| Pv \|^2 = |\langle Pv, Pv \rangle| = | \langle v, Pv \rangle | \le \|v \| \cdot \|Pv\| \implies \frac{\|Pv\|}{\|v\|} \le 1 \implies \|P\|\le 1$$

However, I also noticed the following inequality: $$ \|Pv \| = \|P^2 v\| = \|P(Pv)\| \le \|P\| \cdot \|Pv \| \implies \|P \| \ge 1 $$

So $\|P\| = 1$. But every source I've checked only says $\|P\| \le 1$. Is that second inequality true? Why does it fail, if not?

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    $\begingroup$ Consider $P = 0$. That's the exception. $\endgroup$ May 16, 2014 at 22:27
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    $\begingroup$ Is that the only exception? $\endgroup$
    – Paul Hurst
    May 16, 2014 at 22:29

1 Answer 1

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Yes, if $P^2=P=P^*$, then $P$ is an orthogonal projection to a subspace $U$ of $H$.
(Prove that $H={\rm im\,}P\oplus\ker P\ $ and that $\ {\rm im\,}P\perp\ker P$.)
The elements of $U$ stay fixed under $P$, so $P$ must have norm $\ge 1$ -as you also proved- unless $U=\{0\}$ (i.e. $P=0$).

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    $\begingroup$ If the projection is not orthogonal, the norm is greater than $1$. $\endgroup$ May 16, 2014 at 22:49
  • $\begingroup$ Indeed. I corrected the answer. $\endgroup$
    – Berci
    May 18, 2014 at 21:16
  • $\begingroup$ @Berci When $P$ is an orthogonal projection. Please is it true that $\|Px\|=\|x\|$ for all $x\in H$? $\endgroup$
    – Student
    Aug 7, 2019 at 16:12
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    $\begingroup$ No. It's true for all $x\in im(P)$ only, when we have $Px=x$. For $x\notin im(P)$, we always have $\|Px\|<\|x\|$, basically by the Pythagorean theorem. $\endgroup$
    – Berci
    Aug 7, 2019 at 17:35

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