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Let $\mathcal H$ be a Hilbert space. Let $P : \mathcal H \longrightarrow \mathcal H$ be a self-adjoint idempotent linear operator. Can $P$ be necessarily bounded?

I can't prove boundedness of $P$ from these assumptions. Can anybody please help me in this regard?

Thanks for your time.

Anil Bagchi.
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2 Answers2

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Here is an extended hint. You're given a lot of assumptions to work with here; idempotent, meaning $P^2 = P$, and self adjoint, meaning $\langle Pa, b\rangle = \langle a, Pb\rangle$ for any vectors $a$, $b$. Imagine combining these assumptions - say, consider $\langle P^2a, b\rangle$. On the one hand, this should be equal to $\langle Pa, b\rangle$ by the idempotence. On the other, it should also be equal to $\langle PPa, b\rangle = \langle Pa, Pb \rangle$ by the self-adjointness. If you choose $a = b$, then the latter can be related to $\| Pa \|$. Finally, relate the latter to the former, and try using the cauchy schwartz inequality. Hope this helps!

Rob
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  • So by Cauchy-Schwarz inequality we have $|Pa| \leq |a|,$ for all $a \in \mathcal H.$ Hence $|P| \leq 1.$ In fact if the range of $P$ is a non-zero linear subspace of $\mathcal H$ then by exploiting the idempotence of $P$ we have $|P| = 1.$ – Anil Bagchi. Feb 18 '21 at 08:38
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An alternative proof using that an idempotent operator $P \colon V \to V$ on any inner product space $V$ is self-adjoint if and only if $\operatorname{im} P \perp \ker P$: We always have $V = \operatorname{im} P \oplus \ker P$ when $P$ is idempotent, so if $u \in \operatorname{im} P$ and $v \in \ker P$, then

$$ \lVert P(u + v) \rVert^2 = \lVert u \rVert^2 \leq \lVert u \rVert^2 + \lVert v \rVert^2 = \lVert u + v \rVert^2, $$

since $u$ and $v$ are orthogonal.

Danny
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