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In Rynne & Youngson: Linear Functional Analysis, there is an exercise stated as

Let $\mathcal{H}$ be a complex Hilbert space and let $P$, $Q\in B(\mathcal{H})$ be orthogonal projections. Show that the following are equivalent:

  1. $\text{Im}\,P\subseteq \text{Im}\,Q$;
  2. $QP=P$;
  3. $PQ=P$;
  4. $||Px||\le||Qx||$ for all $x\in\mathcal{H}$;
  5. $P\le Q$.

The set $B(\mathcal{H})=B(\mathcal{H},\mathcal{H})$ denotes the set of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$, and $T\in B(\mathcal{H})$ is an orthogonal projection if $T$ is self-adjoint, and $T^2=T$. The set $\text{Im}\,T$ is the range of $T$, the subspace $T(\mathcal{H})\subseteq\mathcal{H}$.

Now, in order to attempt to show that $3.\Rightarrow 4.$, I use the fact that $P=PQ$, and find that \begin{align} ||Px||=||PQx||\le||P|| \,||Qx||. \end{align} The solution (given in the back of the book) now states that $||P||=1$, so $||Px||\le||Qx||$ follows. But as far as I can tell, the only thing we know about the norm of a general orthogonal projection is that $||P||\le1$, but not necessarily $||P||=1$. And if $||P||\not=1$, then $||Px||\le||Qx||$ doesn't necessarily hold.

Any help in pointing out what I'm missing here would be greatly appreciated!

Morten
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2 Answers2

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Let $V=P(\mathcal{H})$. If $P$ is not the zero projection, there is $x\in V$, $x\ne 0$. Then $Px=x$, implying $\|P\|=1$.

Julián Aguirre
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  • This is basically the same question as in the comments of the other answer, but: Does the existence of an element in $x\in V$, with $x\not=0$, guarantee the existance of an element in $V$ with $||x||=1$? Also; does it matter if there is such an element or not for the sake of the computation of the operator norm? – Morten Nov 25 '14 at 19:34
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    $x/|x|$ has norm 1. – Julián Aguirre Nov 25 '14 at 23:07
  • Yes, I figured out that $V$ is a linear subspace, so if $x\in V$, $x\not=0$, then $x/||x||\in V$. Thank you, sir. :) – Morten Nov 25 '14 at 23:10
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If $P = 0$, then there's nothing to prove, so assume that $P \neq 0$. Then you can find (how?) $0 \neq y \in \mathcal{H}$ such that $Py = y$. What would this imply for the value of $\|P\|$?

Branimir Ćaćić
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  • $Py=y$ holds for any $y\in\text{Im},P\subseteq\mathcal{H}$. And, the definition of the operator norm is (in the book) $||P||=\sup{||Px||:||x||\le 1}$, so this should imply trivially that $||P||=1$. But, using this defintion, do we need the existence of an element in $\text{Im},P$ which norm is $1$? Isn't it possible that none of the elements $x\in\mathcal{H}$ which yield non-zero $Px$ have norm $||x||\ge1$? What would this mean for the norm of $P$? Edit: Apologies if this is a silly question. – Morten Nov 25 '14 at 19:25
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    $H$ is a linear space, if $x\ne 0$ then $x/|x|$ has norm $1$. – daw Nov 25 '14 at 19:55