[Math] Norm of orthogonal projection operator $P$, if $\text{Im}\,P\subseteq\text{Im}\,Q$, with $Q$ also an orthogonal projection

functional-analysis

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!

Best Answer

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

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