Operator norm of orthogonal projection

I was assigned the following homework problem:

"Let $P:\mathcal{H} \to \mathcal{H}$ 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 $v\in\mathcal{H}$, $$\| Pv \|^2 = |\langle Pv, Pv \rangle| = | \langle v, Pv \rangle | \le \|v \| \cdot \|Pv\| \implies \frac{\|Pv\|}{\|v\|} \le 1 \implies \|P\|\le 1$$

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?

Solution 1:

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