Operator norm of orthogonal projection 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?

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. $P=0$).
