text: # 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?
  
  • Consider $P = 0$. That's the exception. May 16, 2014 at 22:27
  • Is that the only exception? May 16, 2014 at 22:29
  
  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$$).
  
  • If the projection is not orthogonal, the norm is greater than $1$. May 16, 2014 at 22:49
  • Indeed. I corrected the answer. May 18, 2014 at 21:16
  • @Berci When $P$ is an orthogonal projection. Please is it true that $\|Px\|=\|x\|$ for all $x\in H$? Aug 7, 2019 at 16:12
  • No. It's true for all $x\in im(P)$ only, when we have $Px=x$. For $x\notin im(P)$, we always have $\|Px\|<\|x\|$, basically by the Pythagorean theorem. Aug 7, 2019 at 17:35