Proof of $\|P\| \leq \|I-P\|$ for L2 norm in Hilbert space when $P$ is a projection

Question

Knowing that $P$ is a projection in Hilbert space $\mathcal{H} \rightarrow \mathcal{H}$, not necessarily orthogonal and that neither $\mathcal{R(P)}$ nor $\mathcal{N(P)}$ equals $\mathcal{H}$, how can we prove the following inequality:

$$\|P\| \leq \| I - P \| $$

for L2 norm using its supremum definition?

$$ ||A|| = \sup_{||x||=1}{||Ax||}$$

I ended up with: $$ \begin{align*}\ \|I-P\| &\geq \|I\| - \|P\| \\ \| I - P \| &\geq 1 - \|P\| \\ 0 \leq \| P \| &\leq 1. \end{align*} $$

Am I missing some obvious property?

Answer 1

[EDIT: This answer was posted (long) before OP changed the conditions to $P\neq I$ and $P\neq 0$.]

The claim is false. $P = I$ is a projection.

Moreover, $0\le\|P\|\le 1$ is only true for orthogonal projections. In fact, from $P^2 = P$ it follows that $\|P\| = \|P^2\|\le\|P\|^2$ and thus $\|P\|\ge 1$.

Answer 2

$$1=\|I\|=\|I-P+P\| \le \|I-P\|+\|P\|$$

If $P$ is a projection in a Hilbert Space, then using $x=(I-P)x+Px$ one sees that $\|x\|^2=\|(I-P)x\|^2+\|Px\|^2$. Taking supremum over all $x \in H$ with $\|x\|=1$, we see that $1=\|I-P\|+\|P\|$.

Like amsmath mentions, $P=I$ is a counterexample to your claim.