Projector has Norm 1

Theorem
An idempotent operator $P$ is a projector on the Hilbert Space $H$ $P$ has  norm 1, i.e.,
 * $\displaystyle \left\Vert{P}\right\Vert \equiv \sup_{x \mathop \in H} \frac{\left\Vert{P x}\right\Vert}{\left\Vert x\right\Vert} = 1$.

Proof
For all $x \in \operatorname {Rng}(P)$
 * $\left \Vert{P \dfrac x { \left\Vert{x}\right\Vert} }\right\Vert = \dfrac {\left\Vert{x}\right\Vert} {\left\Vert{x}\right\Vert} = 1$

so $\left\Vert{P}\right\Vert \ge 1$.

It remains to show that this holds with equality $P$ is a projector.

First, suppose $P$ is a projector.

Let $\{ {p_1, p_2, \ldots}\}$ be an orthonormal basis for $\operatorname {Rng}(P)$, and let $\{ {q_1, q_2,\ldots}\}$ be an orthonormal basis for $\operatorname {Rng}(P)_\perp$.

Then for any $x\mathop\in H$, we can write choose scalars $\{ \alpha_1, \alpha_2, \ldots\}$ and $\{\beta_1, \beta_2, \ldots\}$ so that $x= \sum_{i=1}^\infty \alpha_ip_i+ \sum_{i=1}^\infty \beta_i q_i$.

Because the basis vectors are orthogonal, the Pythagorean theorem shows that $\left\Vert{x}\right\Vert^2 = \sum_{i=1}^\infty \left\vert{\alpha_i}\right\vert^2 + \left\vert{\beta_i}\right\vert^2$.

Then $\left\Vert{Px}\right\Vert$ can be expanded

Hence $\left\Vert{P}\right\Vert \le 1$.

Since it was already shown $\left\Vert{P}\right\Vert \ge 1$, it follows that $\left\Vert{P}\right\Vert = 1$.

Now suppose $P$ is not a projector.

Then there exists $x \in H$ so that $Px - x$ is not orthogonal to $\operatorname{Rng}(P)$.

By writing $x = p + q$ with $p \in \operatorname{Rng}(P)$ and $q \in \operatorname{Rng}(P)_\perp$, it follows that
 * $Px - x = P(p + q) - (p + q) = Pq - q$.

By rescaling $x$, we can assume $\left\Vert{q}\right\Vert=1$.

Since $q \in \operatorname{Rng}(P)_\perp$ but
 * $Pq - q = Px - x=\notin\operatorname{Rng}(P)_\perp$,

it follows that $Pq \neq 0$.

Let $\left\Vert{Pq}\right\Vert = c \ne 0$.

It will be shown that
 * $y = cq+ \dfrac 1 c Pq$

satisfies
 * $\dfrac {\left\Vert{Py}\right\Vert} {\left\Vert{y}\right\Vert} > 1$.

Notice first that

Now