If $X$ is an inner product space, $P:X \rightarrow X$ is a bounded linear operator and $P^2=P$, then $\|P\|=1$ iff $Im(P) \perp Ker(P)$ and $P\neq0$

Question

I have only managed to show that $\langle P(x),y \rangle\leq \|x\| \|y\|$ for every $x \in X$ and $y \in Ker(P)$.
Also, by $\|P\|=1$, I mean $\|P(x)\| \leq \|x\|$ for $x \neq 0$.
I have no idea how to approach this.
Any hints or solutions would be greatly appreciated.

Answer 1

I only know Euclidean spaces so I will assume $X$ is an Euclidean space of dimension $n$. Therefore this is not a complete answer, I hope it can help you though... If it is too far from what you expect, let me know and I'll delete my answer.

As $P^2 = P$ we have $\mathrm{Ker}(P) \bigoplus \mathrm{Im}(P) = X$

If $\mathrm{Ker}(P)\perp \mathrm{Im}(P) $ and $P \neq 0$ then for all $x \in X$, we have $ (x-P(x),P(x)) \in \mathrm{Ker}(P)\times \mathrm{Im}(P) $ therefore,

$$\lVert x \rVert^2 = \lVert x-P(x) \rVert^2+\lVert P(x) \rVert^2 $$ by the Pythagorean theorem.

Hence,$\forall x \in X$, $\lVert P(x) \rVert \leq \lVert x \rVert $ therefore $\lVert P \rVert \leq 1$.

Furthermore, $P(P-Id)=0$ so $\sigma(P) \subset \{0,1\}$ and $P$ is diagonalizable. $P \neq 0$ so $\dim(\mathrm{Ker}(P)) < n$ and therefore $1$ is an eigenvalue of $P$ so we can find $x \in X$ such that $P(x) = x$. Therefore it proves that the upper bound $1$ is reached and $\lVert P \rVert = 1$

---

If $\lVert P \rVert =1$ then $\lVert P \rVert \leq 1$. Let $(x,y) \in \mathrm{Ker}(P)\times \mathrm{Im}(P) $ then forall $\lambda \in \mathbb{R}$, $P(\lambda x+y) = y$, therefore because $\lVert P \rVert \leq 1$ we have,

$$ \lVert y \rVert^2\leq \lVert y+\lambda x \rVert^2 $$

And,

$$\lVert y+\lambda x \rVert^2 = \lVert y\rVert^2+2\lambda \langle x,y\rangle+\lambda^2 \lVert x \rVert^2 $$

So $\varphi :\lambda\mapsto 2\lambda \langle x,y \rangle+\lambda^2 \lVert x \rVert^2$ is a positive polynomial function, so if $x \neq 0$ we have $\lVert x \rVert^2 \neq0$ therefore we have its discriminant $4\langle x,y \rangle^2 \leq 0$ so $\langle x,y \rangle =0$ therefore $x \perp y$.

It proves $\mathrm{Ker}(P)\perp \mathrm{Im}(P) $.

Answer 2

[Axel's original answer was basically already correct. Only one minor observation was needed to prove the stronger condition $\|P\|=1$ instead of $\|P\| \le 1$.]

---

Lemma. $P^2 = P$ implies $X = \text{Im}(P) \oplus \text{Ker}(P)$, i.e. every element of $X$ can be uniquely written as $x+y$ for some $x \in \text{Im}(P)$ and $y \in \text{Ker}(P)$.

Proof sketch. Show that $\dim(\text{Im}(P)) + \dim(\text{Ker}(P)) = \dim(X)$ and that $\text{Im}(P) \oplus \text{Ker}(P) = \{0\}$.

---

Suppose $\text{Im}(P) \perp \text{Ker}(P)$ and $P \ne 0$. For any $x \in X$, we have $x = Px + (x - Px)$, and $Px \in \text{Im}(P)$ and $x-Px \in \text{Ker}(P)$. Since $Px$ and $x-Px$ are orthogonal, $$\|x\|^2 = \|Px\|^2 + \|x-Px\|^2.$$

This shows $\|Px\| \le \|x\|$ for all $x \in X$, which implies $\|P\| \le 1$. To show $\|P\|=1$, note that if $x \ne 0$ is already in $\text{Im}(P)$ (the image contains a nonzero element because $P \ne 0$), then $Px=x$ so $\|Px\| = \|x\|$.

---

Suppose $\|P\|=1$. (Note that this automatically implies $P \ne 0$.)

Let $x \in \text{Ker}(P)$ and $y \in \text{Im}(P)$, both nonzero. Let $\lambda \in \mathbb{R}$. Since $P(y+\lambda x) = y$, the condition $\|P\| \le 1$ implies $$\|y\|^2 = \|P(y+\lambda x)\|^2 \le \|y + \lambda x\|^2 = \|y\|^2 + 2 \lambda \langle y, x \rangle + \lambda^2 \|x\|^2.$$

Thus for any $\lambda \in \mathbb{R}$ and any nonzero $x \in \text{Ker}(P)$ and $y \in \text{Im}(P)$ we must have $-2 \lambda \langle y, x \rangle \le \lambda^2 \|x\|^2$. If $\langle y, x \rangle \ne 0$, then we can choose $\lambda$ to violate this inequality. So we must have $\langle y, x \rangle = 0$.