Let $\mathcal H$ be a Hilbert space and $P$ be a linear operator on $\mathcal H.$ The the following statements are equivalent $:$
$(1)$ $P^2 = P = P^*.$
$(2)$ The space $\mathcal K = \{x\ : Px = x \}$ is a closed subspace of $\mathcal H$ and $\text {Ker}\ P = \mathcal K^{\perp}.$
$(3)$ $P^2 = P$ and $\|P\| \leq 1.$
I have managed to prove $(1) \implies (2)$ and $(1) \implies (3).$
For $(1) \implies (2)$ let us first take $x \in \text {Ker}\ P.$ Then for any $y \in \mathcal K$ we have $$\langle x, y \rangle = \langle x,Py \rangle = \langle Px, y \rangle = 0$$ which shows that $x \in \mathcal K^{\perp}$ and hence $\text {Ker}\ P \subseteq \mathcal K^{\perp}.$ Conversely, let $x \in \mathcal K^{\perp}.$ Then for any $y \in \mathcal K$ we have $\langle x,y \rangle = 0.$ But since $y \in \mathcal K$ we have $Py = y.$ Hence we have $$\langle Px,y \rangle = \langle x, Py \rangle = \langle x,y \rangle = 0,\ \ \text {for all}\ y \in \mathcal K.$$ Now since $P^2 = P$ it follows that $Px \in \mathcal K.$ So by putting $y = Px$ in the above equation we find that $\|Px\|^2 = 0 \implies Px = 0.$ Hence $x \in \text {Ker}\ P.$ Therefore $\mathcal K^{\perp} \subseteq \text {Ker}\ P.$
Combining this with the above inclusion we have $$\mathcal K^{\perp} = \text {Ker}\ P.$$
To prove that $\mathcal K$ is closed we first note that $$\langle x, Px \rangle = \langle x, P^2 x \rangle = \langle Px, Px \rangle = \|Px\|^2.$$ Hence by Cauchy-Schwarz inequality it will follow that $\|Px\| \leq \|x\|.$ Hence $P$ is a bounded linear operator with $\|P\| \leq 1.$ Now given a sequence $\{x_n\}_{n \geq 1}$ from $\mathcal K$ converging to $x \in \mathcal H$ we have $x_n=Px_n \to Px.$ Hence by the uniqueness of the limit it follows that $Px = x$ and hence $x \in \mathcal K.$ This proves that $\mathcal K$ is a closed subspace of $\mathcal H.$ This proves $(1) \implies (2).$
The last paragraph of the above proof also proves $(1) \implies (3)$ as a bi-product.
From here how do I prove the equivalence of the given three statements? Can anybody help me proving $(2) \implies (3)$ and $(3) \implies (1)\ $?
Thanks for reading.
The easier implications to prove are $(3) \Rightarrow (2)$ and $(2)\Rightarrow (1)$.
$(3)\Longrightarrow (2)$ Assume $P^2 = P$ and $\|P\|\leq 1$. The first condition means that $P$ is a projector : we have $\mathcal K = \ker(p-\text{id}) = \operatorname{im}(p)$ and $\operatorname{im}(p) \oplus \ker (p) = \mathcal H$. Since $p-\text{id}$ is bounded, its kernel $\mathcal K$ is a closed subspace of $\mathcal H$.
Let $x\in \mathcal K$ and $y\in \ker p$. Then, for any $\lambda \in \mathbb R$. \begin{align} \|x\|^2 &=\|p(x)\|^2 \\ &= \|p(x+\lambda y)\|^2 \\ &\leq \|x+\lambda y\|^2 \\ & \leq \|x\|^2 + 2\lambda\langle x,y\rangle + \lambda^2\|y\|^2\\ 0& \leq 2 \lambda \langle x,y\rangle + \lambda^2 \|y\|^2 \end{align} This can only be true for all $\lambda$ is $\langle x,y\rangle =0$. Therefore, $\ker p \subset \mathcal K^\perp$. Since they are closed and $\ker p \oplus \mathcal K = \mathcal H$, we get $\ker p = \mathcal K^\perp$
$(2)\Longrightarrow (1)$ Assume $\mathcal K$ is closed and $\mathcal K^\perp = \ker p$. In this case, we have $\ker p \oplus \mathcal K = \mathcal H$. For any $x\in\mathcal H$, we can write it uniquely as $x = y + z$ with $y\in\ker p$ and $z\in\mathcal K$. Then, $p(x) = p(y)+ p(z) = z$. We see that $p^2(x)= p(z) = z = p(x)$ ie $p^2 = p$.
Last, let $x = y+z$ and $x' = y' + z'$ be two vectors written like before. Then : \begin{align} \langle x,p(x')\rangle &= \langle y+z,z'\rangle \\ &= \langle z,z'\rangle \\ &= \langle z,y'+z'\rangle\\ &=\langle p(x),x'\rangle \end{align} so $p^* = p$.