A closed subspace is complemented exactly when it is the range of a bounded projection

Assume $M$ is complemented, and write $x=m+n$ with $m\in M$, $n\in N$ as in [L1]. Let $P:=P_M$. Then $P(x)=m$, so $P^2(x)=P(m)=m=P(x)$ and $\mathrm{ran}(P)=M$. Thus a complemented subspace is the range of a bounded projection.

Conversely, assume $P^2=P$ and $\mathrm{ran}(P)=M$. For every $x\in X$,

The kernel $\ker P$ is a closed linear subspace of $X$. It is linear because $P$ is linear by [L2]. If $x_n\in \ker P$ and $x_n\to x$, let $C$ be a bound for $P$ from [L2]. Then $\mathrm{\parallel }Px\mathrm{\parallel }=\mathrm{\parallel }P(x-x_n)\mathrm{\parallel }\le C\mathrm{\parallel }x-x_n\mathrm{\parallel }\to 0$, so $Px=0$ and $x\in \ker P$.

The sum in step 1.2 is direct: if $z\in M\cap \ker P$, then $z=Pw$ for some $w$ and also $Pz=0$, so $z=Pz=P^{2}w=0$. Therefore $X=M\oplus \ker P$.

The coordinate projections for the direct sum $X=M\oplus \ker P$ are $P$ and $I-P$. The first is bounded by hypothesis, and the second is bounded because $\mathrm{\parallel }(I-P)x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }Px\mathrm{\parallel }$ for every $x$. Together with steps 2.1 and 1.3, this is exactly the complemented-subspace condition of [L1].

Steps 1.1 and 3.1 prove the equivalence.