I'm trying to solve exercises from an older script but I got stuck on this one.
Let X be a $\mathbb C$-Banach Space and let $Y$, $Z$ be two closed subspaces of $X$ s.t. $Y \cap Z=\{0\}$ and $Y + Z =X$.
Prove:
1.) Any element $x$ in $X$ can be uniquely represented as $x=y+z$ where $y \in Y$, $z\in Z$.
2.) The product space $Y \times Z$ is a $\mathbb C$-Banach Space and the liner map $T: Y \times Z \to X$, $(y,z) \mapsto y+z$ is bijective and together with its inverse bounded.
3.) There are bounded linear maps $P: X \to Y$ and $Q: X \to Z$ so that the restrictions of P to Y and Q to Z are the identity operators and so that $ker(P)=Z$ and $ker(Q)=Y$.
I think for 2.) I have to use the Open Mapping Theorem. But for 1.) and 3.) I'm not sure how to even start. Thanks for helping!
1) Let $u\in X$, there exists $f(u)\in Y, g(u)\in Z$ such that $u=f(u)+g(v)$. Suppose that $u=u_1+u_2, u_1\in Y, u_2\in Z$. We have $u_1+u_2=f(u)+g(u)$. This implies that $u_1-f(u)=g(u)-u_2$. We have $u_1-f(u)\in Y$ and $u_2-g(u)\in Z$. This implies that $u_1-f(u), u_2-g(u)\in Y\cap Z=\{0\}$ We deduce that $u_1=f(u)$ and $u_2=g(u)$.
2) We endow $Y\times Z$ with the norm defined by $\|(y,z)\|=\|y\|+\|z\|$ Consider the map $h:Y\times Z\rightarrow X$ defined by $h(y,z)=y+z$. Let $(y_n,z_n)$ be a sequence which converges towards $(y,z)$. Since $\|(y_n,z_n)-(y,z)\|=\|(y_n-y,z_n-z)\|=\|y_n-y\|+\|z_n-z\|$, we deduce that have $lim_ny_n=y$ and $lim_nz_n=z$. This implies that $lim_n(y_n+z_n)=y+z=lim_nh(y_n,z_n)=h(y+z)$. We deduce that $h$ is linear and continue, since 1) implies that $h$ is surjective, the open mapping theorem implies that it is open. We deduce that the inverse of $h$ is continue.
3) Remark that the inverse of $h$ is the map defined by $l(u)=(f(u),g(u))$, since the projection $p_Y:Y\times Z\rightarrow Y$ is continue, we deduce that $p_Y\circ l$ and $p_Z\circ l$ are continue.