If $A$ is a bounded operator on a Banach space $X$, is there a topological decomposition $X=\mathcal N(A)\oplus_1X_2$?

Question

Let $X,Y$ be Banach spaces and $A\in\mathfrak L(X,Y)$.

Under which conditions can we show that

1. $X=\underbrace{\mathcal N(A)}_{=:\:X_1}\oplus_1\underbrace{(1-\pi)X}_{=:\:X_2}$ (topological direct sum) for some projection $\pi\in\mathfrak L(X)$;
2. $X_2$ is complete;
3. $\left.A\right|_{X_2}$ is bijective?

Note that if $X$ is a Hilbert space, all claims are easily established, since we can take $\pi$ as the orthogonal projection of $X$ onto $X_1$. In that case, $X_2=X_1^\perp$.

The problem with the Banach space case is that the annihilator $X_1^\perp$ is a space of functionals (which can be identified with vectors of $X$ by Riesz' representation theorem in the Hilbert space case).

Answer 1

Any closed subspace $X_1$ of $X$ is the null space for some bounded linear map, e.g. the quotient map $$X\to X/X_1.$$

So the question is really whether or not every closed subspace is complemented.

A famous Theorem by Lindenstrauss and Tzafriri [1] asserts that every Banach space which is not isomorphic to a Hilbert space has a non complemented closed subspace.

On the other hand, should there be a closed subspace $X_2$ such that $X$ is the algebraic direct sum of $N(A)$ and $X_2$, then the projections onto both $N(A)$ and $X_2$ are bounded, thanks to the closed graph Theorem, and the restriction of $A$ to $X_2$ is injective.

Finally, if $A$ is surjective, then $A|_{X_2}$ is bijective, hence also an isomorphism, by the open mapping Theorem.

[1] Lindenstrauss, J., Tzafriri, L. On the complemented subspaces problem. Israel J. Math. 9, 263–269 (1971).