Bounded Linear "Change of Basis"

Question

Let $E$ be a separable Banach space and $x,y \in E$. Does there necessarily exist a bijective bounded linear operator $A\in B(E)$ such that $$ Ax =y, $$ such that $A^{-1}$ is also a bounded linear operator? I know this to be true in finite dimensions and on the separable Hilbert space, but what about in general?

Answer 1

If $x$ and $y$ are linearly dependent, then the problem is unsolvable (if exactly one of $x,y$ is zero) or trivial ($A$ is multiple of identity).

Assume that $x$ and $y$ are linearly independent. Define functionals $f,g\in X^*$ by $f(x)=1$, $f(y)=0$, $g(x)=0$, $g(y)=1$, and extension by Hahn-Banach.

Define projection operator $$ Pz:= f(z)x + g(z)y. $$ Define $$ Az := f(z)y + g(z)x + (I-P)(z)= z - (f(z)-g(z))(x-y). $$ The idea of the construction is: exchange $x$ an $y$ on the span of both, on $\ker f\cap \ker g$ act as identity.

Then $Ax=y$ and $Ay=x$. In addition $A^2=I$: We have $$ f(Az) = f(z) -(f(z)-g(z))=g(z), $$ $$ g(Az) = g(z) + (f(z)-g(z))=f(z). $$ Then $$ A^2(z) = Az + (f(z)-g(z))(x-y) = z. $$ This implies that $A$ is bijective. Obviously, $A$ is bounded.