Let $X,Y,Z$ be Banach spaces and let $A : X \to Y$, $B : X \to Z$ be two continuous linear maps such that $\ker B \leq \ker A$

From the vector space homomorphism results I can factor $A$ as $ A = MB$ for $M$ a linear map from the image $im(B) \to A$.

When will this $M$ be continuous as well?

In constructing this $M$, it seems the problem would be solved if I can show that the inverse of the canonical map $\phi: X/\ker B \to im(B)$ was continuous.

Apparently this should be the case if $im(B)$ is a Banach space. I don't have this, but I am wondering if I can somehow use the information that the original map $A$ is continuous to help me or make up for this.

(The reason this $\phi$ seems crucial is that $M$ can be written as $\psi \circ i \circ \phi^{-1}$ where

$i : X/\ker B \to X/\ker A$ is the homomorphism that sends $x + \ker B$ to $x + \ker A$

$\psi$ is the canonical map $X/\ker A \to im(A)$

and the maps $i$ and $\psi$ can be shown to be continuous)


It seems that you are trying to circumvent the completeness hypotheses of the open mapping theorem. Unless you can provide stronger hypotheses, this cannot be avoided.

Consider the following counterexample: Take $X = Y = Z = C[0,1]$ equipped with the sup norm, $A:X\to Y$ the identity map, and $B:X\to Z$ given by $$ B(f)(x) = \int_0^x f(t)dt $$ Now suppose $M: \text{Im}(B) \to Y$ is such that $A = MB$, then $M$ must be the differential map, which is not continuous (Look at where the function $x\mapsto x^n$ must go)

  • $\begingroup$ OK thanks. just to check, if instead the middle space Z was equipped with the norm ||f|| = |f| + |Df| for the sup norm | |, then M would be continuous? $\endgroup$ – ttb Sep 23 '15 at 15:29
  • $\begingroup$ Yes, but by that argument, you could make any linear map continuous by redefining the norm. Not sure if that really helps though. $\endgroup$ – Prahlad Vaidyanathan Sep 23 '15 at 15:58

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