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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)

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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)

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  • $\begingroup$ 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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