Assume that $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded surjective linear map. Is there a Gateaux differentiable function $g:Y\to X$ such that $T\circ g=Id_{Y}$?
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asked Mar 26, 2014 at 20:42
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1 Answer
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No. Consider the case of a surjective bounded linear operator $T:X\to Y$ which is not a (top-linear) left inverse (that is, $\operatorname{ker}(T)$ does not split in $X$). However by classical selection theorems a surjective bounded linear operator $T$ has a continuous right inverse $g$, even $1$-homogeneous; but it can be G-differentiable at no point, otherwise differentiating you would get a bounded linear right inverse to $T$.
Rmk. I refer to the standard definition of the Gâteaux differential of $f:X \to Y$ at $x$, that is, a bounded linear operator $L$ such that for all $v\in X$ there holds $\frac{d}{dt}f(x+tv)\Big|_{t=0}=Lv$. In particular the chain rule holds. (Warning: some adopt a weaker definition, where $L$ is not even linear.)
answered Mar 26, 2014 at 21:44
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$\begingroup$ I can understand from your answer if $T$ is not a left inverse(there is no bounded operator $S:Y\to X$ with TS=id) then there is no a Frechete differentiable map $g$ with $T\circ g=Id$. (using chain rulle) But I do not understand why your argument work for Gateau differentiability? $\endgroup$ Commented Mar 27, 2014 at 0:31
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$\begingroup$ The chain rule still holds with the Gâteaux differential, unless you mean a weaker less standard definition (like e.g. in the wiki article). $\endgroup$ Commented Mar 27, 2014 at 10:40
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1$\begingroup$ In the literuture is there a property "P", stronger than continuity, such that every right inverse "g" to a bounded surjective operator "T" on a banach space $X$, satisfies "P"? Is it true to say that a banach space $X$ is Hilbertable if every bounded surjective map on $X$ has a Frechete diferentiable righth inverse? $\endgroup$ Commented Mar 27, 2014 at 20:59
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1$\begingroup$ Question 1: nice question. I do not know the answer, but once you have a right inverse, I guess many other wild right inverses could possibly be made. The variant of the question with "s.t. at least one right inverse" instead of "s.t. every right inverse" seems more promising (e.g.: does it exist a Hoelder continuous right inverse?) $\endgroup$ Commented Mar 28, 2014 at 11:44
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3$\begingroup$ @PietroMajer I think there may be a Kalton paper (Collect. Math?) on that theme, following from his paper with Godefroy on when there is a Lipschitz right inverse $\endgroup$ Commented Mar 28, 2014 at 12:43