# A differentiable version of the Michael selection theorem

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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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.)
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? – Ali Taghavi Mar 27 '14 at 0:31
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? – Ali Taghavi Mar 27 '14 at 20:59