Bartle-Graves theorem states that a surjective continuous linear operator between Banach spaces has a right continuous inverse that doesn't have to be linear.


Is it possible that this inverse is linear under additional assumption that an operator has a dense kernel?



Can the inverse operator in Bartle-Graves theorem be linear?
In general the right inverse of a linear bounded operator $T:X\to Y$ in Bartle-Graves theorem needs not to be linear, and there can be no linear bounded right inverse. In this case, any right inverse $S:Y\to X$ is everywhere not Fréchet differentiable nor Gateaux differentiable, because if $S$ is differentiable at $x\_0$, by the chain rule $dS(x\_0)$ is a bounded linear right inverse to $T$.


Note however that $\tilde S:Y\to X$ defined by $\tilde S(x):=\frac{\|x\|}2\Big( S\big(\frac x{\|x\|}\big)-S\big(-\frac x{\|x\|}\big) \Big)$ is also a right inverse to $T$, and it is
$1$-homogeneous, $S(tx)=tS(x)$. I'm not aware of other possible improvements one can do on $S$.


*There can be no linear bounded right inverse*: Precisely, a bounded linear operator $T:X\to Y$ between Banach spaces is a left, resp. right (bounded, linear) inverse, iff it is surjective and its kernel splits, resp. it is injective and its range is closed and splits.


In fact, if $T:X\to Y$ and $S:Y\to X$ are bounded linear operators such that $TS=1\_Y$, then $Y=\ker T\oplus \text{ran}\, S,$ with linear projectors $I-ST=[T,S ]$ and $ST$.


