# Can the inverse operator in Bartle-Graves theorem be linear?

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?

• the kernel of a continuous operator is closed, and if it also dense, it is the whole domain, that is, the operator is the null operator (and if it is also surjective, the target space is the null space). – Pietro Majer Oct 31 at 14:06
• I got it, thank you. Actually, the problem I just came across goes like that: assume we have a linear (doesn't need to be bounded and probably even can't in order not to be a null operator) operator $G$ between Hilbert spaces $Z$ and $U$. $Z$ is continuously embedded in another Hilbert space $X$. We know that $ker G$ is dense in $X$ and that $G$ is onto. The statement that is used in a proof I try to understand is "Since $G$ is onto, it has at least one bounded right inverse $H$ which is a linear operator from $Z$ to $U$". – Jarek Sarnowski Oct 31 at 19:25
• OK I see. In Hilbert spaces all subspaces splits, and you don't need selection theorems nor other assumptions for a bounded linear surjective $G$ : If $Z$ is a Hilbert space, $U$ is a Banach space, $G : Z\to U$ is a bounded linear surjective operator, and $V:= (\ker G)^\perp$ with inclusion map $j:V\hookrightarrow Z$, then $G_{|V}:V\to U$ is invertible by the open mapping theorem and a bounded linear right inverse to $G$ is $j(G_{|V})^{-1}$. If $G$ is linear surjective between Hilbert spaces, and not assumed to be continuous, then the problem seems more difficult and I should think... – Pietro Majer Oct 31 at 21:04

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 $$ST$$.