If a linear operator between two Banach spaces is surjective and bounded, can we get any information about a right inverse? For example, is it bounded?
Thanks, trying to understand trace operator stuff on the Sobolev spaces.
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If $T: X \to Y$ is bounded and surjective, the Open Mapping Theorem says there is an isomorphism $S:\; X/\ker(T) \to Y$ such that $T = S \circ \pi$, where $\pi: X \to X/\ker(T)$ is the quotient map. The trouble is, a quotient of $X$ might not be isomorphic to a closed subspace of $X$, so there might not be a bounded right inverse. For example, every separable Banach space is isomorphic to a quotient of $\ell^1$, but not every Banach space is isomorphic to a subspace of $\ell^1$ (e.g. $\ell^2$ is not).
MickG
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Robert Israel
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what if in this particular context, we know that a right inverse exists? I am currently trying to understand any right inverse of the Trace Operator $T:H^1(\Omega) \to H^{1/2}(\partial\Omega).$ – abe.nong Jan 14 '15 at 03:47
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1A right inverse exists: any surjective linear operator has a right inverse (which can be defined e.g. using a Hamel basis), but not necessarily a bounded one. – Robert Israel Jan 14 '15 at 06:11
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If I get it correctly, the open mapping theorem also yields the existence of a bounded right inverse (which may not be linear). Or do I muss something? – gerw Feb 14 '17 at 18:24
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I don't see how the open mapping theorem yields that. – Robert Israel Feb 15 '17 at 17:47
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@RobertIsrael: See my answer. – gerw Feb 15 '17 at 18:07
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Let $T : X \to Y$ be a surjection between Banach spaces $X$, $Y$.
You can get a linear right inverse, cf. Robert Isreal's comment.
You can get a bounded right inverse: The open mapping theorem yields $r > 0$ with $B_r^Y \supset T \, B_1^X$. Hence, for $y \in Y$ you find $x \in X$ with $T \, x = y$ and $\|x\| \le \|y\|/r$.
You cannot always get the existence of a linear and bounded right inverse. In Existence of right inverse. it is mentioned that this is true iff the kernel of $T$ is complemented in $X$.
