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Could someone give a simple example of surjective (not bijective) bounded operator in Banach spaces without bounded right-inverse?

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If you have in mind bijective bounded lienar operators, then the answer is no by the open mapping theorem. However, if you only ask for right-inverses to bounded linear surjections, then they need not exist.

Indeed, every separable Banach space $X$ is an image of a surjective linear map $T\colon \ell_1\to X$. If there is a right-inverse to $T$, that is, a map $S\colon X\to \ell_1$ such that $TS={\rm id}_X$, then $\ell_1$ would have a complemented subspace isomorphic to $X$. To see this, consider the operator $P = ST$. Then $$P^2 = STST=S\,{\rm id}_X\,T=ST=P,$$ so $P$ is a projection with range equal to the range of $S$. However, $S$ must be injective and have closed range, so it is an isomorphism onto its range. Every infinite-dimensional complemented subspace of $\ell_1$ is isomorphic to $\ell_1$, though.

For the sake of the argument, you may take $X=\ell_2$ as every operator $S\colon \ell_2\to \ell_1$ is compact and then you need not know about complemented subspaces of $\ell_1$.


The Bartle-Graves theorem says that you may always find a continuous (but not necessarily linear) right-inverse to a surjective, bounded linear operator.

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Let's take $c_o$. It's uncomplemented subspace of $l_\infty$. Let $T: l_\infty \to l_\infty/c_0$. Both spaces are Banach. T is bounded and surjective (not bijective). According to theorem from Existence of right inverse. if $T$ has a right inverse, $ker~T$, which is $c_0$ in our case, should be complemented, but it's not true. Hence, $T$ has no right inverse at all.

Actually we can take $T: X\to X/Y$, where $X$ is Banach space and $Y$ is closed uncomplemented subspace of X.

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    $\begingroup$ It is much harder to show that $c_0$ is not complemented in $\ell_\infty$ than to prove that every operator from $\ell_2\to \ell_1$ is compact. $\endgroup$ – Tomek Kania May 29 '17 at 21:42

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