Let $X$ be a Banach space and $Y$ a closed subspace of $X$.

Does there exist a bounded linear operator $T \colon X \to X$ such that $\ker T=Y$ ?

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    $\begingroup$ With "on $X$" you mean that the range should be in $X$ as-well, right? $\endgroup$ – Sebastian Bechtel Jan 29 '17 at 20:29
  • $\begingroup$ All right. It means the operator with domaine $X$ and range in $X$. $\endgroup$ – Ali Farzaneh Jan 30 '17 at 2:13
  • $\begingroup$ Compose the quotient map with the inclusion? $\endgroup$ – Sebastian Bechtel Jan 30 '17 at 7:04
  • $\begingroup$ Thanks for your attention. When $Y$ is a complemented subspace it can be simply achieved. My problem is when $Y$ is not complemented. $\endgroup$ – Ali Farzaneh Jan 30 '17 at 8:29
  • $\begingroup$ @SebastianBechtel Which inclusion? There is no canonical one from $X/Y\to X$. $\endgroup$ – Tim B. Jan 30 '17 at 8:59

The answer is yes if $X$ is separable, but no in general. To cast the problem slightly differently, let us recall the fact that for any bounded linear operator $V:W\longrightarrow Z$ there is a factorisation $V=\tilde{V}Q$, where $\tilde{V}$ is one-to-one and $Q:W\longrightarrow W/\ker(V)$ is the quotient map (see, e.g., Theorem 1.7.13 of Megginson's book An Introduction to Banach Space Theory). It is therefore easy to see that OP's question is equivalent to the following question: Given a Banach space $X$ and $Y$ a closed subspace of $X$, does there exist a one-to-one bounded linear operator from $X/Y$ to $X$?

We first show the answer is affirmative when $X$ is separable. To this end let $X$ be a separable Banach space and $Y$ a closed subspace of $X$. To achieve the positive answer in the separable case we shall define one-to-one bounded linear operators $J_1$, $J_2$, $J_3$, $J_4$ and $J_5$ such that the composition $J_5J_4J_3J_2J_1$ is well-defined and a bounded linear operator from $X/Y$ to $X$. Since every separable Banach space is (isometrically) isomorphic to a subspace of $C[0,1]$ (see, e.g., Theorem 5.17 in the book Functional Analysis and Infinite Dimensional Geometry by Fabian et al), there exists a linear isometry $J_1$ from $X/Y$ into $C[0,1]$. Let $J_2: C[0,1]\longrightarrow L_2[0,1]$ be the formal identity map and let $J_3:L_2[0,1]\longrightarrow \ell_2$ be an isometric isomorphism (note that such an operator $J_3$ exists since $L_2[0,1]$ and $\ell_2$ are both infinite dimensional separable Hilbert spaces). Let $J_4$ be the operator from $\ell_2$ to $\ell_1$ that maps each element $(a_n)_{n=1}^\infty$ of $\ell_2$ to the element $(2^{-n}a_n)_{n=1}^\infty$ of $\ell_1$. Let $(x_n)_{n=1}^\infty$ be a bounded Schauder basic sequence in $X$ (see, e.g., Theorem 6.14 of Fabian et al) and let $J_5: \ell_1\longrightarrow X$ be the unique bounded linear operator from $\ell_1$ to $X$ satisfying $J_5((a_n)_{n=1}^\infty) = \sum_{n=1}^\infty a_nx_n$ for each $(a_n)_{n=1}^\infty\in\ell_1$, noting that $J_5$ is one-to-one by the defining property of Schauder basic sequences. As all of the operators $J_1$, $J_2$, $J_3$, $J_4$ and $J_5$ are bounded, linear and one-to-one, and since the composition $J_5J_4J_3J_2J_1$ is well-defined, the proof of the affirmative answer in the separable case is complete.

For a counterexample in the general case, let $\Gamma$ be an uncountable set and let $R:\ell_1(\Gamma)\longrightarrow c_0(\Gamma)$ be a surjective bounded linear operator (for the existence of such an operator $R$, see Exercise 5.46 in Fabian et al). Let $Q:\ell_1(\Gamma)\longrightarrow \ell_1(\Gamma)/\ker(R)$ denote the quotient map and let $\tilde{R}: \ell_1(\Gamma)/\ker(R)\longrightarrow c_0(\Gamma)$ be one-to-one and such that $R=\tilde{R}Q$ (see the reference in the first paragraph above). Let $\iota$ denote the formal identity map from $\ell_1(\Gamma)$ to $\ell_2(\Gamma)$, which is linear, one-to-one and bounded (with $\Vert\iota\Vert=1$).

Since $\tilde{R}$ is one-to-one and onto, the Open Mapping Theorem implies that $\tilde{R}^{-1}$ is a well-defined, linear, one-to-one and bounded operator from $c_0(\Gamma)$ onto $\ell_1(\Gamma)/\ker(R)$. If there were to exist a one-to-one bounded linear operator $S: \ell_1(\Gamma)/\ker(R)\longrightarrow \ell_1(\Gamma)$, then $\iota S\tilde{R}^{-1}$ would be a one-to-one bounded linear operator from $c_0(\Gamma)$ into the Hilbert space $\ell_2(\Gamma)$. However, as noted in this answer, an argument due to Olagunju shows that there is no one-to-one bounded linear operator from $c_0(\Gamma)$ into any Hilbert space, therefore no such operator $S$ can exist. In particular, we have that a counterexample exists when $X=\ell_1(\Gamma)$, for $\Gamma$ an uncountable set.

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  • $\begingroup$ Thanks @Philip Brooker for your answer. I'm going to use this answer in my article. Please, if it's possible, introduce me the reference of this answer in order to refer to it in my article. $\endgroup$ – Ali Farzaneh Mar 13 '18 at 6:53
  • $\begingroup$ @AliFarzaneh, when I wrote this answer I did not use a reference for the first part (about a positive answer for separability) because I did not know of one - I just made up the answer. However, when I came to read your comment I looked again at my answer and noticed immediately that the construction I give in the first part of my answer actually shows that for any separable Banach spaces $Z$ and $X$ there is a one-to-one continuous linear operator from $Z$ to $X$. So I did a Google search for this more general result just now and have found a reference for it (actually for a stronger result): $\endgroup$ – Philip Brooker Mar 14 '18 at 2:54
  • $\begingroup$ In Proposition 3.1 of Cross, Ostrovskii and Shevchik, Operator ranges in Banach spaces, I, Math. Nachr. 173 (1995), p.91--114 they show that for any separable Banach spaces $Z$ and $X$ there exists a continuous linear operator from $Z$ to $X$ that is one-to-one, nuclear and having dense range. The result I provide in my answer above gives one-to-one and nuclearity, but not dense range. $\endgroup$ – Philip Brooker Mar 14 '18 at 3:00
  • $\begingroup$ I forgot to mention above that I do not know a reference for the second part of my answer (the counterexample for the general case). If I find one I'll let you know. $\endgroup$ – Philip Brooker Mar 14 '18 at 3:19

Recently, there was written the following paper on this topic:


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