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Let $X$ be a Banach space and let $U$ be a closed Banach subspace. The inclusion mapping $$ f : U \rightarrow X $$ induces a dual mapping $$ f' : X' \rightarrow U' $$

I am wondering about the existence of a bounded operator $$ g : U' \rightarrow X' $$ such that $f' \circ g = {\rm Id}_{U'}$. This is a generalized right-inverse. Does such an operator always exist?

  • Such a mapping $g$ would enable us to choose a representation of any element of $U'$ by an element of $X'$, in a manner that is linear and bounded.

  • When working with Hilbert spaces, the answer is positive. But this is not as clear for Banach spaces.

  • Note that I can easily construct a linear right inverse after a choice of basis. But it is unclear to me whether it can be bounded.

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  • $\begingroup$ I would think that $g$ exists if and only if $U$ is complemented. That $U$ is complemented is sufficient for the existence of $g$; but I don't see an obvious argument for the necessity, so I might be wrong. $\endgroup$ Commented Jun 24, 2024 at 12:31
  • $\begingroup$ @MartinArgerami It is not necessary. For example, $c_0 \subset \ell^\infty$ is not complemented, but such a $g$ does exist (it is simply the inclusion of $\ell^1$ into its double dual). More generally, if $X = U^{\ast\ast}$ and $f: U \to X$ is the canonical embedding, then this is always satisfied. $\endgroup$ Commented Jun 26, 2024 at 7:53
  • $\begingroup$ Nice! $ \ \ \ \ \ $ $\endgroup$ Commented Jun 26, 2024 at 21:52
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    $\begingroup$ Davie, Bull. London Math. Soc. 5 (1973), 261–266, uses ideas from Enflo's construction to show that $\ell^p$ has a subspace without the approximation property, hence without a Schauder basis. $\endgroup$ Commented Jun 27, 2024 at 3:47
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    $\begingroup$ @MartinArgerami Great! Thanks Martin. That solves it. I’ll write a community wiki answer. $\endgroup$ Commented Jun 27, 2024 at 9:52

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The following is a counterexample:

(In the following, by an isomorphism between Banach spaces, I mean a bounded bijective linear map, which is not necessarily isometric.)

Let $2 < p < \infty$ and $X = \ell^p$. Per a classical result by Davie (Remark (ii) in Davie, A.M. (1973). The approximation problem for Banach spaces. Bull. London Math. Soc., 5, 261-266.), $X$ admits an infinite-dimensional closed subspace $U$ which does not have the approximation property. In particular, $U \not\simeq \ell^p$.

Now, if such a $g$ exists, then as $f’$ is bounded and $f’ \circ g = \text{Id}_{U’}$, we see that $g$ is bounded below, so $g(U’)$ is a closed subspace of $X’$. It is complemented, since $g \circ f’$ is a projection from $X’$ onto $g(U’)$. Thus, $U’$ is isomorphic, via $g$, to a complemented subspace of $X’$, namely $g(U’)$.

Per another classical result of Pełczyński (Theorem 1 in Pełczyński, A. (1960). Projections in certain Banach spaces. Studia Mathematica, 19, 209-228.), any infinite-dimensional complemented subspace of $X’ = \ell^q$, where $\frac{1}{p} + \frac{1}{q} = 1$, must be isomorphic to $\ell^q$. In particular, $U’ \simeq \ell^q$. Since $\ell^q$ is reflexive, so is $U$, whence $U = U’’ \simeq \ell^p$, contradicting our assumptions on $U$. Thus, no such $g$ can exist.

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