Suppose $X$ and $Y$ are two infinite dimensional Banach spaces. What can we say about the set of all injective continuous linear operators between $X$ and $Y$? Is it always nonempty?
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4$\begingroup$ What if $X$ is a non-separable Hilbert space, and $Y$ is a separable one? $\endgroup$– erzCommented May 5, 2020 at 3:55
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4$\begingroup$ Or more generally, if the Hamel dimension of $X$ is larger than that of $Y$. Then there will be no injective linear operators at all, never mind continuous. $\endgroup$– Nate EldredgeCommented May 5, 2020 at 5:59
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$\begingroup$ The answers here show that your second question fails but I think that you have defined an interesting preorder on the family of Banach spaces: $X$ is dominated by $Y$ if such an injection exists. Call the corresponding equivalence relation “weak equivalence”. There is a number of easy pickings (stability properties, etc) that one can quickly deal with but also many interesting questions, in particular, on weak equivalence of specific spaces. I would start with: when is $C(K)$ dominated by (equivalent to) $C(L)$? There is a plethora of such questions for the classical Banach spaces. $\endgroup$– user131781Commented May 5, 2020 at 22:12
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2 Answers
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Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper
A. Avilés, P. Koszmider, A Banach space in which every injective operator is surjective. Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074
the authors constructed an infinitely dimensional Banach space $X$ such that if $T:X\to X$ is bounded and injective, then $T(X)=X$. Therefore if $Y$ is a subspace of $X$, then one cannot find an injective operator $T:X\to Y$.
answered May 5, 2020 at 17:27
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Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density.
Suppose that $X$ fails to have a strictly convex renorming. Thus, there is no injective operator $T$ from $X$ into any space $Y$ that is strictly convex, as if it were $\|x\|^\prime = \|x\| + \|Tx\|$ would be a strictly convex norm on $X$.
Spaces that do not have a strictly convex norm include
- $X = \ell_\infty^c(\Gamma)$, the space of all bounded scalar-valued functions on an uncountable set $\Gamma$ that have at most countable support (Day);
- $X = \ell_\infty / c_0$ (Bourgain).
In the latter case, you may even take $Y= \ell_\infty$.
answered May 5, 2020 at 19:14
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$\begingroup$ Is it absolute that there is an example where the density character of both spaces is $\aleph_1$? Are there examples where both spaces are reflexive and have the same density character? $\endgroup$ Commented May 5, 2020 at 21:46
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$\begingroup$ @BillJohnson, for reflexive it is even easier: take $X = \ell_p(\Gamma)$ and $Y=\ell_q(\Gamma)$ for $q>p$. By Pitt's theorem, every operator from $X$ to $Y$ is compact, hence it has separable range. But an injection will be a homeomorphic embedding of the weakly compact unit balls; a contradiction. $\endgroup$ Commented May 6, 2020 at 5:28
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1$\begingroup$ Duh, now. :) (Actually, I was thinking about not having an injective operator in either direction when I asked that stupid question.) $\endgroup$ Commented May 6, 2020 at 15:27