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Given any (finite or infinite dimensional) Banach space $X$, can we always find a continuous linear map $T:\ell_2\rightarrow X$ with $T(\ell_2)=X$ (where $\ell_2$ is the Hilbert space of square-summable sequences with the usual norm) ?

If $X$ is finite dimensional it is easy to define such a surjective continuous linear map. If $X$ is infinite dimensional we have three cases:

1- If $X$ has a Schauder basis I think we can define such a surjective continuous linear map by using a Schauder basis of $X$ (Can we do it?).

2- If $X$ is separable, perhaps one can use a countable dense subset of $X$, but I do not know how.

3- If $X$ is non-separable, I have no idea.

Can anybody give a hint (or an example, if any) to the above question in each case? Thanks in advance.

serenus
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    If such a map existed then $\ell^2/\ker T$ would be isomorphic to $X.$ Then $X$ would be isomorphic to a Hilbert space. This problem is described here chapter IV – Ryszard Szwarc May 03 '25 at 11:13
  • @Ryszard Szwarc Do you mean topological isomorphism, or algebraic isomorphism (i.e., just linear and bijective) ? It is well known that $\ell_2$ is algebraically isomorphic to $\ell_1$. – serenus May 03 '25 at 16:16
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    I mean topological isomorphism. By the Banach inverse mapping theorem the spaces $\ell^2/\ker T$ and $X$ are isomorphic through the mapping $$\tilde{T}(x+\ker T)=Tx$$ – Ryszard Szwarc May 03 '25 at 17:36

1 Answers1

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To elaborate on the comment by Ryszard Szwarc:

Let $X$ be a Banach space and $T: \ell_2 \to X$ a continuous linear and surjective map. Then $\ker T$ is closed since $T$ is continuous and the quotient space $\ell_2 / \ker T$ is again a Hilbert space (a general fact about quotients of Hilbert spaces by closed subspaces). Now the map $\tilde{T}: \ell_2 /\ker T \to X$ induced by $T$ is bijective and continuous by properties of the quotient space. By a corollary of the open mapping theorem $\tilde{T}$ is an isomorphism. Therefore $X$ is isomorphic to the Hilbert space $\ell_2 / \ker T$.

Let $X$ be a Banach space with a closed subspace $A$ such that there exist no topological complement of $A$ (that is there exists no closed subspace $V\subset X$ with $X\cong V\oplus A$). For the existence of such a space see here. Then $X$ can not be isomorphic to a Hilbert space since otherwise the orthogonal complement of $A$ (modulo the isomorphism) is a topological complement of $A$.

This answers the question with "No" at least in the case of separable (and general) Banach spaces (since $L^1(\mathbb{T})$ is separable and has a non-complemented subspace by the linked post). I am not sure about the Schauder basis case.

Edit: It seems this paper constructs a reflexive Banach space with Schauder basis that has a non-complemented subspace (although i have not checked the details).

jd27
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