
> 
> Are the weak topologies of $\ell\_1$ and $L\_1$ homeomorphic?
> 
> 
> 


Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. And indeed, no linear homomorphism between weak topologies of non-isomorphic spaces exists as prevented by the Closed Graph Theorem. Note that the fact that $\ell\_1$ is a Schur space yet $L\_1$ is not is *a priori* not helpful, unless I am missing something trivial.


However, all infinite-dimensional, separable Banach spaces are homeomorphic. For weak topologies this is not so, as the weak topology of a reflexive space is $\sigma$-compact whereas this does not hold for $L\_1$, say. We could then ask:



> 
> Is there a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic?
> 
> 
> 


31.03.2022: As [noticed](https://mathoverflow.net/questions/419301/distinguishing-topologically-weak-topologies-of-banach-spaces#comment1076935_419301) by Jerzy, the weak topologies of $\ell\_1$ and $C[0,1]$ are not homeomorphic.



Distinguishing topologically weak topologies of Banach spaces
The weak topologies of the spaces $\ell\_1$ and $L\_1$ are not homeomorphic because of the following


**Theorem.** *Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has Shur's property, then $Y$ has Shur's property, too.*


Proof. For a topological space $T$ let $T\_s$ be the sequential coreflection of $T$, which is $T$ endowed with the topology consisting of sequentially open sets. A subset $U\subseteq T$ is *sequentially open* if for any convergent sequence $(x\_n)\_{n\in\omega}$ in $T$ with $\lim x\_n\in U$ there exists $m\in\omega$ such that $x\_n\in U$ for all $n\ge m$.


Assume that $h:X\_w\to Y\_w$ is a homeomorphism between the Banach spaces $X,Y$ endowed with the weak topologies. The homeomorphism $h$ remains a homeomorphism betweem the sequential coreflections $X\_s$ and $Y\_s$ of the spaces $X\_w$ and $Y\_w$, respectively. Since $X$ has Shur's property, the sequential coreflection $X\_s$ of $X\_w$ coincides with $X$. Since $X=X\_s$ is a Baire space, so is the space $Y\_s$. The Baireness of the space $Y\_s$ implies that the closed unit ball $B\_Y$ of $Y$ has non-empty interior in $Y\_s$. Let $y\_\*$ be an interior point of the set $B\_Y$ in $Y$. Assuming that $Y$ fails to have Shur's property, we can find a null sequence $(y\_n)\_{n\in \omega}$ in $Y\_w$ consisting of vectors of norm $\|y\_n\|=3$. Then $(y\_n+y\_\*)\_{n\in\omega}$ is a sequence in $Y\_s\setminus B$ that converges to $y\_\*$, which is not possible as $y\_\*$ is an interior point of $B$ in $Y\_s$. This contradiction implies that the Banach space $Y$ has Shur's property.




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The proof of the above theorem suggests the following characterization of the Shur property.


**Characterization.** A Banach space $X$ has Shur's property if and only if the sequential coreflection $X\_s$ of the weak topology of $X$ is a Baire space if and only if $X\_s=X$.


**Remark.** Concerning the second question, it can be shown that any reflexive separable infinite-dimensional Banach spaces endowed with the weak topologies are sequentially homeomorhic (= the sequential coreflections of their weak topologies are homeomorphic), see this [paper](https://arxiv.org/abs/1908.09115) for more information in this direction.


