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Any information about the following questions would be welcome.

  1. I wonder whether there are (well-known or easy) closed and infinite dimensional subspaces of $L_p([0,1])$ ($1<p<\infty$) whose unit ball is compact for the topology of convergence in measure.

  2. If they do exist, can they be described or characterized? (Do they linearly embed into $\ell_p$ or into $\ell_2$?)

Note that, in the case $p=1$, such subspaces of $L_1([0,1])$ were already considered in the literature. For instance, in a paper called "On subspaces of $L^1$ which embed into $\ell_1$", G. Godefroy, N.J. Kalton and D.Li obtained a description of the subspaces whose unit ball is compact and locally convex in measure (Thereom 3.3 and Corollary 3.5 therein). In their words: "Corollary 3.5 somehow means that the subspaces of $L^1$ whose unit ball is i $\tau_m$-compact locally convex are close to the trivial ones, that is, to w$^*$-closed subspaces of copies of $\ell_1$ generated in $L^1$ by a sequence of disjoint indicator functions."
However, there exist subspaces of $L^1$ whose unit ball is compact but not locally convex in measure (Theorem 4.1 therein).

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    $\begingroup$ Theorem 4.4 in the paper N. Kalton, DW, ``Property (M), M-ideals, and almost isometric structure of Banach spaces.'' J. Reine Angew. Math. 461, 137-178 (1995) says about a subspace $X\subset L_p[0,1]$, $1<p<\infty$, $p\neq2$, that $X$ embeds almost isometrically into $\ell_p$ if and only if $B_X$ is $L_1$-compact. (This paper is the predecessor of the one you are quoting.) For those spaces, the unit ball is compact in measure. Concerning embeddings into $\ell_p$ see also W.B. Johnson and E. Odell's paper. $\endgroup$
    – Dirk Werner
    Commented Jun 15, 2020 at 18:57
  • $\begingroup$ @DirkWerner This is exactly the kind of result I was looking for. Thank you very much for your help (and for the precise references). $\endgroup$
    – user159631
    Commented Jun 16, 2020 at 8:06

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No, it does not: otherwise, the given closed subspace $V$ would be a subspace of $L^0$ (the space of measurable functions, metrised with the convergence in measure) whose unit ball would be compact, and thus $V$ would be a topological vector space in which every ball is compact. This implies that $V$ is finite dimensional, since

Every locally compact topological vector space $X$ has finite dimension: see Theorem 1.22 in

Rudin, Walter. "Functional analysis." (1973).

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    $\begingroup$ That is not a valid argument. Locally compact means that some neighbourhood of $0$ is compact. But the unit ball of the $L^p$-norm on $V$ need not be a $L^0$-neighbourhood. (Your argument would implay that every compact operator between Banach spaces had finite dimensional range.) $\endgroup$
    – Jochen Wengenroth
    Commented Dec 12, 2021 at 14:41
  • $\begingroup$ ah ah! You are totally right, I got confused between the intersection of $V$ with the unit ball in $L^p$, and the unit ball in $L^0$, which are of course very different things $\endgroup$
    – pietro siorpaes
    Commented Dec 13, 2021 at 12:22

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