If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$.

Maybe I would have to use the Rademacher's functions.

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    $\begingroup$ A relevant MathOverflow question: mathoverflow.net/questions/79713/lp-mathbbr-vs-lq-mathbbr/… $\endgroup$ Jan 7, 2012 at 7:52
  • $\begingroup$ Ok, I edited it better. $\endgroup$
    – arawarep
    Jan 7, 2012 at 7:52
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    $\begingroup$ Trivial remark: you certainly need to exclude $p=2$ since all separable Hilbert spaces are isomorphic. $\endgroup$
    – t.b.
    Jan 7, 2012 at 7:53
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    $\begingroup$ You should formulate your questions more clearly. As formulated now, the question can be understood in two ways: A. Fix some $p\in\langle 1,\infty)$. Does it hold that $L^p$ and $\ell_p$ are isomorphic? B. Choose $p,q\in\langle 1,\infty)$, $p\ne q$. Are $\ell_p$ and $\ell_q$ isomorphic? Are $L_p$ and $L_q$ isomorphic? (The MO link posted in Jonas comment and t.b's comment confirm, that people read both these interpretations in your question.) $\endgroup$ Jan 7, 2012 at 7:54

1 Answer 1


Let me get rid of the cases $p = 1$ and $p=2$ first:

  1. The space $\ell^1$ has the Schur property (every weakly convergent sequence is norm-convergent) while $L^1$ doesn't.
    [Alternatively, $\ell^1$ has the Radon–Nikodým property while $L^1$ doesn't, see also this thread]

  2. The spaces $\ell^2$ and $L^2$ are isomorphic because they both are separable Hilbert spaces.

Now I think there's no way around discussing the cases $1 \lt p \lt 2$ and $2 \lt p \lt \infty$ separately.

I'll refer to some results in Albiac-Kalton, Topics in Banach space theory, Springer GTM 233, 2006.

It follows from Pitt's theorem(1) (Theorem 2.1.4, page 32) that every operator $\ell^2 \to \ell^p$ for $1 \lt p \lt 2$ is compact while Proposition 6.4.13 (page 155) shows that $\ell^2$ embeds isometrically in every $L^p$, $1\leq p \lt \infty$ (this isn't hard, it suffices to choose a sequence of independent normalized Gaussians on $[0,1]$).

This shows that $\ell^p$ and $L^p$ aren't isomorphic as Banach spaces if $1 \lt p \lt 2$ because $L^p$ admits a non-compact map from $\ell^2$ while $\ell^p$ doesn't.

The case $2 \lt p \lt \infty$ follows from this by duality: if $\ell^p$ and $L^p$ were isomorphic then their dual spaces would be isomophic and we've just excluded that.

(1) see also this note by Sylvain Delpech as well as this thread.

Later: Pitt's theorem also implies that $\ell^{p}$ and $\ell^{q}$ aren't isomorphic whenever $p \neq q$ and the above argument is easily adapted to show that $L^{p}$ and $\ell^{q}$ are only isomorphic if $p = q = 2$ (exercise). Moreover, $L^{p}$ and $L^{q}$ are non-isomorphic if $p \neq q$, see the MO thread Jonas linked to in a comment. To sum up:

The family $\ell^p, L^q$, $1 \leq p,q \lt \infty$ consists of pairwise non-isomorphic spaces, except for the obvious isomorphism between $\ell^2$ and $L^2$.

  • $\begingroup$ Isn't $L^p[0,1]$ isomorphic to $l^q$ though when $\frac{1}{p} + \frac{1}{q} = 1$? $\endgroup$ Aug 31, 2018 at 18:44
  • $\begingroup$ @SaalHardali What makes you think that? $\endgroup$
    – Not Euler
    Apr 18, 2020 at 21:22
  • $\begingroup$ @HritRoy When I wrote this comment I was under the impression that if you take an L^p function on the circle it has an l^q summable fourier series. That is of course wrong. $\endgroup$ Apr 18, 2020 at 21:27

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