Question

The motivation is simple, as it is trivially right when $p=2$. When considering the duality between $L^p$ ($l^p$) and $L^q$ ($l^q$) when $p$ and $q$ are conjugate in the sense that $1/p+1/q=1$, I wonder if $L^p$ and $l^p$ are the same in the sense of isometry. I tried to use the situation when $p=2$, however I find it difficult to give a linear homeomorphism.

Answer 1

Variants of this question show up often enough here on MO and over at math.SE that it seems worthwhile to collect some facts and links. I say isomorphic for linearly homeomorphic and isometric for isometrically isomorphic.

One main upshot is:

The family of Banach spaces $L^p, \ell^q$ for $1 \leq p, q \leq \infty$ consists of pairwise non-isomorphic spaces, except for two cases:

1. There is the obvious isometry between $L^2$ and $\ell^2$ and
2. There is a non-obvious isomorphism between $L^\infty$ and $\ell^\infty$, due to Pełczyński. However, $L^\infty$ and $\ell^\infty$ are not isometric.

Plenty of references can be found in the following threads on MO and math.SE:

1. Bill Johnson explains in $L^p$ vs. $L^q$ how one can tell these spaces apart for $1 \lt p \neq q \lt \infty$ using type and cotype considerations, and in the comments there is also some discussion of existence of embeddings.

2. A detailed explanation of the non-existence of an isomorphism between the spaces $L^p$ and $\ell^q$ for $1 \leq p,q \lt \infty$ (modulo Bill Johnson's answer in 1.) is in the answer to If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic. The main ingredient in that answer is Pitt's theorem stating that every operator $\ell^q \to \ell^p$ is compact for $1 \leq p \lt q \lt \infty$, the existence of an embedding $L^2 \to L^p$ plus a little bit of duality theory. See also

   • How do you prove that $\ell^p$ is not isomorphic to $\ell^q$? for a discussion how Pitt's theorem implies non-isomorphism of $\ell^p$ and $\ell^q$ and
   • Inclusion of $L^p$-spaces, reloaded for a discussion of embeddings of $L^2$ into $L^p$ via Rademacher functions or Gaussians.

3. Pełczyński's isomorphism between $L^\infty$ and $\ell^\infty$ is discussed in Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? and non-existence of an isometry is discussed in Isometry between $L^\infty$ and $\ell^\infty$.

Let me finish by recommending the very nice book by Albiac and Kalton, Topics in Banach Space Theory, as an alternative to Lindenstrauss-Tzafriri. It contains a gentle introduction to the above ideas and much more.

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Edit: Further links to related topics:

• One crucial point used in establishing Pełczyński's isomorphism is the injectivity of $\ell^\infty$ and $L^\infty$. For $\ell^\infty$ this is a standard exercise in applying Hahn-Banach (coordinatewise) and for $L^\infty$ Bill Johnson gives a quick proof in Direct proof of injectivity of $L^\infty$. See also Direct proof of "K is projective iff C(K) has the Hahn-Banach property"? for a generalization and related results.

• Complemented subspaces of $\ell_p(I)$ for uncountable $I$

• Isometric embeddings of $\ell_q^m$ into $\ell_p$ and $L_p$ for $p,q\in[1,+\infty]$

Answer 2

The standard way to do this is to use the fact that infinite dimensional complemented subspaces of $\ell^p$ are isomorphic to the space itself whereas $L^p$ has a complemented subspace which is isomorphic to Hilbert space (essentially the Khintchine inequality). Probably the most accessible reference is the standard monograph of Lindenstrauss and Tzafriri on classical Banach spaces. The $\infty$ case is a bit more subtle.