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Lindenstrauss has the following paper: http://www.ams.org/journals/bull/1962-68-05/S0002-9904-1962-10787-3/S0002-9904-1962-10787-3.pdf

I would like to see the proof for the following theorem (from the above paper):

Theorem: Let $X,Y,Z$ be Banach spaces with $Y\subset Z$. Suppose $X^*=L_1(\mu)$ for some measure $\mu$. Then every compact operator $T:Y\to X$ has, for every $\epsilon>0$, a compact extension $\widetilde{T}:Z\to X$ with $\|\widetilde{T}\|\leq(1+\epsilon)\|T\|$.

Unfortunately, Lindenstrauss's proof for this theorem consists of the line "due to Grothendieck." He then gives a reference to a paper by Grothendieck, in French.

Grothendieck's paper is here (in French): http://matematicas.unex.es/~navarro/res/ega/Carat%C3%A9risation%20des%20Espaces%20L1.pdf

Unfortunately I only know a little French, and even with Google translate to help, I couldn't even decipher the theorem statements, much less their proofs!

So my question is this: Does anyone know a textbook or a paper in English which contains a detailed proof to the above theorem?


EDIT: By the way, I might as well give some background too. Let $X=c_0(\mathbb{N})$. I am given an operator $T:X\to X$ which satisfies some very funky conditions including (but not limited to) $\sigma_p(T^*)=\emptyset$ and $\partial\sigma(T)\subseteq\sigma_p(T)$, and $\sigma(T)$ uncountable. The set of all $T$-eigenvectors of norm $\leq 1$ is compact. I don't want to list all the conditions due to space, but those are the most striking. I am able to show that there is a closed subspace $Y\subset X$ which is both infinite-dimensional and infinite-codimensional in $X$, and for which the restriction $T|_Y:Y\to X$ is compact. My ultimate goal is to find an operator $S:X\to X$ with countable point spectrum which commutes with $T$ and which is not a multiple of the identity. So if $S$ is compact and commutes with $T$, that will do it. The above theorem allows us to find a compact extension of $T|_Y$ to all of $X$. I hope that this extension is such that it commutes with $T$. Alternatively, perhaps it can give me ideas as to how to construct such an operator.

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Did you look at Lindensrauss' Memoirs? –  Bill Johnson Aug 11 '13 at 22:54

1 Answer 1

This is very mysterious indeed. I remember I have also tried to find this result in Grothendieck's paper but it wasn't there, I think. Fortunately, you can find the proof in:

M. Zippin, Extension of bounded linear operators, Handbook of the Geometry of Banach Spaces, Vol. 2, W.B. Johnson and J. Lindenstrauss, eds., Elsevier, Amsterdam 2003, page 1725.

If you don't care about the norm of an extension of your compact operator, you can give a short proof of this theorem using the fact that the operation of taking the projective tensor product with $L_1(\mu)\cong C(K)^*$ respects quotients.

Regarding your motivations, once you have your subspace $Y$, why can't you pass to a further subspace $Y_0\subset Y$ still having infinite codimension but additionally isomorphic to $c_0$ hence complemented? ($c_0$ is saturated by subspaces isomorphic to $c_0$.) Then you can extend $T|_{Y_0}$ to the whole space by 0.

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Awesome, thank you for the reference! Regarding your comment, it's a good idea and one that I had considered. However the problem is commutation. Suppose $X=Y_0\oplus Z_0$ where $S:=T|_{Y_0}\oplus 0$ is compact. We need to show that $TS=ST$. Then given $y+z\in X$ we have $TS(y+z)=T^2y$, but $ST(y+z)=T^2y$ only if $Tz\in Y_0$. Still, it's a good idea and one I intend to keep in mind as I research. –  Ben Wallis Aug 12 '13 at 13:36
Oh yes, I forgot you need commutation. –  Tomek Kania Aug 12 '13 at 14:26

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