I had the following little question pop up, but I cannot seem to find any reference to it.

Let $X$ be a Banach space and $E\subseteq X$ a proper subspace with $E$ isomorphic to $X$ itself. Is the subspace $E$ necessarily complemented in $X$?

Composing the isomorphism with the inclusion map is a linear surjection of $X$ onto $E$, but this map is not necessarily an idempotent on $X$. So this map will require some tweaking to become a projection if it is actually the case that such subspaces are always complemented. It is not entirely clear if this can always be done, so the answer might be no in general. Is there an example?

There are some (non-Hilbert space) examples for which the answer is yes. Gowers' solution to the Hyper plane problem [1] is one, but just because that space is not isomorphic to any of its proper subspaces. Another example is one from [2] where all subspaces $E$ that are isomorphic to $X$ are exactly those of finite codimension, and hence are complemented.

[1] Gowers, W.T., A Solution to Banach's Hyperplane Problem. Bulletin of the London Mathematical Society, Volume 26, Issue 6, November 1994, Pages 523–530.

[2] Gowers, W. & Maurey, B. Banach spaces with small spaces of operators Math Ann (1997) 307: 543.


2 Answers 2


The first counterexample that comes to my mind, which is probably overkill (memory is easier than thought!) is

J. Bourgain, A counterexample to a complementation problem. Compositio Mathematica, Volume 43 (1981) no. 1, p. 133-144 NUMDAM link

which constructs a subspace $E$ of $L_1$ that is isomorphic (non-isometrically) to $L_1$, but which is not complemented inside $L_1$.

Some years ago, in discussions with Banach-space theorists, one of them sketched to me how one can bootstrap to get uncomplemented copies of $\ell_1$ inside $\ell_1$ but I don't recall the details right now.

  • $\begingroup$ The bibliography in the Bourgain paper makes reference to some other examples as well, so there is material to get some details from. Thank you Yemon. $\endgroup$ Commented Jul 3, 2019 at 14:15
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    $\begingroup$ For every $1\le p \not= 2 < \infty$ there is an uncomplemented subspace of $\ell_p$ that is isomorphic to $\ell_p$. For $1<p<4/3$ and $p>2$ this is due to H. P. Rosenthal (in different papers; the main thing for the first range is W. Rudin's result that there are $|Lambda$-$4$ sets that are not $\Lambda$-$4+\epsilon$ for any $\epsilon > 0$. For $4/3 < p <2$ the result is due to Bennett, Dor, Goodman, Newman, and myself. As noted above, the case $p=1$ is due to Bourgain and probably he references these other papers. $\endgroup$ Commented Jul 3, 2019 at 17:07

Sorry for digging this up but there are way easier examples.

Let $X=C[0,1]$ and let $Y$ be an infinite-dimensional separable space whose no copy is complemented in $X$ (any reflexive space would do). Let $Z$ be a direct sum of $X$ and $Y$. Embed $Z$ into the $X$-summand of $Z$. Certainly, it is not complemented therein.


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