Bounded Extension from Dense Subspace Theorem. Suppose that $Μ$ is a dense subspace of a normed space $X$, that $Y$ is a Banach space, and that $T_0: Μ \to Y$ is a bounded linear operator. Then there is a unique continuous function $T: X \to Y$ that extends $T_0$. This function $Τ$ is a bounded linear operator, and $\|Τ\| = \|T_0\|$.

Megginson's Introduction to Banach Space Theory (and several other books) points out that if $M$ is not dense in $X$, then there might not exist a continuous extension of $T_0$ at all. The example given is this:

If $X=\ell^{\infty}$, $M=Y=c_0$, and $T_0 = Id:c_0 \to c_0$, then $T_0$ cannot be continuously extended to an operator from $X \to Y$.

However, verifying this example is not exactly trivial. Megginson obtains it as a corollary of non-trivial theorem of Philips which says:

$c_0$ is an uncomplemented closed subspace of $\ell^{\infty}.$

Philips original proof is difficult. A shorter but still non-trivial proof was published by Whitley. Megginson's book gives Whitely's proof and a precise reference.

Here is an answer that gives two other examples of closed uncomplemented subspaces of Banach spaces: https://math.stackexchange.com/a/108289/570438 . But again the proofs are not easy.

I wonder if these examples are a bit of overkill.

Question A Is there a simpler example of Banach spaces $X$ and $Y$, a non-dense subspace $M$ in $X$, and a bounded linear operator $T_0:M \to Y$ such that $T_0$ cannot be continuously extended to an operator from $X \to Y$? In particular, is there an example that doesn't require first showing that $M$ is uncomplemented in $X$?

Question B Suppose we have Banach spaces $X$ and $Y$, a non-dense subspace $M$ in $X$, and a bounded linear operator $T_0:M \to Y$ such that $T_0$ cannot be continuously extended to an operator from $X \to Y$. Does that imply $M$ is uncomplemented in $X$?

Note: This question is a modified cross-post of https://math.stackexchange.com/questions/2928488/simple-example-that-density-of-the-subspace-cannot-be-omitted-from-the-bounded-e

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    $\begingroup$ Concerning Question B: If $\pi:X\to M$ is a projection onto $M$ then you can extend and continuous linear $T:M\to Y$ just by setting $\tilde T=T\circ \pi$. $\endgroup$ Oct 10, 2018 at 10:25
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    $\begingroup$ And conversely, if there is no projection from $X$ onto some closed subspace $M$, then taking $Y =M$ and $T = {\rm id}: M \to Y$ gives you a bounded linear map with no bounded extension to $X$. So you have a counterexample if and only if you have an uncomplemented subspace. $\endgroup$
    – Nik Weaver
    Oct 10, 2018 at 16:14

1 Answer 1


The examples are not overkill. There isn't easy way to show that a closed subspace is not complemented. For instance, $\ell_2$ isn't complemented in $L_1$ but again the proof isn't easy. The example you mention is probably the easiest one. There are also examples of Orlicz sequence spaces which contain some $\ell_p$ but not complemented. The proof is 'elementary' (but not easier than your $c_0$ example) if you are familiar with Orlicz spaces, see Example 4.c.6 in Lindenstrauss-Tzafriri's book

However, the only spaces all of whose subspaces are complemented are the ones isomorphic to a Hilbert space. This is a beautiful theorem of Lindenstrauss and Tzafriri. So every non-Hilbertian space has a subspace which is not complemented.

  • $\begingroup$ ... or anything isomorphic to Hilbert space? $\endgroup$
    – Nik Weaver
    Oct 11, 2018 at 6:37
  • $\begingroup$ Yes, meant the isomorphs of Hilbert spaces. $\endgroup$ Oct 12, 2018 at 0:01

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