# Non-existence of continuous extension of continuous linear operator defined on non-dense subspace

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

• 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 $\tilde T=T\circ \pi$. – Jochen Wengenroth Oct 10 '18 at 10:25
• 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. – Nik Weaver Oct 10 '18 at 16:14

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