# Simple example that density of the subspace cannot be omitted from the Bounded Extension from Dense Subspace Theorem.

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 agrees with $$T_0$$ on $$M$$. This function $$Τ$$ is a bounded linear operator, and $$\|Τ\| = \|T_0\|$$.

Megginson's Introduction to Banach Space Theory points out that the hypothesis that $$M$$ is dense in $$X$$ cannot be omitted by this counterexample:

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 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. (See Megginson's book for the proof and precise references).

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

But these examples are a bit of overkill. To show the necessity of $$M$$ being dense in $$X$$ in the Bounded Extension from Dense Subspace Theorem, you don't really need to know that $$M$$ is closed and uncomplemented in $$X$$. (Or maybe you do?)

Question 1 What is a simpler example (if one exists) that shows the hypothesis that $$M$$ is dense in $$X$$ cannot be omitted.

Question 2 What is a simpler example (if one exists) 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 $$X \to Y$$.

Edit: The first question allowed for counterexamples where uniqueness or the $$\| T \| = \| T_0 \|$$ property fails. That was not my original intention. So I added the second question.

• This might not be exactly what you wanted, but the thesis fails without the density assumption because the continuous extension need not be unique, as trivial examples show (if you only require existence in the statement, then the question remains). Sep 24 '18 at 7:56

Let $$X=Y=\mathbb R^2$$ and $$M=\mathbb R\subset \mathbb R^2$$. Note that $$M$$ is not dense in $$Y$$.
Then we define $$T_0:M\to Y, x \mapsto (x,0)$$. Then it can be seen, that the extension is not unique. For example, the maps $$T_1:X\to Y, (x,y)\mapsto (x,y)$$ and $$T_2: X\to Y, (x,y)\mapsto (x,3y)$$ are both extensions of $$T_0$$. In the latter case, the property $$\|T_2\|=\|T_0\|$$ is violated.