# $(1+\epsilon)$-injective Banach spaces, complex scalars

It is well known that a real Banach space which is $(1+\epsilon)$-injective for every $\epsilon >0$ is already 1-injective (Lindenstrauss Memoirs AMS, 1964, download here). Using common terminology, If $E$ is a $\mathcal{P}_{1+\epsilon}$-space for every $\epsilon >0$ then $E$ is a $\mathcal{P}_1$-space.

The proof of Lindenstrauss seems valid only for real scalars. Has a proof of the corresponding statement for complex scalars appeared in the literature?

This result is easier if $E$ is a dual space, and the proof in Semadeni's book seems to work for complex scalars.

[Edit 7/1/2013] After 1 month and 138 views, no answer is posted. Two experts (not on MO) have told me they did not know of a reference. This is likely not in the literature, which is somewhat surprising to me.

With editorial license, I am changing the question: Give a proof of the statement for complex scalars.

[Edit 8/20/2013] Per a reader's suggestion, be warned that the proof I offered below turned out to be incorrect, as mentioned at the end. Maybe the idea can be rescued.

• Sure, because for dual spaces you can use weak$^*$ compactness. A similar proof shows that a dual space that is $\pi_{\lambda}^\infty$ for every $\lambda > 1$ is $1$-injective. Commented Jun 4, 2013 at 15:29
• Why a down vote for this good question? Commented Jun 4, 2013 at 18:17
• @BillJohnson it is sad to waste this bounty in vain. May be you have any ideas how to prove this theorem for complex case? Commented Sep 6, 2013 at 12:37
• I have not had time to think about it, Norbert. Fred is an expert on this topic, so I don't expect it to be easy. Commented Sep 13, 2013 at 18:42

## 1 Answer

Edit: I apologize for not catching the mistake sooner, but at least this can serve as an example of how science evolves. If you don't make any mistakes, you aren't thinking.

I will have to uncheck this answer since the question is still open.

The statement for complex scalars is true. In the mid-to-late 60's, not long after Lindenstrauss' memoir was published, the theory of the "injective hull of a Banach space" was completely worked out. It is pretty well covered in Section 11 of Lacey, "The Isometric Theory of Classical Banach Spaces". There we see that every Banach space $$X$$ is isometrically embedded with an "essential embedding" into a unique $$C(K)$$ space where $$K$$ is compact and extremally disconnected (sometimes called "Stonean"). For our theorem, it therefore suffices to prove that if $$X$$ is a $$P_{1+\epsilon}$$ space for every $$\epsilon > 0$$ then $$X$$ has no proper essential extension.

We use the criterion for essential extensions given by Lacey, p. 89: if $$X\subset Y$$ then $$Y$$ is an essential extension of $$X$$ if and only if the only seminorm on $$Y$$ which is dominated by the norm on $$Y$$ and equal to the norm on $$X$$ is the norm on $$Y$$ itself.

For a $$P_{1+\epsilon}$$ subspace $$X\subset Y$$, we define a seminorm $$\rho$$ on $$Y$$ by $$\rho(y) = \inf\{\|P(y)\|: P \: \text{is a projection of}\: Y \: \text{onto}\: X\}.$$

The proof of the triangle inequality $$\rho(y_1 + y_2)\le \rho(y_1) + \rho(y_2)$$ uses the following lemma: Given $$y_1,\: y_2 \in Y$$ which are linearly independent (mod $$X$$) and projections $$P_1,\: P_2$$ from $$Y$$ onto $$X$$, there exists a projection $$P:Y \twoheadrightarrow X$$ with $$P(y_i) = P_i(y_i),\: i=1,2$$. (Here we use the fact that $$X$$ has the extension property: bounded linear operators into $$X$$ can be extended.) Since $$X$$ is $$P_{1+\epsilon}$$ for all $$\epsilon > 0$$, we have $$\rho(y) \le \|y\|$$ on $$Y$$ and $$\rho(x) = \|x\|$$ for $$x\in X$$. But if $$Y$$ is a proper extension of $$X$$, $$\rho$$ can not be equal to the norm on $$Y$$. Thus $$Y$$ is not an essential extension by the criterion mentioned. QED

Notice that this proof is valid for real and complex scalars.

[Edit 7/6/2013]: Oops! I just reread this and I realize that the function $$\rho$$ as defined above does NOT in general satisfy the triangle inequality. I believe the lemma, as stated, is true, but that is not sufficient to prove the triangle inequality. I can provide a counterexample if requested, but I must withdraw the claim. I still have hopes that a proof showing that $$X$$ has no proper essential extension can be found. In particular, as follows from the above remark about the injective hull, it suffices to show that $$X$$ is not a proper subspace by an essential embedding into any $$C(K)$$ with $$K$$ extremally disconnected. This is equivalent to showing that, for any isometric embedding of $$X$$ onto a proper subspace of such a $$C(K)$$, $$K$$ will have a proper closed subspace which is norming for $$X$$ (see Lacey).