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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.

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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. –  Bill Johnson Jun 4 '13 at 15:29
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Why a down vote for this good question? –  Bill Johnson Jun 4 '13 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? –  Norbert Sep 6 '13 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. –  Bill Johnson Sep 13 '13 at 18:42
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1 Answer

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).

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.

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If you are interested I can send you that paper by Lindenstrauss where he proves the desired theorem for real case –  Norbert Sep 5 '13 at 4:50
    
I had to award this bounty to you, since there were no answers and hope soon you'll find a nice soltion. –  Norbert Sep 8 '13 at 16:47
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