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I'm having a hard time trying to understand a proof of the Principle of Local Reflexivity. I'm following the proofs from

1) Topics in Banach Space Theory (by Fernando Albiac, Nigel J. Kalton) 2) http://www.math.tamu.edu/~schlump/sofar.pdf (some notes from Professor Schlump)

The proof in 2) is basically the same as in 1) , but with further details. I don't quite get why ker $S$ is contained in Ker $S_1$ . And I don't get why that implies that there is a $T$ such that $S_1 = TS$.

Can someone enlight me ? I would really appreciate it!

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How did you happen across Schlumprecht's notes? (BTW: He HATES to be called "Professor Schlump".) –  Bill Johnson Feb 27 '12 at 0:57
It happened by chance -- or, better put, by google. I had no idea he doesn't like being called Professor Schlump :) –  Rafael Feb 28 '12 at 14:53

2 Answers 2

Enlarge $G$, if necessary, so that the restriction mapping from $F$ to $G^*$ is one to one. Then what you want is completely obvious from the equality 4 lines from the bottom of p. 274 of [AK]. (The proof is correct without this step, but enlarging $G$ makes it clear without thinking.)

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I'm sorry, let me be more specific. [AK] doesn't proves that $ker S \subset ker S_1$. But it's proven in Professor Schumprecht notes. Fixing the typos, it goes like this: "If we assume that $\xi \in \mathrm K ^N$ and $\sum^N_{j=1} \xi _j x_j^{**} = 0$, but $\sum^N_{j=1} \xi _j x_j \neq 0$, then there is an $i \in \{ 1,2, \ldots, N\}$ so that \begin{equation} \langle x^*_i, \sum^N_{j=1} \xi _j x_j \neq 0. \end{equation} " And I still don't see why there's such an $i$. –  Rafael Mar 1 '12 at 22:08
i guess i made some typos myself, but i cant edit the post :) –  Rafael Mar 1 '12 at 22:10
bill, can you help me? :) –  Rafael Mar 20 '12 at 0:21

I cannot comment above, so I have to write another answer. To clarify for you Rafael why such an $i$ exists in Bill's comment. Take the same setup, $\xi\in\mathbb{K}^N$ such that \begin{equation}\underset{j=1}{\overset{N}{\sum}}\xi_jx_j^{\*\*}=0 ;\quad\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\neq0 \end{equation}
(i.e. $\xi\in\ker(S)\setminus\ker(S_1)$). But recall that the $x_i$'s were chosen in $S_{X^*}$ so that $$\langle x_j^{**},x_j^*\rangle=\langle x_j,x_j^*\rangle=\langle x_j^*,x_j\rangle$$

Now then by assumption, we have that for all $i$, $$0=\langle\underset{j=1}{\overset{N}{\sum}}\xi_jx_j^{\*\*},x_i^*\rangle=\langle\underset{j=1}{\overset{N}{\sum}}\xi_jx_j,x_i^*\rangle=\langle x_i^*,\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\rangle$$ So if the right hand side is equal to 0, but $\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\neq0$ as we assumed, then we must have that for some $i\in\{1,2,\dots,N\}$, $$\langle x_i^*,\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\rangle\neq0$$ which gives a contradiction.

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