Principle of Local Reflexivity

Question

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!

Answer 1

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.)

Answer 2

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.