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Currently I am reading the book 'Isometries in Banach Spaces: Vector-valued Function Spaces and Operator Spaces, Volume Two', Chapter $7,$ page $6.$

Definitions: A $T$-set is a subset $S$ of a Banach space $X$ with the property that for any finite collection $x_1,x_2,...,x_n \in S$ such that $$\left\Vert \sum_{j=1}^n x_j \right\Vert = \sum_{j=1}^n \|x_j\|,$$ and such that $S$ is maximal with respect to this property.

For any $x\in X,$ let $M_x :=\{ y^*\in B(X^*):y^*(x)=\|x\| \}$ where $B(X^*) = \{ y^*: \|y^*\| \leq 1 \}.$

Question: Because of the norm additive property of the $T$-set $S,$ the collection $\{ M_x:x\in S \}$ has the finite intersection property.

It suffices to show that it (bold statement) for $x_1,x_2\in S,$ as it can be generalized easily.

Suppose that $x_1,x_2\in S.$ We want to show that there exists $z^*\in M_{x_1}\cap M_{x_2}$ such that $\|z^*\|\leq 1, z^*(x_1)=\|x_1\|$ and $z^*(x_2)=\|x_2\|.$

Since $x_1,x_2 \in S,$ we have $\|x_1+x_2\| = \|x_1\| + \|x_2\|.$ Then there exists $z_1^*,z_2^*$ and $z_3^*$ such that $$z_3^*(x_1+x_2) = z_1^*(x_1)+z_2^*(x_2).$$ However, I fail to show that $z_1^*=z_2^*=z_3^*.$

Any hint would be appreciated.

uniquesolution
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Idonknow
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1 Answers1

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Take $x_1,x_2\in S$. There exists a norm-$1$ functional $z^{*}$ such that $$z^*(x_1+x_2)=|\!|x_1+x_2|\!|=|\!|x_1|\!|+|\!|x_2|\!|\enspace\enspace(1)$$ Where the second equality follows from the assumption that $x_1,x_2\in S$. Since $z^*$ has norm $1$ we have $z^*(x_1)\leq|\!|x_1|\!|$ and $z^*(x_2)\leq |\!|x_2|\!|$, and we must have equalities in both these inequalities or else $(1)$ will not hold. This means that $z^*\in M_{x_1}\cap M_{x_2}$.

uniquesolution
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