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Statement: Let $ X $ be a $B^*$-space over the complex field $ \mathbb{C} $. We aim to prove that $ X $ is strictly convex(i.e.for any $ x, y \in X $ where $ x \neq y $ and $ \|x\|, \|y\| = 1 $, implies $ \| \frac{x+y}{2} \| < 1 $), if and only if for any $ x, y \in X $ with $ x \neq y $ and $ \|x\| = \|y\| = 1 $, there exists $ \lambda \in \mathbb{C} $ such that $\| \lambda x + (1 - \lambda)y \| < 1 $.

This problem simplifies in the real case by examining the possible values of $ \lambda $, but it becomes more intricate when considering the complex case. Any insights or assistance would be greatly appreciated.

Matrix AC
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1 Answers1

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One implication is clear. For the other, suppose that negation of the first statement is true, that is suppose there are $x \neq y$ with $\|x\|=\|y\|=1$ with $\|\frac{x+y}{2}\| = 1$. Find a functional $\varphi \in X^*$ with $\|\varphi\|=1$ and $\varphi(\frac{x+y}{2}) = 1$ (it exists by Hahn-Banach theorem). Set $A = \{z \in X: \varphi(z) = 1\}$.

We will show that for any $\lambda \in \mathbb{C}$ we have $\lambda x + (1-\lambda) y \in A$. First note that both $x$ and $y$ are elements of $A$ (apply strict convexity of $\mathbb{C}$ on $\varphi(x)$, $\varphi(y)$ and $\varphi(\frac{x+y}{2}) = \frac{\varphi(x)+\varphi(y)}{2}$). Now fix any $\lambda \in \mathbb{C}$. Then $$\varphi(\lambda x+(1-\lambda)y) = \lambda \varphi(x) + (1-\lambda)\varphi (y) = \lambda + 1 - \lambda = 1$$ and $\lambda x + (1-\lambda)y \in A$. Hence, the negation of the second statement is true as well and we are done.

KeeperOfSecrets
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  • Thank you for your answer. I solved it by myself using the same method, but I'm still curious if there's a more ‘elementary’ approach, given that this serves as an intermediate lemma in proving the uniqueness of the Hahn-Banach Theorem (Taylor-Foguel). – Matrix AC Apr 08 '24 at 04:08