## On moduli of $$k$$-convexity.(English)Zbl 0985.46002

If $$X$$ is a Banach space with unit ball $$B_X$$, the classical modulus of convexity [J. A. Clarkson, Trans. Am. Math. Soc. 40, 396-414 (1936; Zbl 0015.35604)] is defined by: $\delta_X(\varepsilon) := \inf \{ 1 - \|x_0 + x_1\|/2 : x_0,x_1 \in B_X, \|x_1 - x_0\|\geq \varepsilon\}.$ V. I. Gurarij [Mat. Issled. 2, No. 1(3), 141-148 (1967; Zbl 0232.46024)] showed that $$\delta_X(\varepsilon)$$ is continuous on $$[0,2)$$ but not necessarily at 2. The notion of $$k$$-convexity was introduced by F. Sullivan [Can. J. Math. 31, 628-636 (1979; Zbl 0422.46011)] as follows. First let $A(x_0, x_1, \ldots ,x_k) := (k!)^{-1}\sup \{ \det (f_i(x_j - x_0))\}$ where the supremum is taken over all linear functionals $$f_1,\ldots, f_k$$ with norm $$\leq 1$$. This represents a “volume” of $$\text{co}(x_0,x_1, \ldots, x_k)$$ and generalizes $$\|x_1 - x_0\|$$. Now define $\delta_X^k(\varepsilon) := \inf \{ 1 - \|x_0 + x_1+ \cdots +x_k\|/(k+1) : x_0,x_1, \ldots ,x_k \in B_X, A(x_0, x_1, \ldots x_k) \geq \varepsilon)\}.$ In this paper the author constructs a delicate computation to show that $$\delta_X^k(\varepsilon)$$ is continuous on $$[0, \mu_X^k)$$ where $$\mu_X^k:= \sup A(x_0,x_1,\ldots,x_k)$$ where the supremum is over all $$(k+1)$$-tuples from $$B_X$$. This answers a question posed by W. A. Kirk [Proc. Symp. Pure Math., Am. Math. Soc. 45/2, 51-64 (1986; Zbl 0594.47048)].

### MSC:

 46B20 Geometry and structure of normed linear spaces 47H10 Fixed-point theorems

### Citations:

Zbl 0015.35604; Zbl 0232.46024; Zbl 0422.46011; Zbl 0594.47048
Full Text: