Modulus and characteristic of convexity

In mathematics, the modulus of convexity and the characteristic of convexity are measures of "how convex" the unit ball in a Banach space is. In some sense, the modulus of convexity has the same relationship to the ε-δ definition of uniform convexity as the modulus of continuity does to the ε-δ definition of continuity.

Definitions
The modulus of convexity of a Banach space (X, ||&sdot;||) is the function δ : [0, 2] → [0, 1] defined by


 * $$\delta (\varepsilon) = \inf \left\{ 1 - \left\| \frac{x + y}{2} \right\| \,:\, x, y \in S, \| x - y \| \geq \varepsilon \right\},$$

where S denotes the unit sphere of (X, || ||). In the definition of δ(ε), one can as well take the infimum over all vectors x, y in X such that ǁxǁ, ǁyǁ &le; 1 and ǁx &minus; yǁ &ge; ε.

The characteristic of convexity of the space (X, || ||) is the number ε0 defined by


 * $$\varepsilon_{0} = \sup \{ \varepsilon \,:\, \delta(\varepsilon) = 0 \}.$$

These notions are implicit in the general study of uniform convexity by J. A. Clarkson (this is the same paper containing the statements of Clarkson's inequalities). The term "modulus of convexity" appears to be due to M. M. Day.

Properties

 * The modulus of convexity, δ(ε), is a non-decreasing function of ε, and the quotient δ(ε)&thinsp;/&thinsp;ε is also non-decreasing on (0, 2]. The modulus of convexity need not itself be a convex function of ε.  However, the modulus of convexity is equivalent to a convex function in the following sense: there exists a convex function δ1(ε) such that
 * $$\delta(\varepsilon / 2) \le \delta_1(\varepsilon) \le \delta(\varepsilon), \quad \varepsilon \in [0, 2].$$


 * The normed space (X, ǁ&thinsp;&sdot;&thinsp;ǁ) is uniformly convex if and only if its characteristic of convexity ε0 is equal to 0, i.e., if and only if δ(ε) > 0 for every ε > 0.
 * The Banach space (X, ǁ&thinsp;&sdot;&thinsp;ǁ) is a strictly convex space (i.e., the boundary of the unit ball B contains no line segments) if and only if δ(2) = 1, i.e., if only antipodal points (of the form x and y = &minus;x) of the unit sphere can have distance equal to 2.
 * When X is uniformly convex, it admits an equivalent norm with power type modulus of convexity. Namely, there exists q &ge; 2 and a constant c &gt; 0 such that
 * $$\delta(\varepsilon) \ge c \, \varepsilon^q, \quad \varepsilon \in [0, 2].$$

Modulus of convexity of the LP spaces
The modulus of convexity is known for the LP spaces. If $$1<p\le2$$, then it satisfies the following implicit equation:


 * $$\left(1-\delta_p(\varepsilon)+\frac{\varepsilon}{2}\right)^p+\left(1-\delta_p(\varepsilon)-\frac{\varepsilon}{2}\right)^p=2.

$$ Knowing that $$\delta_p(\varepsilon+)=0,$$ one can suppose that $$\delta_p(\varepsilon)=a_0\varepsilon+a_1\varepsilon^2+\cdots$$. Substituting this into the above, and expanding the left-hand-side as a Taylor series around $$\varepsilon=0$$, one can calculate the $$a_i$$ coefficients:
 * $$\delta_p(\varepsilon)=\frac{p-1}{8}\varepsilon^2+\frac{1}{384}(3-10p+9p^2-2p^3)\varepsilon^4+\cdots.

$$

For $$2<p<\infty$$, one has the explicit expression
 * $$\delta_p(\varepsilon)=1-\left(1-\left(\frac{\varepsilon}{2}\right)^p\right)^{\frac1p}.

$$ Therefore, $$\delta_p(\varepsilon)=\frac{1}{p2^p}\varepsilon^p+\cdots$$.