# 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, ||⋅||) 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 ≤ 1 and ǁx − yǁ ≥ ε.
The characteristic of convexity of the space (X, || ||) is the number ε$_0$ defined by
    $\varepsilon _{0}=\sup\{\varepsilon \,:\,\delta (\varepsilon )=0\}.$
These notions are implicit in the general study of uniform convexity by J. A. Clarkson (Clarkson (1936); 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) / 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 δ$_1$(ε) such that
       $\delta (\varepsilon /2)\leq \delta _{1}(\varepsilon )\leq \delta (\varepsilon ),\quad \varepsilon \in [0,2].$
* The normed space (X, ǁ ⋅ ǁ) is uniformly convex if and only if its characteristic of convexity ε$_0$ is equal to 0, i.e., if and only if δ(ε) > 0 for every ε > 0.
* The Banach space (X, ǁ ⋅ 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 = −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 ≥ 2 and a constant c > 0 such that
       $\delta (\varepsilon )\geq c\,\varepsilon ^{q},\quad \varepsilon \in [0,2].$

## Modulus of convexity of the L$^P$ spaces

The modulus of convexity is known for the L$^P$ spaces. If $1<p\leq 2$, 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
    $\delta _{p}(\varepsilon )=1-\left(1-\left({\frac {\varepsilon }{2}}\right)^{p}\right)^{\frac {1}{p}}.$
Therefore, $\delta _{p}(\varepsilon )={\frac {1}{p2^{p}}}\varepsilon ^{p}+\cdots$.