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Modulus and characteristic of convexity

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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.

Contents

• Definitions

• Properties

• Modulus of convexity of the L P spaces

• See also

• Notes

• References

Definitions

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

δ δ ( ε ε ) = inf { 1 − − ‖ x + y 2 ‖ : x , y ∈ ∈ S , ‖ ‖ x − − y ‖ ‖ ≥ ≥ ε ε } , {\displaystyle \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.cite-ref-1[1]

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

ε ε 0 = sup { ε ε : δ δ ( ε ε ) = 0 } . {\displaystyle \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.cite-ref-2[2]

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].cite-ref-3[3] The modulus of convexity need not itself be a convex function of ε.cite-ref-4[4] However, the modulus of convexity is equivalent to a convex function in the following sense:cite-ref-5[5] there exists a convex function δ1(ε) such that

δ δ ( ε ε / 2 ) ≤ ≤ δ δ 1 ( ε ε ) ≤ ≤ δ δ ( ε ε ) , ε ε ∈ ∈ [ 0 , 2 ] . {\displaystyle \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 ε0 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.cite-ref-6[6] Namely, there exists q ≥ 2 and a constant c > 0 such that

δ δ ( ε ε ) ≥ ≥ c ε ε q , ε ε ∈ ∈ [ 0 , 2 ] . {\displaystyle \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 LP spaces.cite-ref-7[7] If 1 < p ≤ ≤ 2 {\displaystyle 1<p\leq 2} , then it satisfies the following implicit equation:

( 1 − − δ δ p ( ε ε ) + ε ε 2 ) p + ( 1 − − δ δ p ( ε ε ) − − ε ε 2 ) p = 2. {\displaystyle \left(1-\delta _{p}(\varepsilon )+{\frac {\varepsilon }{2}}\right)^{p}+\left(1-\delta _{p}(\varepsilon )-{\frac {\varepsilon }{2}}\right)^{p}=2.}

Knowing that δ δ p ( ε ε + ) = 0 , {\displaystyle \delta _{p}(\varepsilon +)=0,} one can suppose that δ δ p ( ε ε ) = a 0 ε ε + a 1 ε ε 2 + ⋯ ⋯ {\displaystyle \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 ε ε = 0 {\displaystyle \varepsilon =0} , one can calculate the a i {\displaystyle a_{i}} coefficients:

δ δ p ( ε ε ) = p − − 1 8 ε ε 2 + 1 384 ( 3 − − 10 p + 9 p 2 − − 2 p 3 ) ε ε 4 + ⋯ ⋯ . {\displaystyle \delta _{p}(\varepsilon )={\frac {p-1}{8}}\varepsilon ^{2}+{\frac {1}{384}}(3-10p+9p^{2}-2p^{3})\varepsilon ^{4}+\cdots .}

For 2 < p < ∞ ∞ {\displaystyle 2<p<\infty } , one has the explicit expression

δ δ p ( ε ε ) = 1 − − ( 1 − − ( ε ε 2 ) p ) 1 p . {\displaystyle \delta _{p}(\varepsilon )=1-\left(1-\left({\frac {\varepsilon }{2}}\right)^{p}\right)^{\frac {1}{p}}.}

Therefore, δ δ p ( ε ε ) = 1 p 2 p ε ε p + ⋯ ⋯ {\displaystyle \delta _{p}(\varepsilon )={\frac {1}{p2^{p}}}\varepsilon ^{p}+\cdots } .