Uniformly convex space

Definition

A uniformly convex space is a normed vector space such that, for every 0 there is some \delta0 such that for any two vectors with |x| = 1 and |y| = 1, the condition

:|x-y|\geq\varepsilon

implies that:

:\left|\frac{x+y}{2}\right|\leq 1-\delta.

Intuitively, the center of a line segment inside the unit ball must lie deep inside the unit ball unless the segment is short.

Properties

• The unit sphere can be replaced with the closed unit ball in the definition. Namely, a normed vector space X is uniformly convex if and only if for every 0 there is some \delta0 so that, for any two vectors x and y in the closed unit ball (i.e. |x| \le 1 and |y| \le 1 ) with |x-y| \ge \varepsilon , one has \left|{\frac{x+y}{2}}\right| \le 1-\delta (note that, given \varepsilon , the corresponding value of \delta could be smaller than the one provided by the original weaker definition). The "if" part is trivial. Conversely, assume now that X is uniformly convex and that x,y are as in the statement, for some fixed 0. Let \delta_1\le 1 be the value of \delta corresponding to \frac{\varepsilon}{3} in the definition of uniform convexity. We will show that \left|\frac{x+y}{2}\right|\le 1-\delta , with \delta=\min\left{\frac{\varepsilon}{6},\frac{\delta_1}{3}\right} .

If |x|\le 1-2\delta then \left|\frac{x+y}{2}\right|\le\frac{1}{2}(1-2\delta)+\frac{1}{2}=1-\delta and the claim is proved. A similar argument applies for the case |y|\le 1-2\delta , so we can assume that 1-2\delta. In this case, since \delta\le\frac{1}{3} , both vectors are nonzero, so we can let x'=\frac{x}{|x|} and y'=\frac{y}{|y|} . We have |x'-x|=1-|x|\le 2\delta and similarly |y'-y|\le 2\delta , so x' and y' belong to the unit sphere and have distance |x'-y'|\ge|x-y|-4\delta\ge\varepsilon-\frac{4\varepsilon}{6}=\frac{\varepsilon}{3} . Hence, by our choice of \delta_1 , we have \left|\frac{x'+y'}{2}\right|\le 1-\delta_1 . It follows that \left|\frac{x+y}{2}\right|\le\left|\frac{x'+y'}{2}\right|+\frac{|x'-x|+|y'-y|}{2}\le 1-\delta_1+2\delta\le 1-\frac{\delta_1}{3}\le 1-\delta and the claim is proved.

• The Milman–Pettis theorem states that every uniformly convex Banach space is reflexive, while the converse is not true.
• Every uniformly convex Banach space is a Radon–Riesz space, that is, if {f_n}_{n=1}^{\infty} is a sequence in a uniformly convex Banach space that converges weakly to f and satisfies |f_n| \to |f|, then f_n converges strongly to f , that is, |f_n - f| \to 0 .
• A Banach space X is uniformly convex if and only if its dual X^* is uniformly smooth.
• Every uniformly convex space is strictly convex. Intuitively, the strict convexity means a stronger triangle inequality |x+y| whenever x,y are linearly independent, while the uniform convexity requires this inequality to be true uniformly.

Examples

• Every inner-product space is uniformly convex.
• Every closed subspace of a uniformly convex Banach space is uniformly convex.
• Clarkson's inequalities imply that Lp spaces (1 are uniformly convex.
• Conversely, L^\infty is not uniformly convex.

References

Citations

General references

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• {{cite journal |author-link=Per Enflo
• Lindenstrauss, Joram and Benyamini, Yoav. Geometric nonlinear functional analysis. Colloquium publications, 48. American Mathematical Society.

References

1. (2011). "Topological Vector Spaces". CRC Press.