In mathematics, uniformly convex spaces (or uniformly rotund spaces) are common examples of reflexive Banach spaces. The concept of uniform convexity was first introduced by James A. Clarkson in 1936.
Definition
A uniformly convex space is a normed vector space such that, for every 0 < ε ≤ 2 {\displaystyle 0<\varepsilon \leq 2} there is some δ > 0 {\displaystyle \delta >0} such that for any two vectors with ‖ x ‖ = 1 {\displaystyle \|x\|=1} and ‖ y ‖ = 1 , {\displaystyle \|y\|=1,} the condition
‖ x − y ‖ ≥ ε {\displaystyle \|x-y\|\geq \varepsilon }implies that:
‖ x + y 2 ‖ ≤ 1 − δ . {\displaystyle \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 {\displaystyle X} is uniformly convex if and only if for every 0 < ε ≤ 2 {\displaystyle 0<\varepsilon \leq 2} there is some δ > 0 {\displaystyle \delta >0} so that, for any two vectors x {\displaystyle x} and y {\displaystyle y} in the closed unit ball (i.e. ‖ x ‖ ≤ 1 {\displaystyle \|x\|\leq 1} and ‖ y ‖ ≤ 1 {\displaystyle \|y\|\leq 1} ) with ‖ x − y ‖ ≥ ε {\displaystyle \|x-y\|\geq \varepsilon } , one has ‖ x + y 2 ‖ ≤ 1 − δ {\displaystyle \left\|{\frac {x+y}{2}}\right\|\leq 1-\delta } (note that, given ε {\displaystyle \varepsilon } , the corresponding value of δ {\displaystyle \delta } could be smaller than the one provided by the original weaker definition).
| Proof |
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The "if" part is trivial. Conversely, assume now that X {\displaystyle X} is uniformly convex and that x , y {\displaystyle x,y} are as in the statement, for some fixed 0 < ε ≤ 2 {\displaystyle 0<\varepsilon \leq 2} . Let δ 1 ≤ 1 {\displaystyle \delta _{1}\leq 1} be the value of δ {\displaystyle \delta } corresponding to ε 3 {\displaystyle {\frac {\varepsilon }{3}}} in the definition of uniform convexity. We will show that ‖ x + y 2 ‖ ≤ 1 − δ {\displaystyle \left\|{\frac {x+y}{2}}\right\|\leq 1-\delta } , with δ = min { ε 6 , δ 1 3 } {\displaystyle \delta =\min \left\{{\frac {\varepsilon }{6}},{\frac {\delta _{1}}{3}}\right\}} . If ‖ x ‖ ≤ 1 − 2 δ {\displaystyle \|x\|\leq 1-2\delta } then ‖ x + y 2 ‖ ≤ 1 2 ( 1 − 2 δ ) + 1 2 = 1 − δ {\displaystyle \left\|{\frac {x+y}{2}}\right\|\leq {\frac {1}{2}}(1-2\delta )+{\frac {1}{2}}=1-\delta } and the claim is proved. A similar argument applies for the case ‖ y ‖ ≤ 1 − 2 δ {\displaystyle \|y\|\leq 1-2\delta } , so we can assume that 1 − 2 δ < ‖ x ‖ , ‖ y ‖ ≤ 1 {\displaystyle 1-2\delta <\|x\|,\|y\|\leq 1} . In this case, since δ ≤ 1 3 {\displaystyle \delta \leq {\frac {1}{3}}} , both vectors are nonzero, so we can let x ′ = x ‖ x ‖ {\displaystyle x'={\frac {x}{\|x\|}}} and y ′ = y ‖ y ‖ {\displaystyle y'={\frac {y}{\|y\|}}} . We have ‖ x ′ − x ‖ = 1 − ‖ x ‖ ≤ 2 δ {\displaystyle \|x'-x\|=1-\|x\|\leq 2\delta } and similarly ‖ y ′ − y ‖ ≤ 2 δ {\displaystyle \|y'-y\|\leq 2\delta } , so x ′ {\displaystyle x'} and y ′ {\displaystyle y'} belong to the unit sphere and have distance ‖ x ′ − y ′ ‖ ≥ ‖ x − y ‖ − 4 δ ≥ ε − 4 ε 6 = ε 3 {\displaystyle \|x'-y'\|\geq \|x-y\|-4\delta \geq \varepsilon -{\frac {4\varepsilon }{6}}={\frac {\varepsilon }{3}}} . Hence, by our choice of δ 1 {\displaystyle \delta _{1}} , we have ‖ x ′ + y ′ 2 ‖ ≤ 1 − δ 1 {\displaystyle \left\|{\frac {x'+y'}{2}}\right\|\leq 1-\delta _{1}} . It follows that ‖ x + y 2 ‖ ≤ ‖ x ′ + y ′ 2 ‖ + ‖ x ′ − x ‖ + ‖ y ′ − y ‖ 2 ≤ 1 − δ 1 + 2 δ ≤ 1 − δ 1 3 ≤ 1 − δ {\displaystyle \left\|{\frac {x+y}{2}}\right\|\leq \left\|{\frac {x'+y'}{2}}\right\|+{\frac {\|x'-x\|+\|y'-y\|}{2}}\leq 1-\delta _{1}+2\delta \leq 1-{\frac {\delta _{1}}{3}}\leq 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 ∞ {\displaystyle \{f_{n}\}_{n=1}^{\infty }} is a sequence in a uniformly convex Banach space that converges weakly to f {\displaystyle f} and satisfies ‖ f n ‖ → ‖ f ‖ , {\displaystyle \|f_{n}\|\to \|f\|,} then f n {\displaystyle f_{n}} converges strongly to f {\displaystyle f} , that is, ‖ f n − f ‖ → 0 {\displaystyle \|f_{n}-f\|\to 0} .
- A Banach space X {\displaystyle X} is uniformly convex if and only if its dual X ∗ {\displaystyle X^{*}} is uniformly smooth.
- Every uniformly convex space is strictly convex. Intuitively, the strict convexity means a stronger triangle inequality ‖ x + y ‖ < ‖ x ‖ + ‖ y ‖ {\displaystyle \|x+y\|<\|x\|+\|y\|} whenever x , y {\displaystyle x,y} are linearly independent, while the uniform convexity requires this inequality to be true uniformly.
Examples
- Every inner-product space is uniformly convex.1
- Every closed subspace of a uniformly convex Banach space is uniformly convex.
- Clarkson's inequalities imply that Lp spaces ( 1 < p < ∞ ) {\displaystyle (1<p<\infty )} are uniformly convex.
- Conversely, L ∞ {\displaystyle L^{\infty }} is not uniformly convex.
See also
Citations
General references
- Clarkson, J. A. (1936). "Uniformly convex spaces". Trans. Amer. Math. Soc. 40 (3). American Mathematical Society: 396–414. doi:10.2307/1989630. JSTOR 1989630..
- Hanner, O. (1956). "On the uniform convexity of L p {\displaystyle L^{p}} and l p {\displaystyle l^{p}} ". Ark. Mat. 3: 239–244. doi:10.1007/BF02589410..
- Beauzamy, Bernard (1985) [1982]. Introduction to Banach Spaces and their Geometry (Second revised ed.). North-Holland. ISBN 0-444-86416-4.
- Per Enflo (1972). "Banach spaces which can be given an equivalent uniformly convex norm". Israel Journal of Mathematics. 13 (3–4): 281–288. doi:10.1007/BF02762802.
- Lindenstrauss, Joram and Benyamini, Yoav. Geometric nonlinear functional analysis. Colloquium publications, 48. American Mathematical Society.
References
Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces (2nd ed.). Boca Raton, FL: CRC Press. p. 524, Example 16.2.3. ISBN 978-1-58488-866-6. 978-1-58488-866-6 ↩