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Uniformly convex space - Wikipedia

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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.

A uniformly convex space is a normed vector space such that, for every $0<\varepsilon \leq 2$ there is some $\delta >0$ 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.

Proof
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 $0<\varepsilon \leq 2$ . Let $\delta _{1}\leq 1$ be the value of $\delta$ corresponding to ${\frac {\varepsilon }{3}}$ in the definition of uniform convexity. We will show that $\left\\|{\frac {x+y}{2}}\right\\|\leq 1-\delta$ , with $\delta =\min \left\{{\frac {\varepsilon }{6}},{\frac {\delta _{1}}{3}}\right\}$ . If $\\|x\\|\leq 1-2\delta$ then $\left\\|{\frac {x+y}{2}}\right\\|\leq {\frac {1}{2}}(1-2\delta )+{\frac {1}{2}}=1-\delta$ and the claim is proved. A similar argument applies for the case $\\|y\\|\leq 1-2\delta$ , so we can assume that $1-2\delta <\\|x\\|,\\|y\\|\leq 1$ . In this case, since $\delta \leq {\frac {1}{3}}$ , both vectors are nonzero, so we can let $x'={\frac {x}{\\|x\\|}}$ and $y'={\frac {y}{\\|y\\|}}$ . We have $\\|x'-x\\|=1-\\|x\\|\leq 2\delta$ and similarly $\\|y'-y\\|\leq 2\delta$ , so $x'$ and $y'$ belong to the unit sphere and have distance $\\|x'-y'\\|\geq \\|x-y\\|-4\delta \geq \varepsilon -{\frac {4\varepsilon }{6}}={\frac {\varepsilon }{3}}$ . Hence, by our choice of $\delta _{1}$ , we have $\left\\|{\frac {x'+y'}{2}}\right\\|\leq 1-\delta _{1}$ . It follows that $\left\\|{\frac {x+y}{2}}\right\\|\leq \left\\|{\frac {x'+y'}{2}}\right\\|+{\frac {\\|x'-x\\|+\\|y'-y\\|}{2}}\leq 1-\delta _{1}+2\delta \leq 1-{\frac {\delta _{1}}{3}}\leq 1-\delta$ and the claim is proved.

Proof

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 $0<\varepsilon \leq 2$ . Let $\delta _{1}\leq 1$ be the value of $\delta$ corresponding to ${\frac {\varepsilon }{3}}$ in the definition of uniform convexity. We will show that $\left\|{\frac {x+y}{2}}\right\|\leq 1-\delta$ , with $\delta =\min \left\{{\frac {\varepsilon }{6}},{\frac {\delta _{1}}{3}}\right\}$ .

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

^ 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.