proof of Hilbert space is uniformly convex space

We prove that in fact an inner product space is uniformly convex. Let $ϵ>0$, $u,v\in H$ such that $\parallel u\parallel \le 1$, $\parallel v\parallel \le 1$, $\parallel u-v\parallel \ge ϵ$. Put $\delta =1-\frac{1}{2}\sqrt{4-{ϵ}^2}$. Then $\delta >0$ and by the parallelogram law

${\parallel u+v\parallel }^2$ $=$ ${\parallel u+v\parallel }^2+{\parallel u-v\parallel }^2-{\parallel u-v\parallel }^2$

$=$ $2{\parallel u\parallel }^2+2{\parallel v\parallel }^2-{\parallel u-v\parallel }^2$

$\le$ $4-{ϵ}^2$

$=$ $4{(1-\delta )}^2.$

Hence, $\parallel \frac{u+v}{2}\parallel \le 1-\delta$.

Since a Hilbert space is an inner product space, a Hilbert space the conditions of a uniformly convex space.

Title	proof of Hilbert space is uniformly convex space
Canonical name	ProofOfHilbertSpaceIsUniformlyConvexSpace
Date of creation	2013-03-22 15:20:48
Last modified on	2013-03-22 15:20:48
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	16
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 46C15
Classification	msc 46H05