# proof of Hilbert space is uniformly convex space

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

 $\displaystyle\|u+v\|^{2}$ $\displaystyle=$ $\displaystyle\|u+v\|^{2}+\|u-v\|^{2}-\|u-v\|^{2}$ $\displaystyle=$ $\displaystyle 2\|u\|^{2}+2\|v\|^{2}-\|u-v\|^{2}$ $\displaystyle\leq$ $\displaystyle 4-\epsilon^{2}$ $\displaystyle=$ $\displaystyle 4(1-\delta)^{2}.$

Hence, $\|\frac{u+v}{2}\|\leq 1-\delta$.

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 ProofOfHilbertSpaceIsUniformlyConvexSpace 2013-03-22 15:20:48 2013-03-22 15:20:48 Mathprof (13753) Mathprof (13753) 16 Mathprof (13753) Proof msc 46C15 msc 46H05