the modulus of smoothness of Hilbert space

Question

Let $(X, ||.||)$ be a Banach space. For $t>0$, the modulus of smoothness of $||.||$ is defined by $$\rho_X(t)=\sup\left\{\frac{||x+ty||+||x-ty||}{2}-1: x,y\in S_X\right\}.$$ Ican easily calculate $$\rho_{H}(t)\leq\sqrt{1+t^2}-1.$$ Bu I can not prove $$\rho_{H}(t)\geq\sqrt{1+t^2}-1.$$ Please help me

thanks

Answer 1

$\def\norm#1{\left\|#1\right\|}\def\<#1>{\left(#1\right)}$ Let $H$ be a Hilbert space (if $H$ is a real Hilbert space, we need $\dim H \ge 2$), $x,y\in H$ and $t \ge 0$. We have for $x,y \in S_H$: \begin{align*} \norm{x+ty} + \norm{x-ty} &= \<x+ty,x+ty>^{1/2} + \<x-ty,x-ty>^{1/2}\\ &= \Bigl( \<x,x> + 2t\Re\<x,y> + t^2\<y,y>\Bigr)^{1/2} + \Bigl( \<x,x> - 2t\Re\<x,y> + t^2\<y,y>\Bigr)^{1/2}\\ &= \Bigl( 1+t^2 + 2t\Re\<x,y> \Bigr)^{1/2} + \Bigl( 1 + t^2 - 2t\Re\<x,y>\Bigr)^{1/2}\\ \end{align*} Now the supremum of these terms is greater or equal to each of them, espacially to those where $\Re\<x,y> = 0$. Hence $$ \sup_{\norm x = \norm y = 1} \norm{x+ty}+\norm{x-ty} \ge 2(1+t^2)^{1/2}. $$

Answer 2

Let H be a hilbert space then for $t>0$ \roh_H(t)=\sup\left{\frac{t ebsilon{2}-1+(1-ebsilon^2\frac{4})^{1/2}:ebsilon\ge 0 and less of 2}=\leq\sqrt{1+t^2}-1.and know \rho_X(t)=\sup\left{\frac{t\ebsilon\frac\2-\delta_X(ebsilon) :ebsilon between 0 and 2} and delta_X(ebsilon)={1-ebsilon^2\frac{4})^{1/2}}