Let $(X,\|\cdot\|)$ be a Banach space. For $t>0$, the modulus of smoothness of $\|\cdot\|$ is defined by $\rho_X(t)=\sup\left\{\dfrac{\|x+ty\|+\|x−ty\|}{2}−1:x,y\in S_X\right\}$. We define an equivalent norm $\|\cdot\|$ on $c_0$ by setting$$\|x\|=\sup_{n\in N}|x_n|+\sum_{n=1}^\infty2^{-n}|x_n|, \text{ for all } x=(x_n)_{n=1}^\infty \in c_0$$. I want to find the modulus of smoothness of $(c_0, \|\cdot\|)$. Thanks you help

  • $\begingroup$ Why? If you want to know whether the norm is uniformly smooth -- it's not, since the space is not reflexive. What use is the modulus of smoothness of a norm that is not uniformly smooth? It's like computing the modulus of continuity for a function that is not uniformly continuous. $\endgroup$ – user147263 Jun 8 '14 at 16:27
  • $\begingroup$ I changed \operatorname{sup} to \sup. In a "displayed" as opposed to "inline" context, that affects positions of subscrips, thus: $\displaystyle\sup_{a\in A}f(a)$. It is standard usage. $\endgroup$ – Michael Hardy Jun 8 '14 at 16:27
  • $\begingroup$ I know this space in not uniformly smooth and so $\lim\limits_{t\to 0}\frac{\rho_X(t)}{t}$ in not 0. I like find this limit. $\endgroup$ – user62498 Jun 8 '14 at 16:49

By the triangle inequality, the modulus of smoothness satisfies $\rho_X(t)\le t$. I claim that in fact $\rho_X(t)=t$ for all $t$.

Consider the two-dimensional space $V_n = \{ \alpha e_{n} +\beta e_{n+1}: \alpha,\beta\in\mathbb R\}$. Observe that $$ \max(|\alpha|,|\beta|) \le \|x \| \le (1+2^{1-n})\max(|\alpha|,|\beta|)$$ Therefore, as $n\to\infty$, the modulus of smoothness of $V_n$ tends to the modulus of smoothness of two-dimensional $\ell_\infty$ space, $\ell_\infty^2$. For the latter the modulus is $\rho(t)=t$, as is shown by $x=e_{n}+e_{n+1}$ and $y = t(e_n-e_{n-1} )$.


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