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Let $X=\{0,1\}^{\mathbb{N}}$ be the sequence space and $T:X\to X$ the left shift mapping. Define the vector space $\text{Bow}(X,T)$ as $$ \text{Bow}(X,T)=\{f\in C^{0}(X);~\sup_{n\in \mathbb{N}}\sup_{d_n(x,y)\leq 1/2} |S_nf(x)-S_nf(y)|<\infty\}, $$ where

  • $S_nf=\sum_{j=0}^{n-1}f\circ T^j$;

  • $d_n(x,y)=\max\{d(T^jx,T^jy),~j=0,\ldots,n-1 \} $;

  • $d(x,y)=2^{-N}$, where $N=\inf\{i\in \mathbb{N}\;; x_i\neq y_i\}$.

Consider in the $\text{Bow}(X,T)$ space the norm $$ |f|_{\text{Bow}} = 2\,\|f\|_{\infty} +\sup_{n\geq 1}\ \max_{d_n(x,y)\leq 1/2}|\,S_nf(x)-S_nf(y)\,|. $$

Question: Is $(\text{Bow}(X,T),|\cdot|_{\text{Bow}} )$ a Banach Space?

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  • $\begingroup$ What do you mean by $C^0(X)$? $\endgroup$
    – Ben Willson
    Commented May 14, 2014 at 6:44
  • $\begingroup$ @BenWillson in this context $C^0(X)$ is usually the space of all continuous functions taking values on $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – Leandro
    Commented May 14, 2014 at 7:37
  • $\begingroup$ It would be sufficient to show to show that the unit ball of your space is closed in $C^0(X)$. Did you try this? $\endgroup$
    – Jochen Wengenroth
    Commented May 15, 2014 at 11:13

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