# Is $\text{Bow}(X,T)$ a Banach Space?

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?

• What do you mean by $C^0(X)$? – Ben Willson May 14 '14 at 6:44
• @BenWillson in this context $C^0(X)$ is usually the space of all continuous functions taking values on $\mathbb{R}$ or $\mathbb{C}$. – Leandro May 14 '14 at 7:37
• 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? – Jochen Wengenroth May 15 '14 at 11:13