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Fedor Petrov
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For non-zero sequence $x=(x_1, x_2, \ldots)$ denote $L(x)=\min\{i:x_i\ne 0\}$ (leader of $x$). By Gauss elimination, $M$ contains a basis $(p_1, \ldots, p_d) $ with distinct leaders $m_1<m_2<\ldots<m_d$ respectively. Let $f_1, f_2, \ldots$ denote consecutive standard basic vectors $e_k$ with $k\notin \{m_1,\ldots,m_d\}$. Consider the span $N$ of $M$ and $f_1, \ldots, f_s$ with very large $s$.On On this $(d+s)$-dimensional space $N$ with the natural basis $\{p_1,\ldots,p_d,f_1,\ldots,f_s\}$, the coordinate functionals corresponding to $m_i$$p_i$'s are uniformly (by $s$) bounded that is proved by induction: actually their expressions via the usual coordinate functionals do not depend on $s$. Now you cut all $m_j$$p_j$'s on the corresponding level $d+s$: denote $n_i=P_{d+s} p_i$ where $P_{d+s} x=(x_1,\ldots,x_{d+s},0,0,\ldots)$. The span $N$ of $n_1,\ldots,n_d,f_1,\ldots,f_s$ is the coordinate subspace $l_\infty^{d+s}\subset c_0$. The map $\Phi\colon M\to N$ which acts as $$\Phi\left( \alpha_1p_1+\ldots+\alpha_sp_s+\beta_1f_1+\ldots+\beta_s f_s\right)= \alpha_1n_1+\ldots+\alpha_sn_s+\beta_1f_1+\ldots+\beta_s f_s$$ satisfies $\|x-\Phi x\|<\varepsilon_s$ for all $x$ on the unit sphere of $M$ and get an $l^\infty$ space very$\varepsilon_s$ tending to 0 for large $s$ (since $\alpha_i's$ are uniformly bounded as observed above, and $n_i$'s become close to $p_i$'s when $s$ is large). This proves that $N$ and $M$ are Banach -- Mazur close.

The existence of a norm-1 projection to $l_\infty^n$ is a general fact, as observed by Dirk Werner in the comments.

For non-zero sequence $x=(x_1, x_2, \ldots)$ denote $L(x)=\min\{i:x_i\ne 0\}$ (leader of $x$). By Gauss elimination $M$ contains a basis $(p_1, \ldots, p_d) $ with distinct leaders $m_1<m_2<\ldots<m_d$ respectively. Let $f_1, f_2, \ldots$ denote consecutive standard basic vectors $e_k$ with $k\notin \{m_1,\ldots,m_d\}$. Consider the span $N$ of $M$ and $f_1, \ldots, f_s$ with very large $s$.On this space $N$, the coordinate functionals corresponding to $m_i$'s are uniformly (by $s$) bounded that is proved by induction. Now you cut all $m_j$'s on the corresponding level $d+s$ and get an $l^\infty$ space very close to $N$.

For non-zero sequence $x=(x_1, x_2, \ldots)$ denote $L(x)=\min\{i:x_i\ne 0\}$ (leader of $x$). By Gauss elimination, $M$ contains a basis $(p_1, \ldots, p_d) $ with distinct leaders $m_1<m_2<\ldots<m_d$ respectively. Let $f_1, f_2, \ldots$ denote consecutive standard basic vectors $e_k$ with $k\notin \{m_1,\ldots,m_d\}$. Consider the span $N$ of $M$ and $f_1, \ldots, f_s$ with very large $s$. On this $(d+s)$-dimensional space $N$ with the natural basis $\{p_1,\ldots,p_d,f_1,\ldots,f_s\}$, the coordinate functionals corresponding to $p_i$'s are uniformly (by $s$) bounded: actually their expressions via the usual coordinate functionals do not depend on $s$. Now you cut all $p_j$'s on the corresponding level $d+s$: denote $n_i=P_{d+s} p_i$ where $P_{d+s} x=(x_1,\ldots,x_{d+s},0,0,\ldots)$. The span $N$ of $n_1,\ldots,n_d,f_1,\ldots,f_s$ is the coordinate subspace $l_\infty^{d+s}\subset c_0$. The map $\Phi\colon M\to N$ which acts as $$\Phi\left( \alpha_1p_1+\ldots+\alpha_sp_s+\beta_1f_1+\ldots+\beta_s f_s\right)= \alpha_1n_1+\ldots+\alpha_sn_s+\beta_1f_1+\ldots+\beta_s f_s$$ satisfies $\|x-\Phi x\|<\varepsilon_s$ for all $x$ on the unit sphere of $M$ and $\varepsilon_s$ tending to 0 for large $s$ (since $\alpha_i's$ are uniformly bounded as observed above, and $n_i$'s become close to $p_i$'s when $s$ is large). This proves that $N$ and $M$ are Banach -- Mazur close.

The existence of a norm-1 projection to $l_\infty^n$ is a general fact, as observed by Dirk Werner in the comments.

Source Link
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

For non-zero sequence $x=(x_1, x_2, \ldots)$ denote $L(x)=\min\{i:x_i\ne 0\}$ (leader of $x$). By Gauss elimination $M$ contains a basis $(p_1, \ldots, p_d) $ with distinct leaders $m_1<m_2<\ldots<m_d$ respectively. Let $f_1, f_2, \ldots$ denote consecutive standard basic vectors $e_k$ with $k\notin \{m_1,\ldots,m_d\}$. Consider the span $N$ of $M$ and $f_1, \ldots, f_s$ with very large $s$.On this space $N$, the coordinate functionals corresponding to $m_i$'s are uniformly (by $s$) bounded that is proved by induction. Now you cut all $m_j$'s on the corresponding level $d+s$ and get an $l^\infty$ space very close to $N$.