In this part, we assume that $n$ is a power of 2: we write $n = 2 ^ { k }$ with $k \in \mathbf { N } ^ { \star }$. Deduce that there exists a vector subspace $F$ of dimension $k$ of $\mathbf { R } ^ { n }$ such that: $$\forall x \in F , \quad \alpha _ { 1 } \sqrt { n } \| x \| _ { 2 } ^ { \mathbf { R } ^ { n } } \leq \| x \| _ { 1 } ^ { \mathbf { R } ^ { n } } \leq \beta _ { 1 } \sqrt { n } \| x \| _ { 2 } ^ { \mathbf { R } ^ { n } } .$$ By ordering the $n$ elements of $\{ - 1,1 \} ^ { k }$ arbitrarily, you may use the map $T$ defined on $\mathbf { R } ^ { k }$ by $T \left( a _ { 1 } , \ldots , a _ { k } \right) = \left( \sum _ { i = 1 } ^ { k } a _ { i } \varepsilon _ { i } \right) _ { \left( \varepsilon _ { 1 } , \ldots , \varepsilon _ { k } \right) \in \{ - 1,1 \} ^ { k } }$.
In this part, we assume that $n$ is a power of 2: we write $n = 2 ^ { k }$ with $k \in \mathbf { N } ^ { \star }$. Deduce that there exists a vector subspace $F$ of dimension $k$ of $\mathbf { R } ^ { n }$ such that:
$$\forall x \in F , \quad \alpha _ { 1 } \sqrt { n } \| x \| _ { 2 } ^ { \mathbf { R } ^ { n } } \leq \| x \| _ { 1 } ^ { \mathbf { R } ^ { n } } \leq \beta _ { 1 } \sqrt { n } \| x \| _ { 2 } ^ { \mathbf { R } ^ { n } } .$$
By ordering the $n$ elements of $\{ - 1,1 \} ^ { k }$ arbitrarily, you may use the map $T$ defined on $\mathbf { R } ^ { k }$ by $T \left( a _ { 1 } , \ldots , a _ { k } \right) = \left( \sum _ { i = 1 } ^ { k } a _ { i } \varepsilon _ { i } \right) _ { \left( \varepsilon _ { 1 } , \ldots , \varepsilon _ { k } \right) \in \{ - 1,1 \} ^ { k } }$.