TITLE: Existence of a basis such that $\|e_i\|=1$ and $\|e_i^{*}\|_*=1$. (dual)
QUESTION [5 upvotes]: Let $E$ a $n$-finite dimensional normed vector space.
Can we find a basis $e_1,e_2,\cdots,e_n$ of $E$ such that $\|e_i\|=1$ and $\|e_i^{*}\|_*=1$ for all $i$ ?

where $\|\|_*$ is the dual norm.
I know that $n=\dim(E)=\dim(E^*)$ and In the case of finite-dimensional vector spaces, the dual set is always a dual basis.
Furthermore,
$$\|f\|_{*}\ = \sup_{x\in E-\{0\}}\frac{|f(x)|}{\|x\|}=\sup_{||x||=1}|f(x)|$$
Anyway, I don't see how can I tackle this exercise.
Any help will be very grateful,
Thank you in advance for your time.

REPLY [4 votes]: Let $(g_i)_{i\in\mathbb{N}_n}$ be arbitrary basis of $E^*$. Since $E$ is finite dimensional, so does the normed space $\ell_\infty^n(E):=E\bigoplus_\infty\ldots \bigoplus_\infty E$. Hence its unit ball $\operatorname{Ball}_{\ell_\infty^n(E)}$ is compact. Then the continuous function
$$
F:\operatorname{Ball}_{\ell_\infty^n(E)}\to\mathbb{C}:(x_j)_{j\in\mathbb{N}_n}\mapsto \det[g_i(x_j)]_{i,j\in\mathbb{N}_n}
$$
attains its maximum at some $(e_j)_{j\in\mathbb{N}_n}$. Vectors $(e_j)_{j\in\mathbb{N}_n}$ forms a basis of $E$, because otherwise $F(e_1,\ldots,e_n)=0$ but this is not a maximum of $F$. Clearly, $\Vert e_j\Vert=1$ for all $j\in\mathbb{N}_n$ (otherwise you can normalize respective $e_j$ with $\Vert e_j\Vert<1$ and get larger value of $F$). Define
$$
e_i^*(x)=\frac{F(e_1,\ldots,e_{i-1},x,e_{i+1},\ldots,e_n)}{F(e_1,\ldots,e_n)}
$$
Then $(e_i^*)_{i\in\mathbb{N}_n}$ is a basis of $E^*$ with $\Vert e_i^*\Vert_*=1$ and $e_i^*(e_j)=\delta_{i,j}$