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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.

Community
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    An orthonormal basis of E fits your needs... – Gabriel Romon Apr 11 '14 at 10:03
  • How exactly are you defining $e^*_i$? Is it the linear functional such that $\sum_j a_j e_j \mapsto a_i$? – Will Nelson Apr 11 '14 at 20:06
  • One method of constructing such a basis would be to choose $e_1$ arbitrarily with $||e_1||=1$, choose $e_2$ such that $e_1^* e_2 = 0$ and $||e_2||=1$, and in general choose $e_i$ such that $e_j^* e_i = 0 $ and $||e_i||=1$ for all $j<i$. – Will Nelson Apr 11 '14 at 20:10

1 Answers1

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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 $e_i^*(e_j)=\delta_{i,j}$

Norbert
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  • I use the following notation $\ell_\infty^n(E)={(x_1,\ldots,x_n):\forall i\in\mathbb{N}n\quad x_i\in E}$. This is a normed space with norm $$\Vert(x_1,\ldots,x_n)\Vert{\ell_\infty^n(E)}=\max\limits_{i\in\mathbb{N}_n} \Vert x_i\Vert$$ – Norbert Apr 12 '14 at 13:09
  • I don't understand finally you have $\Vert e_i^\Vert=1$ not $|e_i^{}|_*=1$ ? – Free X Apr 18 '14 at 13:00
  • @FreeX, fixed${}{}{}$ – Norbert Apr 18 '14 at 19:16