0
$\begingroup$

As the title says, i came across this question and i'm wondering how to prove it.

If $(X,\langle\cdot,\cdot\rangle)$ is an inner product space and it has a countably infinite Hamel basis $(e_n)_{n\in \mathbb N}$ which is an orthonormal set, show that $X$ is isometric to $(c_{00},|| \cdot||_2)$, where $||\cdot||_2$ is the norm: $||x||_2=(\sum_{i=1} ^{\infty} |x_i|^2)^{1/2}$ for $x=(x_1,x_2,...,x_n,0,0,...)\in c_{00}$.

My line of thinking was: The standard Hamel basis for $c_{00}$ is $\{e_1,e_2,...\}$ which is also orthonormal and countably infinite (can we say they are the same?).
Because they both are countably infinite (so they have the same dimension) there exists an isomorphism $T$ between $X$ and $c_{00}$. So we need to prove that it is also an isometry.
Let $||\cdot||$ be the norm that follows from the inner product of $X$, so that $||x||=\langle x,x\rangle^{1/2}$.
We must prove that for each $x\in X$, $||x||=||x||_2$ ( or is it more correct to say prove $||x||=||Tx||_2$ ?).
But $||x||=\langle x,x\rangle ^{1/2}=(\sum_{i=1} ^{\infty} |x_i|^2)^{1/2}=||x||_2$

I guess my question is in the points in the above proof about what it means for these two bases to be isomorphic, and if this isomorphism between between $X$ and $c_{00}$ means that $Tx=x$.

  • 1
    Do you probably mean: It has a countably infinte *Hamel* basis which is additionally orthonormal? In case you don't know: A hamel basis is a basis in the sense of linear algebra, i.e. it is linear independent and each vector is a **finite** linear combination of its elements. – 2017-01-18
  • 0
    Ohh, my bad, you're right! I'm editing the question. – 2017-01-18
  • 0
    The appropriate code for brackets is `\langle` and `\rangle`. Compare $$ with $\langle x,y\rangle$. – 2017-01-18

1 Answers 1

4

The space $\ell^2(\mathbb{N})$ is an inner product space with a countable orthonormal basis, but it is complete (i.e. it is a Hilbert space) whereas $(c_{00},|| \cdot||_2)$ is not. Hence the two spaces cannot be isometrically isomorphic.

Given the changes made to the problem, let $X$ be an IPS with a countably infinite orthonormal Hamel basis $\{f_n\}$, and let $\{e_n\}$ denote the standard basis of $c_{00}$. Denote by $T$ the map from $X$ to $c_{00}$ defined by linear extension of $$ Tf_n=e_n.\qquad \tag{$n\in\mathbb N$} $$ Then $T$ is linear, $1$-$1$, and onto (all of this is easily checked), hence $T$ is an isomorphism. Furthermore, given $x=\sum_{k=1}^n\alpha_kf_k\in X$, we have $Tx=\sum_{k=1}^n\alpha_ke_k$, \begin{align} \|x\|^2&=\langle x,x \rangle \\ &=\left\langle\sum_{k=1}^n\alpha_kf_k,\sum_{k=1}^n\alpha_kf_k \right\rangle \\ &= \sum_{k=1}^n|\alpha_k|^2, \end{align} and thus \begin{align} \|Tx\|^2= \sum_{k=1}^n|\alpha_k|^2=\|x\|^2. \end{align} Hence $\|Tx\|=\|x\|$, and therefore $T$ is an isometric isomorphism of $(X,\langle\cdot,\cdot\rangle)$ onto $(c_{00},|| \cdot||_2)$ .

  • 0
    This is not what i am asking. The question is whether $c_{00}$ with the $||\cdot||_2$ norm is isometrically isomorphic to an inner product space with a countably infinite basis. – 2017-01-18
  • 1
    @ManoPlizzi Rather than downvote, you could just comment and let me know, so I can change the nature of my answer. – 2017-01-18
  • 0
    Of course i'll upvote when you change it, i just found it off topic in it's current form, no hard feelings. – 2017-01-18