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 $Tx=x$.
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 $(c_{00},|| \cdot||_2)$ .