Is it possible to define a euclidean structure on infinite dimensional vector space

Question

It is known that for every infinite dimensional vector space $V$, there exists a Hamel basis $B$, such that each $v\in V$ can be repsentated as $$v=\sum_{i=1}^{n(v)}a_ie_i$$ with $e_i \in B$. Now my question is, can we define a euclidean norm on V in the sense of $$\parallel v \parallel=\sqrt{\sum_{i=1}^{n(v)}\mid a_i\mid^2 } $$?

Answer 1

You've indeed described a norm, and it's also induced by a scalar product such that $\langle e_i,e_j\rangle=\delta_{ij}$ for all $i,j$ (such a scalar product is unique, given an algebraic basis). This norm is complete if and only if the dimension of the space is finite. This is proved by the fact that if $\{e_{i_1},e_{i_2},e_{i_3},\cdots\}$ is a countable subsequence of $B$, then the sequence of vectors $v_n=2^{-n}e_{i_n}$ satisfies $\sum_{j=1}^\infty \lVert v_j\rVert<\infty$, despite there being no vector $w$ such that $\lim_{n\to\infty}\left\lVert w-\sum_{j=1}^n v_j\right\rVert=0$.

In fact, let there be such $w$ and $w=\sum_{h=1}^m w_he_h+\sum_{j=1}^n w_{i_j}e_{i_j}$ - where $e_1,\cdots,e_m$ are not among the $\{e_{i_1},e_{i_2},\cdots\}$. Then, for all $r\ge n+1$ $$\left\lVert w-\sum_{j=1}^r v_j\right\rVert=\sqrt{\sum_{h=1}^m \lvert w_h\rvert^2+\sum_{j=1}^n \lvert w_{i_j}-2^{-j}\rvert^2+\sum_{j=n+1}^{r}2^{-2j}}\ge 2^{-(n+1)}$$

So the sequence of those quantities can't converge to $0$.