Prove that the set of all sequences of real numbers is not a inner product space
The author gave me an outline of the proof. Let $L$ be the set of all sequences of real numbers. For each $i$, let $e_i$ be the sequence $x_1, x_2, \dots, x_i, \dots$ with $x_i = 1$ and $x_r = 0$ for $r\neq i$.
Suppose that an inner product can be defined in $L$, so as to give us an inner product space. Show that for each $i$, and each $M_i>0$, there is an $\alpha_i$ such that $\|\alpha_ie_i\| > M_i$. (Here, as usual, $\|A\| = \sqrt{A\cdot A}$).
Show that there is a sequence $\alpha_i, \alpha_2, \dots$ such that $\|\sum_{i = 1}^v\alpha_ie_i\| > i, \forall i$.
Let $s = \alpha_1, \alpha_2, \ldots \in L$. Let $k = \|s\|$, let $s_v = \sum_{i = 1}^v\alpha_ie_i$, and let $t_v = s = s_v$, so that $s = s_v + t_v$. Show that the $\alpha_i$'s can be chosen so that $s_v - t_v \ge 0$. Now show that this whole situation is impossible.