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Hier is a statement to be proven: A normed space is a pre Hilbert space ( normed space such that the norm is derived from an inner product ), iff all two dimensional subspaces are pre Hilbert spaces, i.e. $\cong l^2(2)$.

  1. I have some difficulty in understanding how $ l^2(2)$ looks like. Given the condition of the inner product on $l^2,$ which is $<(s_i), (s_i)>=\sum_{i=1}^{\infty} s_i\bar s_i < \infty, \,\,$ and the assumption that $l^2(2)$ means two dimensional subspace of $l^2$, I presume that in order for two sequences $(s_i), (t_i)\in l^2(2)$ to be two basis vectors of $l^2(2)$, you need that all entries of the two sequences can be arbitrary ( as long as the sum condition is satisfied ) except for two arbitrary but fixed indices, say $j, k , j\neq k, $ such that the two vectors $(s_i, s_j)\in R^2$ and $(t_j, t_k)\in R^2$ are independent. Is this correct ? If so, given two sequences $(a_n), (b_n) \in l^2(2)$, which I would like to represent through the basis vectors $(s_i), (s_j), $ I could at most chose two scalars to get the entries of the sequences $(a_n), (b_n)$ at the positions $j$ and $k$ but not the other entries ( not even in the simplest case $(s_i, s_j)=(1,0), (t_j, t_k)=(0,1)$ ). Can somebody help me out to understand correctly the two dimensionality of $l^2(2)$.

  2. Given the statement above, I started with the necessary and sufficent condition for the normed space $X$ to be a pre Hilbert space: $$ {\left\lVert x+y\right\rVert}^2 + {\left\lVert x-y \right\rVert}^2=2{\left\lVert x \right\rVert}^2 + 2 {\left\lVert y \right\rVert}^2, \, \, \forall x, y \in X.$$ The first thing I dont understand is, what is meant by ''all'' two dimensional subspaces ? Can somebody offer a proposal how to prove the statement ? Many thanks in advance.

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    $\begingroup$ Isn't the crux of your problem to show that there is an inner product on your normed space, that is, there exists, for any $x$ and $y$ a quantity $B(x,y)$ obeying such-and-such properties? Why not look at the two dimensional space spanned by $x$ and $y$, for which you do know there is an inner product. You have to show that these "little" inner products on all these 2 dimensional subspaces can be fit together into one "global" inner product $B(x,y)$ on the original big space. What's the obstacle preventing this? $\endgroup$ – kimchi lover Feb 4 '18 at 15:44
  • $\begingroup$ Thanks. I talked about the obstacles above. I thought it was important to first understand the structure of $l^2(2).$ Even in the case of $R^3$ I dont see it trivial from the "little" inner products $(R^2)$ to conclude for the inner product in $R^3.$ The other way round will be simpler, by setting zero to one of the coordinates of the inner product in $R^3.$ How you would procede ? $\endgroup$ – user249018 Feb 4 '18 at 16:36
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If you're allowed to use the non-trivial theorem that a normed space is pre-Hilbert if and only if $$ {\left\lVert x+y\right\rVert}^2 + {\left\lVert x-y \right\rVert}^2=2{\left\lVert x \right\rVert}^2 + 2 {\left\lVert y \right\rVert}^2, \, \, \forall x, y \in X\quad(*)$$ then there's a solution much simpler than what kimchi lover suggested. Given $x$ and $y$, let $V$ be the subspace spanned by $x,y$. You're given that $V$ is a pre-Hilbert space, so (*) follows.

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  • $\begingroup$ Thanks. You are right that for all two dimensional subspaces $V$, if assumed pre-Hilbert, then the relation will hold. But, I guess, there must be a further step to prove that the same relation will hold for the bigger space $X.$ What stuck me most in the statement is the assertion that the 2-dimensional subspaces....and not the 1-dimensional, or 3-dimensional subspaces must satisfy the condition of being pre-Hilbert. Why is this? $\endgroup$ – user249018 Feb 4 '18 at 16:13

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