Separability of $L^2$

Asked Apr 07 '18 at 22:21

Active Apr 07 '18 at 22:21

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I am going to reach a contradiction given the following conditions:

Let $f_t$, $t\in I$ be an uncountable family of functions in $L^2([0,1])$ such that $\left \| f_t\right\|=1$ and $\langle f_t,f_s\rangle=c\in (0,1)$ for all $t\neq s$. Then we have a contradiction.

My idea: we know that a separable space cannot have an uncountable orthonormal subset. On the other hand, however, it may have an uncountable subset (for instance, any Hamel basis). How could we turn the above into an orthonormal subset?

asked Apr 07 '18 at 22:21

Tongou Yang

2,055

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Hint: $||f_t-f_s||_2^2=\langle f_t-f_s,f_t-f_s\rangle\ge 2-2c$. – David C. Ullrich Apr 07 '18 at 22:31
We have uncountably many unit vectors that are positively separated, and why does that contradict the separability? – Tongou Yang Apr 08 '18 at 00:53
I have found the solution in this post https://math.stackexchange.com/questions/450106/uncountable-basis-and-separability?rq=1. Thanks! – Tongou Yang Apr 08 '18 at 01:10
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There is a countable dense subset. If $||f_t-f_s||\ge 2r>0$ then the balls $B(f_t,r)$ are disjoint. Each one contains an element of that countable dense subset. – David C. Ullrich Apr 08 '18 at 01:18

Separability of $L^2$

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