# A question about multiresolution analysis (from a topological point of view)

#### lajka

Hi,

I have a problem understanding something

This is a snapshot of a book I am reading Point no. 2 concerns me, because it looks to me like it contradicts itself, with "this or this"

The first part says

$$\displaystyle \sum_{j}V_j = {L^2(R)}$$ which, to me, looks completely equivavalent to
$$\displaystyle \lim_{j \rightarrow \infty}V_j = {L^2(R)}$$
given the nested nature of these subspaces.

However, the paper says so what troubles me is this: is this countable union $$\displaystyle \sum_{j}V_j$$ equal to $$\displaystyle {L^2(R)}$$ or is it only dense in $$\displaystyle {L^2(R)}$$?

I personally think it's the former, and I don't understand this "dense" part. Could someone perhaps clarify this for me?

Much obliged!

#### Rebesques

It is equal.
What the author wants to state by saying dense, is that for every element $$\displaystyle u$$ of $$\displaystyle L^2$$, there exists a series $$\displaystyle (u_n)\subset \cup_j V_j$$ with $$\displaystyle u=\sum_j u_j$$.

Enter the wavelets!
That is, a Schauder basis for $$\displaystyle L^2$$ with exactly one element in each $$\displaystyle V_j$$.

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