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

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

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

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

However, the paper says

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

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

Much obliged!

2. ## Re: A question about multiresolution analysis (from a topological point of view)

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

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