Dealing with Linear Spans in an Infinite-Dimensional Space

Question

When dealing with the concept of linear span, which is defined as the set of all finite linear combinations of a set of vectors in a vector space $V$ $$ \operatorname{span}(V)=\left\{ \sum^n_{i=1} \lambda_i v_i:v_i \in V \right\} $$ for $n=1,2,\ldots,n$ and $\lambda_{i} \in \mathbb{R}$ (or more generally, any field). Obviously $\operatorname{span}(V)$ is a linear subspace of $V$. Now let's say that $V$ is a Banach space and $V_n$ is a sequence of infinite-dimensional subspaces of $V$. Clearly, $S_n=\operatorname{span}(V_n)$ is a finite-dimensional subspace of $V_n$. So my question is whether we can take an infinite direct sum of a sequence of linear spans such that we can recover the original space $V$: $$\bigoplus^\infty_{n=1}S_n=V$$

To further motivate my question in the context of a multi-resolution analysis, let $V=L^{2}(\mathbb{R})$ be a Hilbert space and let $V_{n}$ be a nested sequence of subspaces of $V$ such that $$V_{-\infty}=\{0\}\subset \cdots \subset V_{-n} \subset \cdots\subset V_{0}\subset \cdots \subset V_{n} \subset \cdots V_{\infty}=L^{2}(\mathbb{R})$$ with the conditions that $f(t) \in V_n \implies f(2t) \in V_{n+1}$ and for a sequence of (scaling) functions $\varphi_{j,k}(t)=2^{j/2}\varphi(2^j t-k)$, we have that $$ V_j=\operatorname{span}_k\{\varphi_{j,k}(t)\} $$ which seems to imply that any $f \in V_j$ can be written as a finite sum $$ f(t)=\sum^n_{k=-n}\alpha_k \varphi(2^j t+k) $$ where $j$ is fixed. Here is where my understanding breaks down. Obviously each $V_j$ is a finite-dimensional subspace of $L^2(\mathbb{R})$ (by the definition of span) and as $j$ increases, finer and finer functions are admitted to the space and each space is spanned by it's finite integer translates. My intuition tells me that if you take an indicator function and half the support (dilation), you can recover the original function by adding the dilated function translated by one to the dilated function. For example, let $f(t)=\mathbb{1}_{[0,1]}$, then $f(2t)$ is given by

$$ f(2t)= \begin{cases} 1 & \text{if } t \in [0,1/2] \\ 0 & \text{otherwise } \end{cases} $$ and $$ f(t)=f(2t)+f(2t-1) $$ However, I am not sure if this works for more complicated functions. Can someone clarify this for me?

Answer 1

You wrote:

Obviously each $V_j$ is a finite-dimensional subspace of $L^2(\mathbb{R})$ (by the definition of span)

That is wrong. And I wonder how you propose to create an infinite descending chain of finite-dimensional spaces, unless it's an eventually constant sequence.

Take a simple case: the space of all sequences $(x_1,x_2,x_3,\ldots)$ of real numbers. That is an infinite-dimensional vector space over the reals. And $$ \operatorname{span}\left\{ \begin{array}{c} (1,0,0,0,0,\ldots), \\ (0,1,0,0,0,\ldots), \\ (0,0,1,0,0,\ldots), \\ (0,0,0,1,0,\ldots), \\ \vdots \end{array} \right\} $$ is the subspace consisting of all sequences with only finitely many non-zero terms. That subspace, the span, is an infinite-dimensional space. Each of its members is a linear combination of finitely many of the members of the set whose span was taken.

Answer 2

For the wavelet part, you are somewhat on the right track. If you don't know this already, the wavelet function $\psi$ is given by $$\psi(t)=\sum^{\infty}_{k=-\infty} \alpha_{k} \sqrt{2} \varphi(2t-k)$$ which follows from the fact that $W_{0} \subset V_{1}$ and $V_{1}=V_{0} \oplus W_{0}$ BUT the value of the coefficient $\alpha_{k} \neq 0$ only for a finite number of values of $k$. Let us consider the Haar wavelet system with scaling function $\varphi(t)$ be given by $$ \varphi(t)=\begin{cases} 1 & \text{if } x \in [0,1]\\ 0 & \text{otherwise} \end{cases} $$ then our wavelet system can be given by $$ \psi(t)=\sum^{\infty}_{k=-\infty} \alpha_{k}\sqrt{2} \phi(2t-k) $$ which with coefficients $\alpha_{0}=\frac{1}{\sqrt{2}}$ and $\alpha_{1}=\frac{-1}{\sqrt{2}}$ and the rest of the coefficients zero, gives us $$ \psi(t)=\varphi(2t)-\varphi(2t-1) $$ which is the mother wavelet for the Haar family. The basic idea is that as your functions get finer and finer (i.e. $V_{j}$ as $j \to \infty$) fewer of the coefficients for integer translates by $k$ are zero (but each $V_{j}$ is still infinite dimensional) until you get to the entire space $L^{2}(\mathbb{R})$ where any function $f(t) \in L^{2}(\mathbb{R})$ can be represented by $$ f(t)=\sum^{\infty}_{k=-\infty} c_{k}\varphi_{k}(t)+\sum^{\infty}_{j=0}\sum^{\infty}_{k=-\infty} d_{j,k}\psi_{j,k}(t)$$ where $$ c_{k}=\langle f(t),\varphi_{k}(t)\rangle $$ and $$ d_{j,k}=\langle f(t), \psi_{j,k}(t) \rangle $$ with the caveat that $j=0$ was chosen arbitrarily as our starting scale.