What's going on here - density in $L^2$ of $\bigcup_{j\in\mathbf{Z}}V_j$ vs just Haar wavelets

Question

I'm looking at Haar wavelets, and I'm confused about something. We define $V_j$ as the set of all square integrable functions of the form $g(x)=\sum_k a_k\phi(2^jx-k)$. A basic theorem for wavelets states that $\bigcup_{j\in\mathbf{Z}}V_j$ is dense in $L^2(\mathbf{R})$. My question is why is it important that this particular set is dense in $L^2$, vs just the set of wavelets? What's going on here?

Answer 1

Note that for $f\in L_{2}(\mathbb{R}),$ the closest function to $f$ in $V_{j}$ is $P_{V_{j}}(f)=\sum_{k\in\mathbb{Z}}\langle f,\phi_{j,k}\rangle \phi_{j,k},$ where $\phi_{j,k}=2^{j/2}\phi(2^{j}x-k),$ and $\langle f,g\rangle=\int_{\mathbb{R}}f(x)g(x)\mathrm{d}x,$ the usual $L_{2}(\mathbb{R})$ inner product. Since the coefficients depend on $j,$ it is true that we have $g_{j}=P_{V_{j}}(f)\rightarrow f$ in $L_{2}(\mathbb{R}),$ $g_{j}\in V_{j}$ for all $j\geq 1,$ so $g_{j}(x)=\sum_{k\in\mathbb{Z}}a_{k}\phi(2^{j}x-k)$ for some coefficients $a_{k}\in\mathbb{R},$ for each $j,$ but we cannot fix the $a_{k}$ and then take a limit with respect to $j$ and still have convergence to $f.$ In particular, $\langle f,\phi_{j,k}\rangle$ doesn't converge as $j\rightarrow\infty$ for every $f\in L_{2}(\mathbb{R})$, so this won't be possible.