The book I read, give an excise to show that not every function can be a scaling function to form a MDA.
Let
$\phi(t) = \begin{cases} 1 - 2|t| & \text{if $|t|\leq 1/2$} \\ 0 & \text{otherwise} \\ \end{cases} $
So, if $V_{0} \not\subset V_{1}$, there should be at least one function, that in $V_{0} $ but not in $V_{1}$. Assume, that $V_j = \overline {span_{k\in Z}(\phi_{j,k}(t))}$
$k \in Z$ - translation parameter
$\phi_{j,k} = 2^{j/2}\phi(2^jt-k)$
How to find this function?
Unless I'm missing something finding such function is actually trivial. Consider $$ \phi_{0,k}(t) = \phi(t-k) \Rightarrow {supp(\phi_{0,k})} = \left[-\frac{1}{2}+k,\frac{1}{2}+k\right] $$ For any integer $l$ you have $$ supp(\phi_{1,k+l}) =\left[- \frac{1-2l}{4} + \frac{k}{2},\frac{1+2l}{4} +\frac{k}{2}\right]. $$
If you fix $k=0$ you have $$ \begin{array}{l} {supp(\phi_{0,0})} = \left[-\frac{1}{2},\frac{1}{2}\right] \\ supp(\phi_{1,l}) =\left[- \frac{1-2l}{4},\frac{1+2l}{4}\right] \end{array} $$
Which implies for $l \leq -2$ and $l \geq 2$ $supp(\phi_{0,0}) \cap supp(\phi_{0,l}) = \emptyset$. Therefore you can potentially generate $\phi_{0,0}$ using $\phi_{0,-1}, \phi_{1,0}$ and $\phi_{1,1}$.
Writing down the linear combination $$ \phi_{0,0}(t) = \sum_{-1\leq k \leq 1} \alpha_k \phi_{1,k}(t) $$
You will see, that the only solution $\alpha_{-1},\alpha_0,\alpha_1$ is the only $(0,0,0)$ implying that the functions area linearly independent. Therefore there's no chance you can generate $V_0$ from $V_1$ because there's at least one element in $V_0$ that cannot be written as linear combination of elements in $V_1$.
