= Solution
Let $P_j$ be the <orthogonal projection> onto $V_j$. The dilation property and the <orthonormal basis> property give the <orthonormal basis> $\varphi_{j,k}(x)=2^{j/2}\varphi(2^jx-k)$ of $V_j$. The periodized nonnegative function
$$
F(u)=\sum_{k\in\mathbb Z}|\varphi(u-k)|^2
$$
has period one and $\int_0^1F=\|\varphi\|_2^2=1$, by <Tonelli theorem>. For a function $g\in L^2(\mathbb R)$ with <compact support> contained in a bounded interval $I$, <Cauchy-Schwarz inequality> and the <Parseval identity> give
$$
\|P_jg\|_2^2=\sum_k|\langle g,\varphi_{j,k}\rangle|^2
\leq \|g\|_2^2\int_I2^jF(2^jx)\,dx
=\|g\|_2^2\int_{2^jI}F(u)\,du\longrightarrow0
$$
as $j\to-\infty$. The last step uses absolute continuity of the integral of a locally $L^1$ function, since the interval's length tends to zero. If $f\in\bigcap_jV_j$, then
$$
|\langle f,g\rangle|=|\langle f,P_jg\rangle|\leq\|f\|_2\|P_jg\|_2\longrightarrow0.
$$
Functions with <compact support> are dense in $L^2$, so $f=0$. Consequently
$$
\boxed{\bigcap_{j\in\mathbb Z}V_j=\{0\}.}
$$
This proof needs neither the density axiom nor any <compact support> assumption on the <scaling function>.
Back to article page
