= Solution
At zero, the partial products are $m(0)^n$. Their assumed convergence and $|m(0)|=1$ force $m(0)=1$ and $\Theta(0)=1$. Uniform convergence on compact sets makes $\Theta$ continuous. The identity with $k=0$ gives $\Theta\in L^2$ with squared norm $2\pi$. Define $\varphi$ by the inverse <Fourier transform> in $L^2$. The <Plancherel theorem> and the other given integral identities give
$$
\langle\varphi(\cdot-k),\varphi(\cdot-l)\rangle=\delta_{kl}.
$$
Define $V_j$ as the closed span of $\varphi_{j,k}(x)=2^{j/2}\varphi(2^jx-k)$. These functions form an <orthonormal basis> of $V_j$, and dyadic dilation gives the scale axiom. Shifting the infinite product gives
$$
\Theta(2\xi)=m(\xi)\Theta(\xi),
$$
so the finite Fourier series of $m$ gives a <scaling refinement equation> and $V_j\subset V_{j+1}$. The coarse-scale projection argument of part 1(b) proves the trivial-intersection axiom.
For density, take a function $f$ with bounded Fourier support. The <MRA projection Fourier identity> gives, for large $j$,
$$
\|P_jf\|_2^2=\frac1{2\pi}\int|\widehat f(\xi)|^2|\Theta(2^{-j}\xi)|^2\,d\xi\longrightarrow\|f\|_2^2.
$$
Uniform convergence of $\Theta(2^{-j}\xi)$ to one on that support proves the limit. Since $P_j$ is an <orthogonal projection>, $\|f-P_jf\|_2^2=\|f\|_2^2-\|P_jf\|_2^2\to0$. Such functions are dense in $L^2$, giving
$$
\boxed{\overline{\bigcup_jV_j}=L^2(\mathbb R).}
$$
Thus all the <multiresolution analysis> axioms hold. Under the strong convergence and orthogonality assumptions supplied here, the additional nonvanishing condition is not needed in this last verification; it is useful when establishing those assumptions from a filter.
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