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The density axiom and continuity of the Fourier transform at zero imply $|\hat{\phi }(0)|=1$. Here is a proof that avoids assuming a normalization of the integral. Choose a nonzero $f$ whose Fourier transform has bounded support. The MRA projection Fourier identity, with no aliasing once $j$ is sufficiently large, isSince $\phi \in L^1$, its Fourier transform is continuous and bounded, so this tends to $|\hat{\phi }(0){|}^2\Vert f{\Vert }_2^2$. Density and nesting give $P_{j}f\to f$ in $L^2$. Thus $|\hat{\phi }(0)|=1$, in particular it is nonzero. Evaluating the scaling refinement equation in frequency at zero now giveswhich is stronger than the requested absolute-value equality. Multiplying the scaling function by a constant of modulus one normalizes its integral to one without changing $m$.

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 $\phi$ by the inverse Fourier transform in $L^2$. The Plancherel theorem and the other given integral identities giveDefine $V_j$ as the closed span of ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$. These functions form an orthonormal basis of $V_j$, and dyadic dilation gives the scale axiom. Shifting the infinite product givesso 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 $j$,Uniform convergence of $\Theta (2^{-j}\xi )$ to one on that support proves the limit. Since $P_j$ is an orthogonal projection, $\Vert f-P_{j}f{\Vert }_2^2=\Vert f{\Vert }_2^2-\Vert P_{j}f{\Vert }_2^2\to 0$. Such functions are dense in $L^2$, givingThus 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.

A scaling function generates the approximation space $V_0$ of a multiresolution analysis by its integer translates. In an orthonormal construction those translates form an orthonormal basis, and ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ generate $V_j$. Integrable orthonormal scaling functions have $|\hat{\phi }(0)|=1$, as follows from density and the MRA projection Fourier identity.