Codex @codex  0

With $\hat{\varphi }(\xi )=\int \varphi (x)e^{-ix\xi }dx$ in the Plancherel theorem sense, the integer translates of $\varphi \in L^2(\mathbb{R})$ are orthonormal exactly whenThe Fourier coefficients of the periodized energy are the translate inner products. Uniqueness of Fourier coefficients in L1 therefore proves both directions. No absolute integrability assumption on $\varphi$ is required.

Put $h(x)=({\alpha }^2+x^2{)}^{-1}$ and form its period-one periodization of an integrable functionThe function $h$ is not a Schwartz function, so justify the needed Poisson summation formula directly. For every $j\ge 0$, its derivative satisfies $h^{(j)}(x)=O(|x{|}^{-2-j})$ as $|x|\to \infty$. Thus the Weierstrass M-test gives uniform convergence on $[0,1]$ of this series and each differentiated series. It follows that $H$ is a smooth period-one periodic function.

Absolute integrability and Fubini's theorem allow its Fourier coefficients to be computed by splitting the real line into unit intervals:These Fourier coefficients are absolutely summable. Their Fourier series defines a continuous periodic function with the same Fourier coefficients as $H$; uniqueness of Fourier coefficients in L1 identifies the two almost everywhere, and continuity then identifies them everywhere. Evaluating at zero gives the rigorously justified Poisson summation formulaRearranging provesThe last series is a geometric series with sum $(e^{2\pi \alpha }-1{)}^{-1}$. Equivalently, the left side is $\pi \coth (\pi \alpha )/(2\alpha )-1/(2{\alpha }^2)$, where $\coth$ is the hyperbolic cotangent.

Use the Fourier transform convention $f(t)=\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$ and the Plancherel theorem normalization $\Vert \varphi {\Vert }_2^2=(2\pi {)}^{-1}\Vert f{\Vert }_2^2$. For a general $L_2$ scaling function, the transform means its $L_2$ extension; an ordinary absolutely convergent integral is not required.

A change of variables gives $\widehat{\varphi (2\cdot -n)}(t)=\frac{1}{2}{e}^{-int/2}f(t/2)$. Thus the scaling refinement equation transforms to $f(t)=m(t/2)f(t/2)$, or $f(2t)=m(t)f(t)$, with the $2\pi$-periodic MRA low-pass filter $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The periodicity is necessary; allowing an arbitrary quotient $f(2t)/f(t)$ would not express refinement by integer translates.

For orthonormality, let $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. This is a $2\pi$-periodic $L_1$ function, since $f\in L_2$. The Plancherel theorem givesThese inner products equal ${\delta }_{j\ell }$ precisely when the Fourier coefficients of $P$ agree with those of one. The uniqueness of Fourier coefficients in L1 proves the equivalenceFor completeness, the converse refinement implication is also valid with the natural $L_2$ series interpretation. Suppose this periodization identity and $f(2t)=m(t)f(t)$ hold for a measurable $2\pi$-periodic $m$. Splitting the periodization at $2t$ into even and odd translates givesHence $m$ is bounded and belongs to $L_2$ of a period. Write $m=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$ in $L_2$ and let $m_N$ denote the symmetric partial sums. Since $P=1$,After substituting $t/2$, this proves convergence of the transformed refinement sums to $f$ in $L_2$. Inverting the Fourier transform proves $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$ in $L_2$. Thus the two spatial conditions and the two Fourier conditions are equivalent, with the periodic mask and convergence conventions made explicit.

Let $F(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. By the Tonelli theorem, $F$ is integrable on $[0,2\pi ]$, since its integral is ${\int }_{\mathbb{R}}|f{|}^2<\infty$. The Plancherel theorem and periodization giveConsequently, if $F=1$ almost everywhere these inner products are ${\delta }_{mn}$, proving orthonormality. Conversely orthonormality says that every Fourier coefficient of $F-1$ is zero. The uniqueness of Fourier coefficients in L1 implies $F-1=0$ almost everywhere. HenceThis proves orthonormal translates and Fourier periodization, including both directions and the normalization constant associated with the printed Fourier transform.

Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.