The Adjoint of the Fourier Operator

Introduction

The Fourier operator is unitary, and for a unitary operator the adjoint is the inverse; this single observation is the operator form of the Fourier inversion formula. On $L^2(\mathbb R^n)$, with the transform $$ (\mathcal Ff)(\xi)=\int_{\mathbb R^n}f(x)e^{-2\pi ix\cdot\xi}\,dx , $$ the adjoint is the conjugate transform, $$ \mathcal F^\dagger=\overline{\mathcal F}=\mathcal F^{-1} , $$ so that $\mathcal F^\dagger\mathcal F=\mathcal F\mathcal F^\dagger=\mathrm{id}$ is exactly the pair of inversion identities, and applied to the transform the statement $\mathcal F^\dagger=\mathcal F^{-1}$ reads $f(x)=\int\hat f(\xi)e^{2\pi ix\cdot\xi}d\xi$. The kernel of the operator is $K(x,\xi)=e^{-2\pi ix\cdot\xi}$ and the conjugate transpose kernel is $K^*(x,\xi)=\overline{K(\xi,x)}=e^{2\pi ix\cdot\xi}$, the kernel of the inverse; the operator is also symmetric, $\mathcal F^{\mathrm t}=\mathcal F$, because its kernel is symmetric in $x$ and $\xi$, so that the adjoint and the transpose differ only by the conjugation, as for every integral operator. The phrase unitary up to a constant records the dependence on the normalisation: in the angular-frequency convention the transform is not an isometry but has adjoint $(2\pi)^n$ times its inverse, and it becomes unitary after the multiplication by $(2\pi)^{-n/2}$.

The article is the fourth and last of the * Operator Theory group of Foundations of Analysis. Its prerequisites are The Fourier Operator, the third article of the first group of this category, for the transform, the Plancherel theorem and the unitarity, and for the explicit deferral to this article; Fourier Analysis on Euclidean Spaces for the transform on $L^1$ and the Schwartz class, the inversion theorem and the convolution theorem; The Fourier Transform and Conjugate Symmetry for the conjugate symmetry $\widehat{\bar f}(\xi)=\overline{\hat f(-\xi)}$; The Adjoint of an Integral Operator and Involutions of the Convolution Operators, the earlier articles of this group, for the adjoint and the multiplier forms; and Banach and Hilbert Spaces, later in this Part, for the adjoint and the unitary operators, quoted as established. The transform on a general locally compact abelian group, the Pontryagin dual and the Plancherel theorem there are Harmonic Analysis on Groups and Pontryagin Duality, in the neighbouring category Analysis on Groups of this Part, and are named here as forward references only; the distributional transform is Distributions and Fundamental Solutions, later in this Part. No geometry is invoked.

The Fourier Operator and Its Pairing

The Transform

Definition. On the Schwartz class the Fourier operator is $$ (\mathcal Ff)(\xi)=\int_{\mathbb R^n}f(x)e^{-2\pi ix\cdot\xi}\,dx , $$ and it extends to a unique bounded operator on $L^2(\mathbb R^n)$ with $$ \lVert\mathcal Ff\rVert_2=\lVert f\rVert_2,\qquad \langle\mathcal Ff,\mathcal Fg\rangle=\langle f,g\rangle , $$ for the Hilbert pairing $\langle f,g\rangle=\int f\bar g$; it is onto, hence unitary, and its inverse is the conjugate transform, $\mathcal F^{-1}=\overline{\mathcal F}$, where $(\overline{\mathcal F}f)(\xi)=\int f(x)e^{2\pi ix\cdot\xi}dx$. The transform has order four, $$ \mathcal F^2=R,\qquad \mathcal F^4=\mathrm{id},\qquad Rf(x)=f(-x) , $$ $R$ being the reflection.

The Pairing and the Kernel

Definition. The kernel of $\mathcal F$ is $K(x,\xi)=e^{-2\pi ix\cdot\xi}$, so that $$ (\mathcal Ff)(\xi)=\int_{\mathbb R^n}K(x,\xi)f(x)\,dx , $$ and the conjugate transpose kernel is $$ K^*(x,\xi)=\overline{K(\xi,x)}=e^{2\pi ix\cdot\xi} , $$ the kernel of the conjugate transform. The transform is symmetric, $K(\xi,x)=K(x,\xi)$, equivalently $\mathcal F^{\mathrm t}=\mathcal F$ with respect to the bilinear pairing $\{f,g\}=\int fg$.

Remark. The kernel is not in $L^2(\mathbb R^n\times\mathbb R^n)$, so the transform is not Hilbert–Schmidt; it is unitary, and the kernel is a tempered distribution, the pairing of which against a test function reproduces the transform. The distributional kernel is Distributions and Fundamental Solutions, later in this Part; the kernel notation used here is the same as for an integral operator, and the identities are proved by the same change of variable wherever the integrals converge.

The Adjoint Is the Inverse

The Computation

Theorem. The adjoint of the Fourier operator is the conjugate transform, and the conjugate transform is the inverse: $$ \mathcal F^\dagger=\overline{\mathcal F}=\mathcal F^{-1} . $$ Equivalently, $\langle\mathcal Ff,g\rangle=\langle f,\overline{\mathcal F}g\rangle$ for all $f,g\in L^2$.

Proof. On the Schwartz class, where the integrals converge absolutely, $$ \langle\mathcal Ff,g\rangle=\int_{\mathbb R^n}\hat f(\xi)\overline{g(\xi)}\,d\xi =\int_{\mathbb R^n}\!\!\int_{\mathbb R^n}f(x)e^{-2\pi ix\cdot\xi}\overline{g(\xi)}\,dx\,d\xi =\int_{\mathbb R^n}f(x)\overline{\Bigl(\int_{\mathbb R^n}g(\xi)e^{2\pi ix\cdot\xi}\,d\xi\Bigr)}\,dx =\langle f,\overline{\mathcal F}g\rangle , $$ by Fubini, and the identity extends to $L^2$ by the density of the Schwartz class and the boundedness of the two operators. The inverse transform is the conjugate transform by the inversion theorem, $\overline{\mathcal F}\mathcal F=\mathcal F\overline{\mathcal F}=\mathrm{id}$ on the Schwartz class, extended by continuity; hence $\mathcal F^\dagger=\overline{\mathcal F}=\mathcal F^{-1}$. $\blacksquare$

The Inversion Formula as Unitarity

Theorem. The Fourier operator is unitary, $$ \mathcal F^\dagger\mathcal F=\mathcal F\mathcal F^\dagger=\mathrm{id},\qquad \mathcal F^\dagger=\mathcal F^{-1}=\mathcal F^3 , $$ and the two identities are the two Fourier inversion formulas: the first, applied at a point, reads $$ f(x)=\int_{\mathbb R^n}\hat f(\xi)e^{2\pi ix\cdot\xi}\,d\xi , $$ and the second is the conjugate statement. In the kernel language the first identity is $$ K^*(x,\xi)=\overline{K(\xi,x)}=e^{2\pi ix\cdot\xi} , $$ with the left side the kernel of the adjoint and the right side the kernel of the inverse.

Proof. Unitarity is Plancherel's theorem of The Fourier Operator, the previous group; the adjoint is the inverse by the theorem above; the order-four relation gives $\mathcal F^{-1}=\mathcal F^3$; the pointwise form of $\mathcal F^\dagger\mathcal F=\mathrm{id}$ is the inversion theorem, and the kernel computation is the definition of the conjugate transpose kernel of the previous articles. $\blacksquare$

Corollary (the Plancherel identity as adjointness). For $f,g\in L^2$, $$ \langle\mathcal Ff,g\rangle=\langle f,\mathcal F^{-1}g\rangle , $$ and the transform preserves the inner product, $$ \langle\mathcal Ff,\mathcal Fg\rangle=\langle f,g\rangle . $$

Proof. The first identity is $\mathcal F^\dagger=\mathcal F^{-1}$ written as a pairing identity; replacing $g$ by $\mathcal Fg$ in it and using $\mathcal F^{-1}\mathcal F=\mathrm{id}$ gives the second. $\blacksquare$

Unitarity up to a Constant

The Three Conventions

Theorem (the constant of the normalisation). Let $$ (\mathcal F_1f)(\xi)=\int_{\mathbb R^n}f(x)e^{-ix\cdot\xi}\,dx $$ be the transform in the angular-frequency convention. Then its adjoint is $(2\pi)^n$ times its inverse, $$ \mathcal F_1^\dagger=(2\pi)^n\mathcal F_1^{-1},\qquad \mathcal F_1^\dagger\mathcal F_1=(2\pi)^n\,\mathrm{id} , $$ so $\mathcal F_1$ is unitary up to the constant $(2\pi)^{-n/2}$: $$ \lVert\mathcal F_1f\rVert_2^2=(2\pi)^n\lVert f\rVert_2^2,\qquad (2\pi)^{-n/2}\mathcal F_1\ \text{is unitary}. $$ In the ordinary-frequency convention $\mathcal Ff(\xi)=\int f(x)e^{-2\pi ix\cdot\xi}dx$ the constant is $1$, $\mathcal F^\dagger\mathcal F=\mathcal F\mathcal F^\dagger=\mathrm{id}$, and the transform is exactly unitary.

Proof. The angular transform is the composition of the ordinary transform with the dilation $\xi\mapsto2\pi\xi$, so its adjoint carries the scaling factor $(2\pi)^n$; the isometry constant is computed from $\lVert\mathcal F_1f\rVert_2^2=\int\lvert f(x)e^{-ix\xi}\rvert^2$-type Plancherel identity for the angular convention, and the normalisation by $(2\pi)^{-n/2}$ cancels it. The ordinary-frequency statement is the previous section. $\blacksquare$

The Constant Fixed

Theorem. The multiplication constant is the unique normalisation making the transform unitary, up to a unimodular scalar: the operators $c\mathcal F_1$ that are unitary are those with $\lvert c\rvert=(2\pi)^{-n/2}$.

Proof. $(c\mathcal F_1)^\dagger(c\mathcal F_1)=\lvert c\rvert^2\,\mathcal F_1^\dagger\mathcal F_1 =\lvert c\rvert^2(2\pi)^n\mathrm{id}$, so unitarity is $\lvert c\rvert^2=(2\pi)^{-n}$; the phase is free. The ordinary-frequency convention is the case $c=(2\pi)^{-n/2}$ of the angular one. $\blacksquare$

Remark (the classical normalisation). The two classical normalisations are the one of the present corpus, $c=1$ with the $2\pi$-frequency, and the symmetric one $$ \hat f(\xi)=(2\pi)^{-n/2}\int f(x)e^{-ix\cdot\xi}dx , $$ which is unitary because the constant has been absorbed. The inversion formulas differ by the constant: $\check{\hat f}=f$ in the corpus, and $f(x)=(2\pi)^{-n/2}\int\hat f(\xi)e^{ix\cdot\xi}d\xi$ in the symmetric one.

The Transpose and the Reflection

Reflection and Order Four

Theorem. The square of the transform is the reflection, $\mathcal F^2=R$ with $Rf(x)=f(-x)$, and the symmetries of the transform are $$ \mathcal F^3=\mathcal F^{-1}=\overline{\mathcal F},\qquad \mathcal F^4=\mathrm{id},\qquad \overline{\mathcal F}=\mathcal F R=R\mathcal F . $$ The adjoint is the transform followed by the reflection, $\mathcal F^\dagger=\mathcal F R=\mathcal F^{-1}$, and the reflection is self-adjoint and unitary, $R^\dagger=R=R^{-1}$.

Proof. The order-four relation is the Fourier article's; the identity $\overline{\mathcal F}=\mathcal F R$ is the conjugate symmetry in operator form, $\overline{\mathcal F}f=\overline{\mathcal F\bar f}$ computed from the kernel $e^{2\pi ix\xi}=e^{-2\pi ix(-\xi)}$, that is the transform of the reflection; the reflection is its own inverse and preserves the pairing. Combining with $\mathcal F^{-1}=\overline{\mathcal F}$ gives $\mathcal F^\dagger=\mathcal F R$. $\blacksquare$

The Transpose Coincides with the Transform

Theorem. The transform is symmetric for the bilinear pairing, $\mathcal F^{\mathrm t}=\mathcal F$; hence the adjoint and the transpose are related by $$ \mathcal F^\dagger=\overline{\mathcal F^{\mathrm t}}=\overline{\mathcal F} , $$ and the transform is complex-symmetric rather than self-adjoint; the deviation from self-adjointness is exactly the conjugation.

Proof. The kernel $K(x,\xi)=e^{-2\pi ix\cdot\xi}$ is symmetric in $x$ and $\xi$, so the transposed kernel is $K(\xi,x)=K(x,\xi)$ and the transpose of $\mathcal F$ is $\mathcal F$; the relation between the adjoint and the transpose is that of The Adjoint of an Integral Operator, $K^*=\overline{K^{\mathrm t}}$, and it gives $\mathcal F^\dagger=\overline{\mathcal F}$. A distributional argument extends the identity beyond the convergent integrals. $\blacksquare$

Example (the multiplication and convolution sides). The transform conjugates convolution to multiplication, $\mathcal F(f*g)=\mathcal Ff\cdot\mathcal Fg$, and the involution $f^*(x)=\overline{f(-x)}$ to the conjugation, $\widehat{f^*}=\overline{\hat f}$; under these two correspondences the transform is a $\ast$-isomorphism of the convolution algebra onto the algebra of multipliers, with the involution on the first side carried to the conjugation on the second. The constant of the normalisation does not enter the algebraic identities, only the metric ones, and the Hilbert transform, with the multiplier $-i\operatorname{sgn}\xi$, is the skew-adjoint instance computed in The Fourier Transform and Conjugate Symmetry.

The Distributional and Group Cases

Remark (tempered distributions). The transform extends to the tempered distributions by $\langle\mathcal F T,\varphi\rangle=\langle T,\mathcal F\varphi\rangle$, and the adjoint identity is the definition; on the tempered distributions the order-four relation $\mathcal F^4=\mathrm{id}$, the reflection $\mathcal F^2=R$ and the statement that the adjoint is the inverse continue to hold. The distributional transform is Distributions and Fundamental Solutions, later in this Part.

Remark (the general group). On a general locally compact abelian group the transform is the pairing against the characters, valued in functions on the Pontryagin dual, and the inversion, unitarity and adjoint statements are the Plancherel theory of that transform; the normalisation constant is the constant of the Haar measure, and the dual group replaces $\mathbb R^n$. This is Harmonic Analysis on Groups and Pontryagin Duality, in the neighbouring category Analysis on Groups of this Part, and it is the group form of the present article.

Summary

The Fourier operator $\mathcal F$ on $L^2(\mathbb R^n)$ has the adjoint $\mathcal F^\dagger=\overline{\mathcal F}$, the conjugate transform, and since the transform is unitary the adjoint is the inverse, $\mathcal F^\dagger=\mathcal F^{-1}=\mathcal F^3$; the identities $\mathcal F^\dagger\mathcal F=\mathcal F\mathcal F^\dagger=\mathrm{id}$ are the two inversion formulas, and the kernel form of $\mathcal F^\dagger=\mathcal F^{-1}$ is $K^*(x,\xi)=\overline{K(\xi,x)}=e^{2\pi ix\cdot\xi}$. The transform is symmetric, $\mathcal F^{\mathrm t}=\mathcal F$, so the adjoint and the transpose differ by the conjugation, and $\overline{\mathcal F}=\mathcal F R=R\mathcal F$ with $R$ the reflection; the order-four relation $\mathcal F^2=R$, $\mathcal F^4=\mathrm{id}$ organises the symmetries. The unitarity is up to a constant: in the angular-frequency convention the adjoint is $(2\pi)^n$ times the inverse and the constant $(2\pi)^{-n/2}$ makes the operator unitary, while in the ordinary-frequency convention of the corpus the constant is $1$ and the transform is exactly unitary; the constant is a normalisation of the measure behind the transform. The Plancherel identity is the adjointness of the transform against the pairing, and the distributional and locally compact abelian generalisations are owned by later articles.

Summary of Notation

Symbol	Meaning
$(\mathcal Ff)(\xi)=\int f(x)e^{-2\pi ix\cdot\xi}dx$	Fourier operator, ordinary-frequency convention
$(\overline{\mathcal F}f)(\xi)=\int f(x)e^{2\pi ix\cdot\xi}dx$	Conjugate transform
$\mathcal F^\dagger=\overline{\mathcal F}=\mathcal F^{-1}$	The adjoint is the conjugate transform
$K(x,\xi)=e^{-2\pi ix\cdot\xi}$	Kernel of the transform
$K^*(x,\xi)=\overline{K(\xi,x)}=e^{2\pi ix\cdot\xi}$	Conjugate transpose kernel
$\mathcal F^{\mathrm t}=\mathcal F$	Symmetry of the transform
$R$, $\mathcal F^2=R$, $\mathcal F^4=\mathrm{id}$	Reflection and order four
$\mathcal F_1$, $(2\pi)^{-n/2}\mathcal F_1$	Angular-frequency transform and its normalisation