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Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

Use $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$. For a Schwartz function, Fourier inversion is $f(x)=(2\pi {)}^{-1}\int e^{ix\xi }\hat{f}(\xi )d\xi$. For a tempered distribution define $⟨\hat{T},\varphi ⟩=⟨T,\hat{\varphi }⟩$ for Schwartz functions; this agrees with the function transform in the stated convention. The half-indicator kernel has transformwith value $t$ at zero. Transforming the wave equation gives ${\hat{u}}_{tt}+{\xi }^2\hat{u}=0$, zero initial value and initial derivative $\hat{g}$. Hence $\hat{u}=\hat{g}\sin (t\xi )/\xi$, and convolution yieldsFinite propagation speed means that the solution at $(t,x)$ depends only on initial data in this interval, so disturbances cannot travel faster than one. The formula also makes sense for any smooth $g$, regardless of its growth at infinity, because the integration interval is compact. Direct differentiation verifies the equation and both initial conditions. Equivalently, locally replace $g$ by a compactly supported smooth function agreeing throughout the relevant domain of dependence.

The real matrix $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with determinant one preserves $u\wedge v$. Hence $W_g(u)=W(gu)$ obeys the same Weyl relations. It is strongly continuous and irreducible because $g$ is invertible and its arguments range over the whole plane. Part (b) gives a unitary $V$ intertwining $W$ and $W_g$, so $VW(u)V^{\ast }=W(gu)$. If $V_1,V_2$ both work, $V_2^{\ast }{V}_1$ commutes with all $W$ and is scalar by part (a); its scalar has modulus one. Thus the implementer is unique up to phase.

For the displayed quarter-turn, choose the scaled inverse Fourier transformThe formula starts on Schwartz functions and extends unitarily to $L^2$ by Plancherel and a change of scale. It conjugates $T_{x}f(t)=f(t+x)$ to multiplication by $e^{-2ixs}$, and multiplication by $e^{2iyt}$ to $T_y$. Keeping the Weyl phase therefore gives $VW(x,y)V^{\ast }=W(y,-x)$, as required. With the opposite Fourier-kernel sign one obtains the opposite quarter-turn, so that convention must be checked.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

For $g\in \mathcal{S}({\mathbb{R}}^3)$ define convolution byThis is smooth and has at most polynomial growth, hence defines a tempered distribution. Its equivalent pairing is $⟨T\ast g,\varphi ⟩=⟨T,\tilde{g}\ast \varphi ⟩$ with $\tilde{g}(y)=g(-y)$. Fourier duality and the usual Schwartz-function convolution calculation giveThe product on the right is distributional multiplication by the smooth Schwartz function $\hat{g}$. With the convention in part (i), no extra factor occurs. One can verify it first for functions by changing $x=y+z$ inside the two integrals, then use the displayed pairing to perform the same calculation for arbitrary $T$.

With the streamfunction convention $(u,v)=({\psi }_y,-{\psi }_x)$, disturbance vorticity is $\omega =-\Delta \psi$. The basic Couette flow has constant vorticity $-1$. Linearizing the two-dimensional Euler equations for an inviscid fluid in vorticity form givesThe second term vanishes, soThis derivation uses incompressibility and an inviscid homogeneous fluid, without introducing a viscous decay term.

For the initial streamwise Fourier mode, write $\psi =e^{i\alpha x}a(y,t)$. Its transported Laplacian has amplitudeThe decaying Green function on the line isThis is the negative of the one-dimensional modified Helmholtz Green function for $-{\partial }_y^2+{\alpha }^2$: it is continuous and its first derivative with respect to $y$ has jump one at $y=s$. Consequently $({\partial }_y^2-{\alpha }^2)G=\delta (y-s)$, and inversion yieldsUnder the usual decay of the initial profile and its derivatives, the initial integral returns $\varphi$. There is no nonzero homogeneous solution of $a^{\prime \prime }-{\alpha }^{2}a=0$ decaying at both infinities.

For the large-time calculation, put $q=\alpha t$ and $H(s)=G(y,s)F(s)$, keeping the observation point $y$ fixed. Assume sufficient localization to justify the following integrations by parts; for example, $F$ may be a Schwartz function. The function $H$ is continuous, but its derivative in $s$ has jumpsince $G_s(y,y+)-G_s(y,y-)=1$. The Fourier decay from a Green-function derivative jump givesHere the regular second derivative is computed separately on either side of $s=y$. If it is integrable, its Fourier transform tends to zero by the Riemann-Lebesgue lemma. HenceThe cusp in the Green function, rather than a stationary point of the linear phase, supplies this leading contribution.

To obtain the streamwise velocity, differentiate the integral itself. This avoids an unjustified differentiation of a pointwise remainder. The kernel $G_y$ has a jump of $-1$ as a function of $s$ at $s=y$, so one integration by parts givesThe transverse velocity follows exactly from $v=-i\alpha \psi$, while the Laplacian equation gives vorticity without an asymptotic approximation:Thus the generic orders are $u=O(t^{-1})$, $v=O(t^{-2})$, and $\omega =O(1)$, at each fixed $y$. If $F(y)=0$, the leading coefficients vanish and faster decay can occur. For real physical disturbances, take the real part of these complex Fourier modes. The vorticity generally keeps oscillating rather than approaching a pointwise limit, even while the velocity decays. This is linear inviscid damping in unbounded Couette flow caused by phase mixing.

For arbitrary initial vorticity, the characteristics satisfy $\dot{x}=y$, $\dot{y}=0$, and $\dot{\omega }=0$. Therefore the exact solution isPutting ${\omega }_0=-F(y)e^{i\alpha x}$ recovers the preceding nondecaying mode. In general the shear is area preserving, so its transport preserves every finite Lp norm of vorticity. Local pointwise behavior for arbitrary ${\omega }_0$ depends on its spatial profile: an initially localized profile can move away from a fixed point. There is no universal pointwise nonzero limit or $t^{-2}$ decay of vorticity for arbitrary initial data.

The stated smoothness and decay of $\varphi$ alone do not guarantee the claimed algebraic rates. The intended integration by parts calculation requires control of derivatives and tails; convergence of the displayed integral alone is insufficient. Here is a counterexample to the literal unrestricted rate, with $\alpha =1$. Choose a nonnegative smooth function $r$ supported in $(-1/4,1/4)$ with $\int r=1$, and a sufficiently rapidly increasing sequence $t_n$. SetChoose $\sum e^n/t_n<\infty$. Then $F$ is smooth and integrable, $\varphi$ is smooth and tends to zero at both infinities, and every evolution integral converges absolutely. At the observation point $(x,y)=(0,0)$, all supports have $s>0$, soAt $t=t_n$, the $n$th summand is $-1/(2t_n)$. The Fourier transform of $r$ decays faster than any inverse power. Inductively choose $t_n$ so large that all earlier summands at $t_n$ together have modulus less than $1/(8t_n)$, and choose subsequent times so that ${\sum }_{m>n}1/t_m<1/(4t_n)$. The later summands then also have modulus less than $1/(8t_n)$. Thus $|\psi (0,0,t_n)|\ge 1/(4t_n)$, contradicting $O(t^{-2})$. This construction meets the printed assumptions; what it lacks is uniform derivative localization. The preceding boxed decay formulas are the correct intended result under the explicitly stated sufficient localization hypothesis.

Taking the Fourier transform gives $(1+|\xi {|}^2)\hat{u}(\xi )=\hat{\varphi }(\xi )$. Thus the canonical solution isA smooth compactly supported function is a Schwartz function, as is its Fourier transform. The multiplier $(1+|\xi {|}^2{)}^{-1}$ and all its derivatives have polynomial bounds, so their product with $\hat{\varphi }$ is again a Schwartz function. The inverse integral can therefore be differentiated to verify $-\Delta u+u=\varphi$ classically. By part (b), its Fourier-space integrand factor $\hat{\varphi }/(1+|\xi {|}^2)$ is radial. Changing variables by any orthogonal transformation in the inverse integral gives $u(Qx)=u(x)$, proving radiality.

This is also the unique tempered-distribution solution: the nonvanishing multiplier $1+|\xi {|}^2$ can be inverted on that space. Unrestricted classical solutions need not be unique, since one may add, for example, $e^{x_1}$; the Fourier construction selects the rapidly decaying solution.

Use $D_j=-i{\partial }_{x_j}$ and $⟨\xi ⟩=(1+|\xi {|}^2{)}^{1/2}$, so the Fourier transform of $P(D)u$ is $P(\xi )\hat{u}(\xi )$. Write $P_N$ for the homogeneous degree-$N$ part. An elliptic differential operator has $P_N(\xi )\ne 0$ for every real $\xi \ne 0$. By continuity on the unit sphere, $m={\min }_{|\omega |=1}|P_N(\omega )|>0$. The lower-degree terms are bounded by $C|\xi {|}^{N-1}$ for $|\xi |\ge 1$, whencefor sufficiently large $|\xi |$. This proves the high-frequency lower bound for an elliptic polynomial. Degree zero just means a nonzero constant and has no derivative gain.

With the Fourier convention in Question 3, the Sobolev space isThe condition includes that the weighted transform is represented by an $L^2$ function. For open $X$, the Local Sobolev space consists of $u\in {\mathcal{D}}^{\prime }(X)$ for which $\chi u$, extended by zero, belongs to $H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$. We use the elementary Sobolev multiplication by a smooth cutoff fact for every real $s$. It follows from Fourier convolution with the rapidly decreasing $\hat{\chi }$, the weighted inequality $⟨\xi {⟩}^s\lesssim ⟨\xi -\eta {⟩}^{|s|}⟨\eta {⟩}^s$, and the $L^1\ast L^2\to L^2$ convolution bound.

For a compactly supported distribution, continuity on test functions supported in a fixed compact neighborhood gives finite order: for some integer $M$,Insert a compact smooth cutoff function equal to one near the support, times $e^{-ix\cdot \xi }$. The resulting Fourier transform is smooth and satisfies $|\hat{u}(\xi )|\le C^{\prime }⟨\xi {⟩}^M$. The distribution is also tempered by this same finite-order bound. ThusIndeed the square of the weighted bound is integrable exactly when $2(t+M)<-n$. This is negative Sobolev regularity of a compactly supported distribution.

We next prove local regularity by a high-frequency reciprocal parametrix kernel, including its off-diagonal smoothing property. Choose $\psi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one on a ball containing all real zeros of $P$, and put $b(\xi )=(1-\psi (\xi ))/P(\xi )$, defining it smoothly as zero in the inner ball. Differentiation of the reciprocal and the elliptic lower bound giveThe Fourier multiplier $E=b(D)$ maps $H^s$ to $H^{s+N}$ by its zeroth-order bound, andThe kernel $K={\mathcal{F}}^{-1}b$ is smooth away from the origin: for $x\ne 0$, repeatedly integrate by parts using $e^{ix\cdot \xi }=(i|x{|}^2{)}^{-1}x\cdot {\partial }_{\xi }{e}^{ix\cdot \xi }$. After sufficiently many integrations, the differentiated symbol is integrable. For any desired $x$ derivative, repeat the argument with the additional polynomial ${\xi }^{\beta }$. A large-radius cutoff function justifies each step and its removal uniformly on compact sets away from zero. Thus convolution by $K$ carries a compactly supported distribution to a smooth function at points separated from its support. Also $R$ carries a compactly supported distribution to a Schwartz function, because $\psi \hat{u}$ is smooth with compact support.

For a target compact set in $X$, choose $\chi \in C_c^{\infty }(X)$ equal to one on a neighborhood of it. Then $\chi u$ and $[P(D),\chi ]u$ have compact support, and the commutator is supported where derivatives of $\chi$ occur, away from the target. The parametrix identity givesThe first term is in $H^{s+N}$ since $\chi P(D)u\in H^s$. The other two terms are smooth near the target by the proved kernel property. Since the target was arbitrary,This proof does not discard the cutoff function commutator; it places its support away from the set where regularity is sought.

For the final polynomial, select a multi-index ${\alpha }_0$ with $|{\alpha }_0|=N$ and ${\partial }^{{\alpha }_0}Q$ a nonzero constant. The derivative-ratio hypothesis immediately gives $|Q(\xi )|\ge c|\xi {|}^{\delta N}$ at large frequency. In particular $Q$ has no real zero there. For $N>0$, its degree bound also forces $\delta \le 1$. Differentiating $1/Q$ gives products of ratios ${\partial }^{\beta }Q/Q$, whose total derivative order is $|\alpha |$. Hence the corresponding high-frequency reciprocal satisfiesIts multiplier maps $H^s$ to $H^{s+\delta N}$. Its kernel is still smooth off zero: each frequency integration by parts now lowers the order by $\delta$, so more iterations may be needed, but $\delta >0$ supplies arbitrarily much decay. The same separated-support commutator identity applies without a change. The derivative-ratio Sobolev gain for a polynomial operator is thereforeSimply estimating the commutator by its differential order would lose this sharper gain; the off-diagonal kernel argument uses the full derivative-ratio hypothesis.

A Schwartz function is a smooth function for which every seminormis finite. These seminorms define the Fréchet topology of the Schwartz space $\mathcal{S}({\mathbb{R}}^n)$. The tempered distributions form its continuous linear dual ${\mathcal{S}}^{\prime }$. We use bilinear distribution pairing, with no complex conjugation on the test function.

Differentiation under the Fourier integral and integration by parts giveTherefore each Schwartz seminorm of $\hat{\phi }$ is bounded by a finite sum of $L^1$ norms of polynomially weighted derivatives of $\phi$. Insert the integrable weight $⟨x{⟩}^{-n-1}$ to bound those norms by finitely many Schwartz seminorms. This proves that $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

To justify inversion without a merely formal exchange of integrals, insert $e^{-\epsilon |\xi {|}^2/2}$ in the inverse integral. The Gaussian Fourier integral givesThe normalized Gaussian is an approximate identity, so the right side tends to $\phi (x)$; the left side converges by dominated convergence theorem since $\hat{\phi }$ is integrable. It follows thatReflection is continuous in Schwartz seminorms, so ${\mathcal{F}}^{-1}=(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$ is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the distributional Fourier transform byIt agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on ${\mathcal{S}}^{\prime }$, and the same squared-transform identity holds. For the usual strong dual topology, a seminorm is $p_B(u)={\sup }_{\phi \in B}|⟨u,\phi ⟩|$ for a bounded set $B\subset \mathcal{S}$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\hat{u})=p_{\mathcal{F}B}(u)$ proves continuity, and similarly for the inverse. Continuity also holds in the weak dual topology. Thus the transform is a continuous isomorphism on tempered distributions, with the stated normalization.

For a real symmetric positive-definite matrix $G$, there is $c>0$ with $g(x)=x^{T}Gx\ge c|x{|}^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2{r}^{-2}dr$. At infinity, rapid decay of a Schwartz test function makes it integrable. More quantitatively,This single seminorm bound proves a tempered distribution. No principal-value extension is needed at the origin.

First calculate the isotropic transform. For $f_{\epsilon }(x)=e^{-\epsilon |x|}/|x{|}^2$ and $k=|\xi |>0$, spherical integration and the supplied sine-integral identity giveThe original functions converge in ${\mathcal{S}}^{\prime }$ to $|x{|}^{-2}$ by dominated convergence theorem against tests. The transforms are bounded by $2{\pi }^2/|\xi |$, which is locally integrable in three dimensions and integrable against Schwartz functions at infinity. HenceLet $y=G^{1/2}x$. The Jacobian is $(\det G{)}^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The Fourier transform of a reciprocal positive quadratic form is consequentlyThe frequency-origin value is understood distributionally; there is no additional delta term.

For the complex extension use a symmetric matrix $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a quadratic form; a skew-symmetric part contributes nothing. The bound $|x^{T}Ax|\ge x^{T}Gx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the matrix while its real part is positive: compact parameter sets give a common $|x{|}^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive matrices along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive matrices, so the one-variable identity theorem supplies the equality; connected continuation along $0\le z\le 1$ reaches $A$.

The analytic determinant square root for accretive symmetric matrices must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}B{G}^{-1/2}$ and let its real eigenvalues be $b_j$. Thenwhere each factor has positive real part. This is the determinant square root normalized positively on real positive matrices. AlsoThus the quadratic-form square root has the unambiguous branch with positive real part. The accretive complex quadratic reciprocal Fourier transform isUniform local integrability of the right side justifies its analytic continuation as a tempered distribution, not merely pointwise away from the origin.

For the particular form,Substitution gives the required normalized expressionFor every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular tempered distributions, despite their locally integrable singularities at zero.

The real-space expression $1/x$ means the principal-value reciprocal distribution. Apply even exponential damping. Oddness givesDifferentiate the integral with respect to $k$; its derivative is ${\int }_0^{\infty }{e}^{-ax}\cos (kx)dx=a/(a^2+k^2)$. Its value at $k=0$ is zero, so the integral equals $\arctan (k/a)$. Taking $a\downarrow 0$ givesThe limit is a tempered distribution limit: the regularized transforms are uniformly bounded and converge away from zero, allowing dominated convergence against a Schwartz function. This Fourier transform of a principal-value reciprocal argument also establishes the needed Dirichlet integral ${\int }_0^{\infty }\sin u/udu=\pi /2$ in its Abel interpretation.

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

For $u\in {\mathcal{E}}^{\prime }$, choose a cutoff function $\chi =1$ on a neighborhood of its support. Its extension to smooth functions makesindependent of the choice of cutoff. For real frequency, the definition of the distributional Fourier transform gives $F(\lambda )=\hat{u}(\lambda )$: interchange the pairing with the integral of a Schwartz function, using the finite-order estimate on the cutoff's compact support. On each compact set of complex frequencies the exponential and all its $x$ derivatives have uniformly convergent power series. Continuity of the distribution allows termwise differentiation, withConsequently $F$ is entire on ${\mathbb{C}}^n$.

For the precise exponential type, a fixed enlarged support would give an unnecessarily enlarged radius. Instead use the shrinking-cutoff exponential-type estimate. On one fixed compact neighborhood of the closed radius-$\delta$ ball, $u$ has finite order of a distribution $m$. Choose ${\chi }_{\epsilon }=1$ near that ball, supported in the radius-$(\delta +\epsilon )$ ball, with $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1/2$. The finite-order estimate givesTake $\epsilon =[2(1+|z|){]}^{-1}$. Its extra exponential is bounded by $e^{1/2}$, while the derivative cost is polynomial. ThusThe entire function was defined with a fixed cutoff; only its bound uses a frequency-dependent cutoff, so no holomorphic dependence is lost. This proves the required estimate with some $N\ge 0$ without enlarging $\delta$.

For the converse, set $V(z)=e^{iz\cdot y}U(z)$. Its restriction to real frequencies has polynomial growth. Define the inverse tempered distribution $v$ byRapid decay of the Schwartz function transform proves convergence and continuity, and Fourier inversion gives $\hat{v}=V$ on real frequencies.

We prove its support directly by contour shifting, rather than assuming the support conclusion of the Paley–Wiener–Schwartz theorem. Let $\omega$ be a real unit vector and take a test function $\varphi$ supported in $x\cdot \omega \ge \delta +ϵ$ for some $ϵ>0$. For every integer $M\ge 0$, integration by parts in the real-frequency Fourier integral givesIndeed, move $(1-{\Delta }_x{)}^M$ from the oscillatory exponential to $e^{-t\omega \cdot x}\varphi$; its derivatives supply at most $2M$ powers of $t$.