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Given any positive $k(n)\to \infty$, choose increasing integers $n_j$ with $k(n_j)\ge j{2}^j$. The absolutely uniformly convergent series $g(t)={\sum }_j{2}^{-j}{e}^{i{n}_{j}t}$ is continuous, and its Fourier coefficient at $n_j$ is exactly $2^{-j}$. The displayed unbounded limsup follows, although every individual continuous function still has coefficients tending to zero by the Riemann-Lebesgue lemma.

For an integrable periodic function with one-sided limits $L_{+},L_{-}$ at $t$, the displayed condition implies that its classical Fourier partial sums converge to $(L_{+}+L_{-})/2$. To prove it use the Dirichlet kernel $D_n(s)=\sin ((n+1/2)s)/\sin (s/2)$ and its integral $2\pi$. Subtract the target value and combine the positive and negative halves of the convolution integral. The resulting difference isThe coefficient of the sine is integrable near zero by the Dini condition, and away from zero by integrability of $f$. The Riemann-Lebesgue lemma therefore makes this expression tend to zero. Step functions satisfy the condition because the numerator vanishes in a sufficiently small punctured neighborhood. This is a local Fourier convergence test, distinct from Dini's theorem on uniform convergence of monotone sequences of continuous functions.

In $L^{\infty }(\mathbb{R})=(L^1(\mathbb{R}){)}^{\ast }$, the sequence $f_n(x)=\sin (nx)$ converges weak-star to zero by the Riemann-Lebesgue lemma, while $f_n^2$ converges weak-star to the constant $1/2$. Nonlinear pointwise operations therefore need not preserve weak-star limits.

As distributions on the line, $m\sin (m|t|)\to 2{\delta }_0$. Pairing with a test function $A$ folds the integral to ${\int }_0^{\infty }m\sin (ms)[A(s)+A(-s)]ds$. One integration by parts yields $2A(0)$ plus a cosine integral tending to zero by the Riemann-Lebesgue lemma. Pointwise convergence is unnecessary.

If $H$ is continuous, piecewise twice differentiable, and has derivative jump $[H^{\prime }]=H^{\prime }(y+)-H^{\prime }(y-)$ at $y$, then its distributional second derivative is $H_{\mathrm{reg}}^{\prime \prime }+[H^{\prime }]{\delta }_y$. If the regular part is integrable and the boundary terms vanish, two integrations by parts give the displayed asymptotic expansion by the Riemann-Lebesgue lemma. A jump $[H]$ in $H$ itself instead produces $[H]e^{-iqy}/(iq)$ at leading order. This distinguishes the different decay rates of the streamfunction and streamwise velocity in linear inviscid damping in unbounded Couette flow.

The Riemann-Lebesgue lemma and Fourier inversion give a bounded injective map $L^1(\mathbb{R})\to C_0(\mathbb{R})$ with dense image. It is not onto: inverse transforms of ${\chi }_{[-n,n]}\ast {\chi }_{[-1,1]}$ have unbounded $L^1$ norms while the targets have uniform norm $2$, contradicting the bounded inverse that the open mapping theorem would give.

For a continuously differentiable periodic kernel, integration by parts and the Riemann-Lebesgue lemma give $c_n=o(1/|n|)$. Nonperiodic data $e^{\alpha x}$ with $\alpha \notin i\mathbb{Z}$ have Fourier coefficients asymptotic to a nonzero multiple of $1/n$. Their inverse coefficients $y_n/c_n$ therefore fail square summability, and there is no Hilbert-space inverse or least-squares minimizer. Periodic exponential data $\alpha =im$ instead give the single-mode solution $e^{imx}/c_m$.

For smooth $h$ on $[0,2\pi ]$ and positive integer $n$, repeatedly integrating $\int h(t)\cos (nt)dt$ twice eliminates the sine endpoint terms and yields only even inverse powers of $n$. The exact remainder after $M$ pairs is $(-1{)}^M{n}^{-2M}\int h^{(2M)}\cos (nt)$, which is $o(n^{-2M})$ by the Riemann-Lebesgue lemma. Without integer frequencies an $n^{-1}$ sine endpoint term generally remains.

For a Schwartz function $f$, the Hilbert transform satisfies $\mathcal{H}f(x)=(\int f)/(\pi x)+o(|x{|}^{-1})$ at both infinities. Splitting the inverse Fourier transform at the sign multiplier's jump and integrating each half-line by parts extracts the zero-frequency value $\hat{f}(0)$; the remainder vanishes by the Riemann-Lebesgue lemma. A nonzero zeroth moment therefore obstructs Schwartz decay.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

If $\alpha >1$, then $f_{\alpha }$ is an integrable function, so the Riemann-Lebesgue lemma supplies a continuous representative of its Fourier transform. Equivalently, in the mixture for $h_{\alpha }$ the majorant $t^{(\alpha -3)/2}{e}^{-t}$ is integrable at zero precisely when $\alpha >1$. In this case dominated convergence also givesIf $0<\alpha \le 1$, Fatou's lemma applied along any sequence of nonzero $\xi$ tending to zero givesThe divergence comes from zero, where the exponent is at most $-1$. A continuous representative $H$ would agree with $h_{\alpha }$ almost everywhere. Since both are continuous on each open half-line, they must agree everywhere on those half-lines: a nonzero continuous difference would be nonzero on an interval of positive Lebesgue measure. The displayed blow-up then contradicts continuity of $H$ at zero. Changing a value at one point cannot remove this singularity. Consequently

Splitting the Fourier transform integral at zero gives two absolutely convergent integrals:Both $g_t$ and $\widehat{g_t}$ are integrable functions. Let $h_t(x)=(t^2+x^2{)}^{-1}$, so $\widehat{g_t}=2t{h}_t$. The duality property of the Fourier transform, or Fourier inversion in tempered distributions, givesThese Fourier transforms have continuous representatives by the Riemann-Lebesgue lemma, so the identity holds at every $\xi$. Since $g_t$ is an even function,

For the Riemann-Lebesgue lemma, a half-period shift of the oscillation gives a direct proof. For a nonzero integer $r$, set $h=\pi /r$. Periodicity and a change of variable show that the Fourier coefficient of $t\mapsto f(t+h)$ at $r$ is $e^{irh}\hat{f}(r)=-\hat{f}(r)$. ThusThe right side tends to zero as $|r|\to \infty$ by uniform continuity on the circle. Consequently $\boxed{\hat{f}(r)\to 0\text{ as }|r|\to \infty }$.

There is nevertheless no prescribed rate shared by all continuous functions. Choose increasing positive integers $n_j$ with $k(n_j)\ge j{2}^j$, which is possible since $k(r)\to \infty$. DefineThe series converges absolutely and uniformly, so $g$ is continuous. Termwise integration is justified by uniform convergence, and the orthogonality of distinct exponential modes gives $\hat{g}(n_j)=2^{-j}$. HenceThis arbitrarily slow Fourier coefficient decay is compatible with the Riemann-Lebesgue lemma: for this particular function all Fourier coefficients still tend to zero, but along the selected subsequence they beat the proposed decay scale by an unbounded factor.

Choose a step function $s$ with $\Vert f-s{\Vert }_1<\epsilon$, by part (i). The Fourier transform satisfiesBy part (ii), ${\mathrm{limsup}}_{|t|\to \infty }|\hat{f}(t)|\le \epsilon$. Since $\epsilon$ is arbitrary,This proves the Riemann-Lebesgue lemma with the transform convention in the question.

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.

The asymptotic expansion means that for every fixed integer $M\ge 0$,It concerns the successive finite truncations; the infinite series need not converge. Because the assertion holds at every order, one may equivalently bound the remainder after order $M$ by $O(n^{-2M-2})$.

Put $h(t)=(1+t^2{)}^{-1}$ and $b=2\pi$. For integer $n$, $\sin (nb)=\sin 0=0$ and $\cos (nb)=\cos 0=1$. Two integrations by parts giveRepeating exactly $M$ times givesAll derivatives are integrable on this compact interval. The Riemann-Lebesgue lemma says their oscillatory integrals tend to zero, proving the expansion at every order, including $a_0=0$. Since $h$ is even, its odd derivatives vanish at zero. Now $h^{\prime }(t)=-2t/(1+t^2{)}^2$ and $h^{\prime \prime \prime }(t)=24t(1-t^2)/(1+t^2{)}^4$, soThe restriction to integer $n$ is essential to this even-power endpoint expansion: otherwise the first integration by parts has a generally nonzero $n^{-1}$ boundary term.

Work first with $q$ in the Schwartz space, so that all Fourier manipulations and spectral contour integrals are justified; weaker classes follow by the usual density or distribution arguments. DefineThe phase is purely imaginary. Since ${\partial }_{\bar{z}}E=kE$, setting $F=E\Phi$ reduces the spectral equation to ${\partial }_{\bar{z}}\Phi =E^{-1}q$. The whole-plane Cauchy-Pompeiu formula therefore constructs the solution decaying spatially at infinity:The freedom to add $E$ times an entire function is removed by this decay condition. For every fixed $k$, the integral is a spatial Cauchy-Green operator applied to a modulated source.

Now differentiate in the conjugate spectral parameter. The two exponential derivatives produce $\zeta -z$, canceling the Cauchy denominator, soThis is the spectral dbar equation: its right-hand side is the forward transform of $q$ multiplied by a known plane wave. Apply the whole-plane Cauchy-Pompeiu formula again, now in $k$:The spatial spectral equation also gives $F(z,k)=-q(z)/k+o(k^{-1})$ as $|k|\to \infty$. One way to justify this is to integrate by parts in the first Cauchy integral: $F=-q/k+T_k(q_{\bar{z}})/k$, where the modulated Cauchy integral $T_k(q_{\bar{z}})$ tends to zero by the Riemann-Lebesgue lemma. Comparing the $1/k$ coefficient of the spectral contour integral therefore givesThis derives the transform pair from two uses of the Cauchy-Pompeiu formula, not from an assumed inversion formula.

To identify the usual normalization, write $z=x+iy$ and $k=k_1+i{k}_2$. Then $\bar{k}z-k\bar{z}=2i(k_{1}y-k_{2}x)$. Set ${\xi }_1=2k_2$, ${\xi }_2=-2k_1$, so $d{\xi }_{1}d{\xi }_2=4dA(k)$. The result is exactly the two-dimensional Fourier transform pairThe factor four in the real-frequency change of variables is essential.

Use the orthonormal Fourier series basis $e_n(x)=(2\pi {)}^{-1/2}{e}^{inx}$, indexed by $n\in \mathbb{Z}$, and the Hilbert space inner product $⟨f,g⟩={\int }_0^{2\pi }\overline{f(x)}g(x)dx$. The integral operator is a periodic convolution operator. Changing variables $s=x-x^{\prime }$ and using periodicity givesThere is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x^{\prime }-x)}$, so $A^{\ast }{e}_n=\overline{c_n}{e}_n$.

A singular system of a periodic convolution operator is thereforeIndeed $A{v}_n={\sigma }_n{u}_n$ and $A^{\ast }{u}_n=(c_n/|c_n|)\overline{c_n}{e}_n={\sigma }_n{v}_n$. The unit factor determined by the complex argument of $c_n$ in $u_n$ is necessary when the complex Fourier coefficients are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne 0$. One may enumerate $\mathbb{Z}$ by $\mathbb{N}$, or order the positive singular values by decreasing magnitude.

The continuous kernel on a finite square makes $A$ a Hilbert-Schmidt operator, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne 0$,by the Riemann-Lebesgue lemma. In particular the singular values tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^{\ast }=0$: the range is dense, but the inverse is unbounded and the range is not closed.

Let $f=\hat{\phi }$, a Schwartz function. Splitting the inverse Fourier transform at the jump in the Hilbert-transform Fourier multiplier givesFor $x\ne 0$, integration by parts on each half-line showsBoth restricted derivatives are in $L^1$. The Riemann-Lebesgue lemma makes the bracket tend to zero as $|x|\to \infty$. Hence the two-sided tail isHere $\hat{\phi }(0)=\int \phi$. The large-distance tail of the Hilbert transform thus depends on the zeroth moment of the input. If that moment is nonzero, the $1/x$ tail proves that the output is not a Schwartz function, despite being smooth and square-integrable.

The test-function space is $\mathcal{D}({\mathbb{R}}^n)=C_c^{\infty }({\mathbb{R}}^n)$. Its topology is the strict inductive limit topology of the spaces ${\mathcal{D}}_K$ of smooth functions supported in a fixed compact set $K$, each with the seminorms $p_{K,j}(\phi )={\max }_{|\alpha |\le j}{\sup }_K|{\partial }^{\alpha }\phi |$. In particular, a sequence converges in $\mathcal{D}$ precisely when its supports lie in one compact set and all its derivatives converge uniformly. The indices $\alpha$ use multi-index notation.

The distribution space ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ consists of continuous complex-linear forms on $\mathcal{D}$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\ge 0$ such thatWe use distributional convergence: $u_j\to u$ means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{D}$. The pairings are bilinear, with no complex conjugation. A test function $f$ is identified with its regular distribution $⟨f,\phi ⟩=\int f\phi$.

For a distribution $u$ and a test function $\psi$, their smoothing convolution with a test function isIt is a smooth function, with ${\partial }_x^{\alpha }(u\ast \psi )(x)=⟨u_y,{\partial }^{\alpha }\psi (x-y)⟩$. Indeed, for $x$ in a compact neighborhood all translated tests have support in one compact set, and their difference quotients converge in the space of test functions. Notice that $u\ast \psi$ need not be compactly supported.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }(B_1)$ with $\int \rho =1$, and write ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $\tilde{\rho }(x)=\rho (-x)$, thenThe right-hand test tends to $\phi$ in $\mathcal{D}$: its supports lie in $\mathrm{supp}\phi +{\overline{B}}_1$ for $\epsilon \le 1$, and every derivative converges uniformly by the approximate-identity argument. Thus the smooth regularizations converge to $u$ as distributions.

To obtain actual compactly supported approximants, choose a smooth cutoff function $\chi$ equal to one on $B_1$ and supported in $B_2$, and setFor each fixed test function $\phi$, the cutoff is identically one on its support once $j$ is large. Hence $⟨f_j,\phi ⟩=⟨u,{\tilde{\rho }}_{1/j}\ast \phi ⟩\to ⟨u,\phi ⟩$. This proves $\mathcal{D}$ is dense in ${\mathcal{D}}^{\prime }$, and in fact establishes the density of test functions in distributions and gives a convergent approximating sequence for each distribution. The expanding cutoff is essential when $u$ has noncompact support of a distribution.

For the radial limit, take a test function $\phi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $rdr=dt/2$ givesWhen $\phi$ is supported away from the origin, $A$ vanishes near $t=-1$ and extends to a compactly supported smooth function on the whole line. The folded sine approximation to a Dirac delta is exposed by setting $B(s)=A(s)+A(-s)$: integration by parts givesby the Riemann-Lebesgue lemma. ThereforeThis surface delta distribution is arclength measure on the unit circle. By the level-set normalization of a surface delta, it is $2\delta (|x{|}^2-1)$: the factor two cancels the gradient magnitude $|\nabla (|x{|}^2-1)|=2$ on the circle. It is not twice arclength measure.

For a test function that can meet the origin, the endpoint cannot be discarded. Now $A(-1)=\pi \phi (0)$. Its right derivative is integrable, with $A^{\prime }(-1+s)=(\pi /4)\Delta \phi (0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi \phi (0)+(\pi /4)(1+t)\Delta \phi (0)+O((1+t{)}^2)$ near $t=-1$. Applying integration by parts separately on the negative and positive intervals givesThe last two integrals tend to zero by the Riemann-Lebesgue lemma. Thus the radial quadratic oscillation defect in two dimensions gives the stronger oscillating point-mass defect formulaChoose $\phi$ supported in $B_{1/2}$ with $\phi (0)=1$. Its pairing is $-\pi \cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos (m+1)+\cos (m-1)=2\cos 1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos (2m)=2{\cos }^{2}m-1$ would then force $L=-1$. Consequently there is no limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.