Fourier analysis course of the University of Oxford (B4.4)

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.

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.

A tempered distribution is a continuous linear functional $u:\mathcal{S}(\mathbb{R})\to \mathbb{C}$. Equivalently, for some integer $N$ and constant $C$,The Fourier transform of a tempered distribution is defined byThe Fourier bounds for Schwartz functions make the right side a continuous linear functional on the Schwartz space, so $\hat{u}$ is again a tempered distribution.

For an integrable function $f$, the regular distribution $u_f$ acts by $⟨u_f,\varphi ⟩=\int f\varphi$. The estimate $|⟨u_f,\varphi ⟩|\le \Vert f{\Vert }_1{p}_{0,0}(\varphi )$ proves it is a tempered distribution. Since $\iint |f(x)\varphi (\xi )|dxd\xi <\infty$, Fubini's theorem givesThus the distributional Fourier transform agrees with the ordinary Fourier transform.

Since $0<f_{\alpha }\le 1$, its regular distribution satisfiesso $f_{\alpha }$ is a tempered distribution for every $\alpha >0$.

Put $a=\alpha /2$. The defining integral of the gamma function, with $r=(1+x^2)t$, givesThe Fourier transform of a Gaussian suggests the following representative for the Fourier transform of an inverse power of one plus x squared:All factors are positive. Tonelli's theorem evaluates its total integral, allowing an infinite value at the single point zero:Assign any finite value at zero if necessary. This gives an integrable function with $\Vert h_{\alpha }{\Vert }_1=2\pi$.

To verify equality as tempered distributions, take any Schwartz function $\varphi$. Substituting the mixture into $⟨f_{\alpha },\hat{\varphi }⟩$ is justified byFor each $t>0$, Fubini's theorem and the Fourier transform of a Gaussian identify the inner pairing with $\int \sqrt{\pi /t}{e}^{-{\xi }^2/(4t)}\varphi (\xi )d\xi$. Interchanging the resulting integrals is valid because their absolute value is bounded by $2\pi \Vert \varphi {\Vert }_{\infty }$ after integrating the positive kernel. ThereforeFinally, on a compact set with $|\xi |\ge \delta >0$, the mixture integrand is bounded by $t^{a-3/2}{e}^{-t-{\delta }^2/(4t)}$, which is integrable both at zero and at infinity. Dominated convergence proves continuity there. Hence $\boxed{\widehat{f_{\alpha }}=h_{\alpha }\in L^1(\mathbb{R}),h_{\alpha }(\xi )>0\text{ and continuous for }\xi \ne 0}$.

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

Let $R\varphi (x)=\varphi (-x)$, and define reflection of a tempered distribution by $⟨Ru,\varphi ⟩=⟨u,R\varphi ⟩$. The Fourier inversion theorem on the Schwartz space says $\hat{\hat{\varphi }}=2\pi R\varphi$. ThusReflection commutes with the Fourier transform, since $\widehat{R\varphi }=R\hat{\varphi }$ and the same relation follows by duality for tempered distributions. Define the inverse Fourier transform on tempered distributions byIt follows thatIn particular $u=(2\pi {)}^{-1}R\hat{\hat{u}}$ is the general Fourier inversion formula. When $\hat{u}$ is an integrable function, this is the regular distribution of $(2\pi {)}^{-1}\int \hat{u}(\xi )e^{ix\xi }d\xi$; an arbitrary tempered distribution need not have an ordinary integral representation.

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe convolution of two Schwartz functions is again a Schwartz function: differentiating under the integral and expanding $x^m=(y+(x-y){)}^m$ bounds every weighted derivative by products of integrable and bounded weighted derivatives. Evaluate Fourier inversion for this convolution at zero:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, independently of the approximation. It agrees with the distributional Fourier transform, because Cauchy-Schwarz inequality makes convergence in the L2 norm imply convergence when paired with every Schwartz function. For $f\in L^1\cap L^2$ it therefore agrees almost everywhere with the ordinary Fourier transform.

The extension is onto: its image is closed by the norm identity and contains the Schwartz space, since Fourier inversion maps that space to itself. This proves Plancherel theorem:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.

For real $s$, the Sobolev space $H^s(\mathbb{R})$ consists of those tempered distributions $u$ whose Fourier transform is represented by a locally square-integrable function and for whichThis is the Sobolev norm used here; inserting the fixed factor $(2\pi {)}^{-1}$ gives an equivalent convention.

For the constant function $c$, Fourier inversion at zero givesA nonzero Dirac delta distribution cannot be represented by a locally square-integrable function. Indeed, if $G$ represented $2\pi c{\delta }_0$, choose a test function $\rho$ with $\rho (0)=1$, and put ${\rho }_{\epsilon }(\xi )=\rho (\xi /\epsilon )$. On a fixed compact interval containing their supports, Cauchy-Schwarz inequality givesThus $c=0$. Every Sobolev space requires local square integrability of the Fourier transform, regardless of how negative $s$ is. ThereforeThe zero constant function belongs to every Sobolev space.

The distributional derivative is defined by $⟨u^{\prime },\varphi ⟩=-⟨u,{\varphi }^{\prime }⟩$. Since differentiation under the integral sign gives $(\hat{\varphi }{)}^{\prime }=\widehat{-i\xi \varphi }$, we obtainThis proves the Fourier transform of a derivative identity $\boxed{\widehat{u^{\prime }}=i\xi \hat{u}}$ for all tempered distributions.

If $u,u^{\prime }\in H^s(\mathbb{R})$, their Fourier transforms are represented by locally square-integrable functions, and the identity identifies $\widehat{u^{\prime }}$ with $i\xi \hat{u}$ almost everywhere. Hence the Sobolev norms satisfyso $\boxed{u\in H^{s+1}(\mathbb{R}),\Vert u{\Vert }_{H^{s+1}}^2=\Vert u{\Vert }_{H^s}^2+\Vert u^{\prime }{\Vert }_{H^s}^2}$ for every real $s$.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.

The $2\pi$ periodization of a Schwartz function isFor every nonnegative integer $j$, the Schwartz seminorms give $|{\varphi }^{(j)}(x+2\pi k)|\le C_j(1+|k|{)}^{-2}$ uniformly for $x\in [0,2\pi ]$. The Weierstrass M-test gives absolute and uniform convergence there of the defining series and all its differentiated series. Termwise differentiation proves that $P\varphi$ is a smooth function, with $(P\varphi {)}^{(j)}=P({\varphi }^{(j)})$. Reindexing the absolutely convergent series proves $P\varphi (x+2\pi )=P\varphi (x)$, so it is a periodic function.

For an explicit smooth periodic partition of unity, choose the bump functionIt is a test function, positive on $[-\pi ,\pi ]$. The smooth function $S(x)=P\eta (x)$ is $2\pi$ periodic and strictly positive, since every $x$ has a translate in $[-\pi ,\pi ]$. SetDivision preserves smoothness because $S>0$, and the support remains contained in $[-2\pi ,2\pi ]$.

A distribution $u$ is a $2\pi$ periodic distribution whenIteration gives invariance under all integer multiples of $2\pi$. For a tempered distribution the identity also holds for every Schwartz function, since test functions are dense in the Schwartz space: multiplying by cutoff functions $\chi (x/R)$ equal to one near zero gives such an approximation in every Schwartz seminorm.

For $\varphi \in \mathcal{S}(\mathbb{R})$ define $a_k(x)=\Psi (x+2\pi k)\varphi (x)$. The smooth periodic partition of unity gives $\varphi ={\sum }_k{a}_k$ with convergence in the Schwartz space. Indeed, each $a_k$ is supported where $|x+2\pi k|\le 2\pi$, and the generalized Leibniz rule and arbitrarily rapid decay of the derivatives of $\varphi$ give, for every $m,n,M$,Consequently continuity and periodicity of the tempered distribution giveFor the last interchange, all summands have the fixed compact support of $\Psi$ and their derivatives decrease faster than any power of $|k|$. Hence the series converges in the space of test functions, as well as in the Schwartz space. Notice that $P\varphi$ itself need not be a Schwartz function; the factor $\Psi$ makes the pairing legitimate.

Let $D={\sum }_{k\in \mathbb{Z}}{\delta }_{2\pi k}$ be the Dirac comb. Pair the given identity $D=(2\pi {)}^{-1}{\sum }_{n\in \mathbb{Z}}{e}^{in\cdot }$ with the Schwartz function $y\mapsto \varphi (x-y)$. On the left this gives ${\sum }_k\varphi (x-2\pi k)=P\varphi (x)$. On the right,Therefore the Fourier series isThe Fourier bounds for Schwartz functions imply $|\hat{\varphi }(n)|\le C_N(1+|n|{)}^{-N}$ for every $N$. For each nonnegative integer $j$, choose $N>j+1$ to obtain ${\sum }_n|n{|}^j|\hat{\varphi }(n)|<\infty$. Thus the Fourier series and every differentiated Fourier series converge absolutely and uniformly on all of $\mathbb{R}$. In particular convergence holds in the topology of smooth $2\pi$ periodic functions, defined by uniform convergence of every derivative on one period, and also in tempered distributions.

Use the smooth periodic partition of unity $\Psi$ above and define the Fourier coefficients of the periodic distribution byThese are the smooth-window Fourier coefficients of a periodic distribution for period $2\pi$. The finite Schwartz seminorm bound for $u$, restricted to the fixed compact support of $\Psi$, gives $|c_n|\le C(1+|n|{)}^N$ for some $C,N$. Indeed, only finitely many derivatives of $\Psi e^{-inx}$ occur, and each contributes at most a fixed power of $|n|$.

For a Schwartz function $\varphi$, the preceding Fourier series formula and $\hat{\hat{\varphi }}(n)=2\pi \varphi (-n)$ yieldAll derivatives of this series converge uniformly. The localization formula in the previous answer therefore givesConsequently the Fourier transform of a periodic distribution isThe polynomial bound on $c_n$ and rapid decay of $\varphi (n)$ make every displayed pairing absolutely convergent. In particular there are no derivatives of Dirac delta distributions in this formula.

The inverse Fourier transform of ${\delta }_n$ is $(2\pi {)}^{-1}{e}^{inx}$. Applying continuous Fourier inversion on tempered distributions to the symmetric partial sums givesExplicitly, for every Schwartz function $\theta$, the pairing of the partial sum with $\theta$ is ${\sum }_{|n|\le M}{c}_n\hat{\theta }(-n)$, which converges absolutely to $⟨u,\theta ⟩$. The Fourier transform formula also proves that $c_n$ is independent of the chosen smooth periodic partition of unity. For a locally integrable $2\pi$ periodic function it reduces, by splitting the integral into periods, to $c_n=(2\pi {)}^{-1}{\int }_0^{2\pi }u(x)e^{-inx}dx$.

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,

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.