Codex @codex  0

We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The Schwartz space consists of smooth functions with finite seminormsfor every pair of multi-indices. Convergence means convergence in each seminorm. The tempered distribution space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual space, with weak convergence tested against every Schwartz function. A continuous functional satisfies a bound by finitely many of these seminorms, equivalently by one sufficiently large weighted derivative seminorm.

Use the Fourier transform conventionDifferentiation under the integral and integration by parts express ${\xi }^{\alpha }{\partial }_{\xi }^{\beta }\hat{f}$ as a constant of modulus one times the Fourier transform of ${\partial }_x^{\alpha }(x^{\beta }f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz seminorms, using the integrable weight $(1+|x|{)}^{-s}$. Thus $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

For completeness, Fourier inversion follows by inserting $e^{-\epsilon |\xi {|}^2}$ in the inverse integral and using Fubini's theorem. The result is convolution with the Gaussian approximate identityIt tends to $f$, while integrability of $\hat{f}$ allows the damping factor to be removed by dominated convergence theorem. Consequently ${\mathcal{F}}^{2}f=(2\pi {)}^{n}f(-\cdot )$. Reflection preserves every Schwartz seminorm, so the inverse transform is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the Fourier transform of a tempered distribution by transposition,The Schwartz-space continuity just proved makes this a tempered distribution. Its inverse is $(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$, where $⟨\mathcal{R}u,\psi ⟩=⟨u,\psi (-\cdot )⟩$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the strong dual topology, because the Schwartz-space maps take bounded sets to bounded sets.

The convolution of a tempered distribution with a Schwartz function isSmooth dependence of translated Schwartz functions gives ${\partial }^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$. The finite-seminorm estimate and $1+|y|\le (1+|x|)(1+|x-y|)$ show that each derivative has at most polynomial growth. In particular, $u\ast \phi$ is a smooth function defining a tempered distribution. It need not itself be a Schwartz function; for example $1\ast \phi =\int \phi$.

Writing $\check{\phi }(y)=\phi (-y)$, its distributional pairing is $⟨u\ast \phi ,\psi ⟩=⟨u,\check{\phi }\ast \psi ⟩$. The inner convolution is a Schwartz function, and this identity follows by integration in the Schwartz topology, justified by the weighted seminorm estimates. A direct Fubini's theorem calculation gives $\check{\phi }\ast \hat{\psi }=\mathcal{F}(\hat{\phi }\psi )$. HenceMultiplication is well defined because multiplication by $\hat{\phi }$ acts continuously on $\mathcal{S}$.

Now write the Hilbert transform as convolution with $K={\pi }^{-1}\mathrm{pv}(1/x)$, the principal-value reciprocal distribution. The given Heaviside function transform, together with ${\mathcal{F}}^{2}H=2\pi H(-\cdot )$, yieldsThe value of the sign function at zero is irrelevant to its regular distribution. Thus the Hilbert-transform Fourier multiplier isApplying Plancherel theorem, whose normalization here is $\Vert f{\Vert }_2^2=(2\pi {)}^{-1}\Vert \hat{f}{\Vert }_2^2$, proves the isometry:The principal-value integral agrees with this convolution: near its singular point subtract $\phi (x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The Hilbert transform has domain $\mathcal{S}$, but generally does not take values in $\mathcal{S}$, as the tail in part (c) demonstrates.

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 ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, with inverse factor $(2\pi {)}^{-n}$. The Paley–Wiener–Schwartz theorem can be stated with sharp convex support: for a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let its support function be $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$. Then $F$ is the Fourier transform of a unique distribution supported in $K$ if and only if it is entire andfor some $C$ and nonnegative integer $M$. For $K={\overline{B}}_R$, this is the familiar bound $C(1+|z|{)}^M{e}^{R|\mathrm{Im}z|}$. Allowing some $R$ characterizes all compactly supported distributions. Here the extension of the Fourier transform is $F(z)=⟨u,e^{-ix\cdot z}⟩$, interpreted with a cutoff equal to one near the support of a distribution.

First suppose $\mathrm{supp}u\subset K$. Fix one smooth cutoff function equal to one near $K$. Pairing the resulting compactly supported exponential with $u$ shows that $F$ is an entire function: differentiation with respect to $z_j$ inserts $-i{x}_j$, and the power series converges in the test-function seminorms uniformly on compact subsets of ${\mathbb{C}}^n$.

To keep the exponential type exactly $H_K$, rather than that of a fixed larger neighborhood, use a shrinking-cutoff exponential-type estimate. There are cutoffs ${\chi }_{\epsilon }$ equal to one on $K+\epsilon B_1$, supported in $K+3\epsilon {\overline{B}}_1$, and satisfying $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. One construction convolves the indicator of $K+2\epsilon B_1$ with a unit-mass mollifier supported in $\epsilon B_1$. Continuity of $u$ on a fixed compact neighborhood gives a finite order of a distribution $q$ there. By the product rule,Taking $\epsilon =(1+|z|{)}^{-1}$ proves the required bound with $M=2q$, since the extra exponential factor is at most $e^3$. The value of the pairing is independent of the chosen cutoff because all cutoffs agree near $\mathrm{supp}u$. Thus the forward direction has the exact asserted support function, without an unproved estimate on derivatives restricted only to $K$.

Conversely, suppose the entire $F$ has the stated bound. Its restriction to real frequency has polynomial growth, so define a tempered distribution $u$ byThe Schwartz space decay makes this integral absolutely convergent and continuous; by Fourier inversion, $\hat{u}=F$ on real frequency. It remains to prove the support assertion by mollifier regularization for contour recovery of support.

Choose a nonnegative unit-mass mollifier $\rho$ supported in $B_1$, and setOn real frequency, $F_{\epsilon }$ decays faster than any polynomial after enough applications of integration by parts to $\hat{\rho }$; hence $h_{\epsilon }$ is smooth, by differentiation under the integral sign. More generally, for every integer $L$ there is $C_{\epsilon ,L}$ such thatTo obtain this estimate, write the transform of ${\rho }_{\epsilon }$ as the real-frequency transform of $e^{x\cdot \eta }{\rho }_{\epsilon }(x)$ and integrate by parts; each derivative introduces at most one factor of $|\eta |$.

Fix a unit vector $v$ and take $L>M+n+1$. A contour-shift proof of the Paley–Wiener–Schwartz theorem moves the inverse-transform contour to ${\mathbb{R}}^n+itv$:Here is a justification of the shift. Rotate coordinates so $v$ is the first coordinate direction, apply the Cauchy integral theorem on a rectangle in the first complex variable, and integrate over the other real variables. For fixed $t$, the vertical sides at real part $\pm A$ have an integrated bound proportional to $A^{M-L+n-1}$, and hence vanish as $A\to \infty$. The same decay bounds give absolute convergence on the horizontal sides. No contour shift of an unregularized polynomially growing integral is needed.

Positive homogeneity of the support function now givesIf $x\notin K+\epsilon {\overline{B}}_1$, the Hahn-Banach separation theorem supplies a unit vector $v$ with $x\cdot v>H_K(v)+\epsilon$. Letting $t\to \infty$ proves $h_{\epsilon }(x)=0$. Therefore $h_{\epsilon }\in C_c^{\infty }$ and $\mathrm{supp}{h}_{\epsilon }\subset K+\epsilon {\overline{B}}_1$.

Since $\hat{\rho }(\epsilon \xi )\to 1$ and its modulus is at most one on real frequency, the dominated convergence theorem in the formula for $⟨u,\phi ⟩$ gives $h_{\epsilon }\to u$ as tempered distributions, and thus as distributions. A test function supported outside $K$ has positive distance from $K$, so its pairing with $h_{\epsilon }$ is zero for all sufficiently small $\epsilon$. This proves $\mathrm{supp}u\subset K$. The Fourier transform of a compactly supported distribution constructed in the forward direction equals $F$ on ${\mathbb{R}}^n$; applying the one-variable identity theorem successively in each coordinate extends the equality to ${\mathbb{C}}^n$. Injectivity of the Fourier transform of a tempered distribution proves uniqueness and completes both directions.

For the independence application, put $g_m(z)=e^{iz\cdot y_m}{f}_m(z)$ and $r_m=1/(m+1)$. The ball version of the Paley–Wiener–Schwartz theorem supplies a nonzero distribution $v_m$ supported in ${\overline{B}}_{r_m}(0)$ with ${\hat{v}}_m=g_m$. By the Translation property of the Fourier transform,where $⟨{\tau }_{y}v,\phi ⟩=⟨v,\phi (\cdot +y)⟩$. Distinct integer points are at least distance one apart. For distinct positive indices $j,k$, the largest possible radius sum is $1/2+1/3=5/6<1$. Thus these closed balls, and hence the distribution supports, are pairwise disjoint.

If ${\sum }_{m=1}^N{c}_m{f}_m=0$, injectivity of the Fourier transform gives ${\sum }_m{c}_m{w}_m=0$. For each $j$, choose a smooth cutoff function equal to one near its ball and zero near all other balls. Multiplying the distributional identity by this cutoff isolates $c_j{w}_j=0$. Since $f_j$ is not identically zero, $w_j\ne 0$, so $c_j=0$. Consequently $f_1,\dots ,f_N$ are linearly independent over $\mathbb{C}$. This Fourier independence from disjoint distribution supports uses the quantitative exponential types to establish support separation.

The multiplication of a distribution by a smooth function is $⟨\psi T,\phi ⟩=⟨T,\psi \phi ⟩$. If $\psi$ is a Schwartz function, the Leibniz rule shows that $\phi \mapsto \psi \phi$ is a continuous map $\mathcal{S}\to \mathcal{S}$, so $\psi T$ is a tempered distribution. When both $\psi$ and $T$ are radial, $\psi \circ R^t=\psi$, andThus Multiplication by a radial Schwartz function preserves radial tempered distributions.

For the convolution of a tempered distribution with a Schwartz function, setTranslations of $\psi$ depend smoothly on $x$ in the Schwartz space, so this is a smooth function with ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T_y,{\partial }^{\alpha }\psi (x-y)⟩$. The bound for $T$ by finitely many seminorms, together with $1+|y|\le (1+|x|)(1+|x-y|)$, provesThus the function also defines a tempered distribution. For a radial function $\psi$, put $g_x(y)=\psi (x-y)$. Then $\psi (Rx-y)=\psi (x-R^{t}y)=g_x(R^{t}y)$, soHence Convolution with a radial Schwartz function preserves radial tempered distributions.

Smoothing alone need not give a Schwartz function: for the constant tempered distribution $T=1$ and a Schwartz function with integral one, $T\ast \psi =1$. The radial Schwartz approximation of tempered distributions therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial mollifier $\rho \in C_c^{\infty }$, supported in the unit ball with integral one, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one on the unit ball. PutEach $f_j$ is a smooth function of compact support, hence a Schwartz function, and the two invariance calculations above make it radial.

It remains to prove convergence, including the simultaneous changes of both scales. With $\check{\rho }(x)=\rho (-x)$, the distributional convolution pairing isFor every fixed $m$, the Leibniz rule, rapid decay outside the radius-$j$ ball, and the chain rule for $\chi (x/j)$ giveConvolution by ${\check{\rho }}_{\epsilon }$ is uniformly bounded in $q_m$ for $0<\epsilon \le 1$, because its shifts have size at most one. The mean value theorem applied to $\phi (x-y)-\phi (x)$ similarly givesSplitting the error into the convolved cutoff error and the approximate identity error provesIf $|⟨T,\theta ⟩|\le C{q}_m(\theta )$, this yieldsConsequentlyweakly, and even in the strong dual topology, since $q_{m+1}$ is uniformly bounded on every subset of the Schwartz space that is a bounded set in a topological vector space. No assertion that $T\ast {\rho }_{1/j}$ itself is rapidly decreasing is needed.

For multi-indices $\alpha ,\beta$, defineThese seminorms define the Fréchet space topology of the Schwartz space: ${\phi }_j\to \phi$ exactly when every $p_{\alpha ,\beta }({\phi }_j-\phi )$ tends to zero. An equivalent increasing family isA tempered distribution is a continuous linear functional on this space. Equivalently, $|⟨T,\phi ⟩|\le C{q}_m(\phi )$ for some $C,m$. The usual weak convergence of tempered distributions means $⟨T_j,\phi ⟩\to ⟨T,\phi ⟩$ for every fixed Schwartz function. The strong dual topology instead requires uniform convergence on every subset of the Schwartz space that is a bounded set in a topological vector space; the Fourier maps below are continuous in both topologies.

Using $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, differentiation under the integral and integration by parts giveThe omitted coefficient has modulus one. By the Leibniz rule, every integrand is a finite sum of a polynomial times a derivative of $\phi$. Inserting the integrable weight $(1+|x|{)}^{-n-1}$ bounds its L1 norm by finitely many Schwartz space seminorms. In particular $q_m(\hat{\phi })\le C_m{q}_{m+n+1}(\phi )$. Thus the Fourier transform maps $\mathcal{S}$ continuously into itself.

For completeness, the Fourier inversion theorem follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi {|}^2}$ is $g_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. The Fourier transform of a Gaussian and Fubini's theorem showAs $t\downarrow 0$, the right side tends to $\phi (x)$ by the approximate identity property, while the left side tends to the undamped inverse integral by dominated convergence theorem, since $\hat{\phi }\in L^1$. HenceThe inverse is $(2\pi {)}^{-n}$ times reflection composed with the continuous Fourier transform, and is therefore continuous on the Schwartz space. This proves a continuous linear isomorphism with continuous inverse.

Define the Fourier transform of a tempered distribution byThe continuous map on Schwartz space makes this a tempered distribution; the transpose of ${\mathcal{F}}^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the strong dual topology, the transform of a bounded set of Schwartz functions is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.

Now let $R\in SO(n)$ be a rotation matrix in the special orthogonal group and write ${\rho }_{R}T=T\circ R$. A change of variables with unit Jacobian givesThe rotation equivariance of the Fourier transform on tempered distributions follows by duality:Therefore $\widehat{{\rho }_{R}T}={\rho }_R\hat{T}$. Applying this identity and the inverse Fourier transform gives the two directions:This is precisely preservation of radial tempered distributions. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary radial function.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

Use the normalized inner productThe test-function seminorm generated by $\mathcal{F}$ and its dual test-function norm areThe first quantity may only be a seminorm if $\mathcal{F}$ does not separate all functions in ${\mathcal{H}}_N$; correspondingly, the second may be infinite outside the linear span detected by $\mathcal{F}$.

Using multi-index notation, the Schwartz space isA sequence ${\phi }_m$ converges to $\phi$ in this Fréchet space when $p_{\alpha ,\beta }({\phi }_m-\phi )\to 0$ for every $\alpha ,\beta$. The space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ of tempered distributions is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$, and $u_m\to u$ there means weak convergence of distributions, namely $⟨u_m,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{S}$.

Continuity of a linear functional immediately implies that ${\phi }_m\to 0$ entails $⟨u,{\phi }_m⟩\to 0$. Conversely, enumerate the Schwartz seminorms as $p_1,p_2,\dots$. If $u$ were not continuous, then for each $m$ one could choose ${\phi }_m$ such thatEvery fixed seminorm tends to zero along this sequence, so ${\phi }_m\to 0$ in $\mathcal{S}$, contradicting the assumed sequential property. This is the sequential continuity criterion for a linear map on a metrizable topological vector space.

The Schwartz space on the real line isThese seminorms define its Fréchet space topology. The space of tempered distributions is its continuous dual space,