The [Fourier transform](/page/Fourier%20Transform) of a function $f$ is defined by the integral \begin{align*} \hat{f}(\xi) &= \int_{\mathbb{R}^n} f(x) e^{-i\xi \cdot x} \, d\mathcal{L}^n(x). \end{align*} For this integral to converge, $f$ must be integrable. But many objects in PDE theory and harmonic analysis — the Dirac delta, constant [functions](/page/Function), plane waves, polynomials, solutions with distributional regularity — are not integrable. We need a framework in which the Fourier transform applies to all of these and remains a bijection, so that every equation $P(D)u = f$ can be transformed into the algebraic relation $P(\xi)\hat{u} = \hat{f}$ and back. The space of **tempered distributions** $\mathcal{S}'(\mathbb{R}^n)$ is precisely this framework. [motivation] ## Motivation ### The Schwartz Space: Perfect but Too Small We want the Fourier transform to *interact well with calculus*. The exchange identities \begin{align*} \widehat{\partial_j f}(\xi) &= i\xi_j \hat{f}(\xi), \\ \widehat{x_j f}(\xi) &= -i\partial_{\xi_j}\hat{f}(\xi) \end{align*} convert differentiation into polynomial multiplication and polynomial weights into differentiation on the frequency side. For both to work simultaneously, $f$ and all its derivatives must be integrable, and multiplying $f$ by any polynomial must preserve [integrability](/page/Integral). This forces $f$ to decay faster than every polynomial — which is precisely the condition defining the [Schwartz space](/page/Schwartz%20Space) $\mathcal{S}(\mathbb{R}^n)$. The Schwartz space has the remarkable property that \begin{align*} \mathcal{F}: \mathcal{S}(\mathbb{R}^n) &\xrightarrow{\;\sim\;} \mathcal{S}(\mathbb{R}^n) \end{align*} is a [topological automorphism](/theorems/228) — the Fourier transform maps Schwartz functions to Schwartz functions, continuously and bijectively. But $\mathcal{S}$ is too small: the Dirac delta $\delta_0$, constant functions, plane waves $e^{ik \cdot x}$, polynomials, and PDE solutions with distributional regularity all lie outside $\mathcal{S}$. ### Distributions: Large Enough but No Fourier Transform The space of [distributions](/page/Distribution) $\mathcal{D}'(\mathbb{R}^n)$ contains all these objects, but the Fourier transform cannot be defined on all of $\mathcal{D}'$ — distributions like $T_{e^{e^x}}$ grow too fast (see the example in the section on the [boundary](/page/Boundary) of temperedness). We need a space *large enough* to contain all the objects listed above, yet *small enough* that the Fourier transform extends to it. ### The Duality Extension The extension works by **duality**. For $f, \phi \in \mathcal{S}(\mathbb{R}^n)$, Fubini's theorem gives the transpose identity \begin{align*} \int_{\mathbb{R}^n} \hat{f}(x) \, \phi(x) \, d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n} f(x) \, \hat{\phi}(x) \, d\mathcal{L}^n(x), \end{align*} which shows that the Fourier transform of the functional $T_f$ can be defined by \begin{align*} \widehat{T_f}(\phi) &:= T_f(\hat{\phi}). \end{align*} For this to make sense for a general continuous linear functional $u$ on $\mathcal{S}$, we need $\hat{\phi} \in \mathcal{S}$ whenever $\phi \in \mathcal{S}$ — and this is exactly what $\mathcal{F}(\mathcal{S}) = \mathcal{S}$ guarantees. This leads to the space of **tempered distributions** $\mathcal{S}'(\mathbb{R}^n)$: the [topological dual](/page/Topological%20Dual) of $\mathcal{S}(\mathbb{R}^n)$. Because $\mathcal{F}$ is a [topological](/page/Topology) automorphism of $\mathcal{S}$, the composition $u \mapsto u \circ \mathcal{F}$ is automatically a bijection $\mathcal{S}' \to \mathcal{S}'$ — the Fourier transform extends as a [topological automorphism](/theorems/230), and this is the *raison d'être* of the space. The word "tempered" means "of moderate growth." A functional on $\mathcal{S}$ must be controlled by finitely many Schwartz semi-norms \begin{align*} \|\phi\|_{\alpha,\beta} &= \sup_{x \in \mathbb{R}^n} |x^\alpha \partial^\beta \phi(x)|, \end{align*} and this forces the distribution to have at most polynomial growth at infinity. Distributions that grow faster are continuous on $\mathcal{D}(\mathbb{R}^n)$ (where compact support kills any growth) but not on $\mathcal{S}$ (where only polynomial decay is available to compensate). The temperedness condition is the maximal growth rate compatible with the duality extension. [/motivation] ## Definition [definition: Tempered Distribution] A **tempered distribution** is a continuous linear functional on $\mathcal{S}(\mathbb{R}^n)$. Explicitly, a map \begin{align*} u: \mathcal{S}(\mathbb{R}^n) &\to \mathbb{C} \end{align*} is a tempered distribution if it is linear — meaning $u(\alpha\phi + \beta\psi) = \alpha u(\phi) + \beta u(\psi)$ for all $\phi, \psi \in \mathcal{S}(\mathbb{R}^n)$ and $\alpha, \beta \in \mathbb{C}$ — and continuous with respect to the Schwartz topology defined on the [Schwartz Space](/page/Schwartz%20Space) page. The set of all tempered distributions is denoted $\mathcal{S}'(\mathbb{R}^n)$. [/definition] We need a notion of convergence on $\mathcal{S}'(\mathbb{R}^n)$. The natural choice is the weakest topology that makes every evaluation functional continuous, ensuring that $u_k \to u$ if and only if $u_k(\phi) \to u(\phi)$ for every test function $\phi$. [definition: Weak Star Topology On Tempered Distributions] For each $\phi \in \mathcal{S}(\mathbb{R}^n)$, the **evaluation map** is the function \begin{align*} \mathrm{ev}_\phi: \mathcal{S}'(\mathbb{R}^n) &\to \mathbb{C} \\ u &\mapsto u(\phi). \end{align*} The **weak-$*$ topology** on $\mathcal{S}'(\mathbb{R}^n)$ is the coarsest topology such that $\mathrm{ev}_\phi$ is continuous for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. It is generated by the sub-basis of [sets](/page/Set) \begin{align*} \{u \in \mathcal{S}'(\mathbb{R}^n) : |u(\phi) - u_0(\phi)| < \varepsilon\} \end{align*} for $u_0 \in \mathcal{S}'(\mathbb{R}^n)$, $\phi \in \mathcal{S}(\mathbb{R}^n)$, and $\varepsilon > 0$. [/definition] This is an instance of an initial topology, characterised by the property that a net $(u_\lambda)$ converges to $u$ if and only if $u_\lambda(\phi) \to u(\phi)$ for every $\phi$. For [sequences](/page/Sequence): \begin{align*} u_k \to u \text{ in } \mathcal{S}'(\mathbb{R}^n) \quad &\Longleftrightarrow \quad u_k(\phi) \to u(\phi) \text{ for every } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} Since $\mathcal{S}(\mathbb{R}^n)$ is a [Fréchet space](/page/Fr%C3%A9chet%20Space) (metrizable and complete), [continuity](/page/Continuity) of a linear functional admits two equivalent reformulations. [quotetheorem:456] The equivalence (1) $\Leftrightarrow$ (2) uses the Fréchet structure: for linear functionals on metrizable locally convex spaces, continuity and sequential continuity coincide. The equivalence (1) $\Leftrightarrow$ (3) is the characterisation used in practice. The integers $N$ and $M$ measure the *order* and *growth* of the distribution: $M$ controls how many derivatives of the test function $u$ "uses", and $N$ controls how much polynomial decay of $\phi$ is consumed. This should be contrasted with the [characterisation of distributions](/theorems/449) on $\mathcal{D}(\Omega)$. For $u \in \mathcal{D}'(\Omega)$, the semi-norm bound can use *different* constants $C_K$ and orders $N_K$ on different compact sets $K$: \begin{align*} |u(\phi)| &\le C_K \sum_{|\beta| \le N_K} \sup_{x \in K} |\partial^\beta \phi(x)| \quad \text{for } \operatorname{supp}(\phi) \subseteq K. \end{align*} For tempered distributions, a *single* bound with *fixed* constants $C$, $N$, $M$ works for all $\phi \in \mathcal{S}(\mathbb{R}^n)$ simultaneously — this is the price of temperedness. [example: Verifying Temperedness Of The Dirac Delta] The Dirac delta $\delta_0: \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}$ defined by $\delta_0(\phi) := \phi(0)$ is a tempered distribution. The semi-norm bound from the [characterisation theorem](/theorems/456) is verified by \begin{align*} |\delta_0(\phi)| &= |\phi(0)| \le \sup_{x \in \mathbb{R}^n} |\phi(x)| = \|\phi\|_{0,0}, \end{align*} giving $C = 1$, $N = 0$, $M = 0$. The derivative $\partial_j \delta_0$, defined by $(\partial_j \delta_0)(\phi) = -\partial_j \phi(0)$, satisfies \begin{align*} |(\partial_j \delta_0)(\phi)| &\le \sup_{x \in \mathbb{R}^n} |\partial_j\phi(x)| = \|\phi\|_{0, e_j}, \end{align*} so $M = 1$: the derivative $\partial_j\delta_0$ "uses" one derivative of the test function. More generally, $\partial^\alpha\delta_0$ has \begin{align*} |(\partial^\alpha\delta_0)(\phi)| &= |\partial^\alpha\phi(0)| \le \|\phi\|_{0,\alpha}, \end{align*} giving $N = 0$ and $M = |\alpha|$ — the order of the distribution equals the number of derivatives it consumes. [/example] ## Tempered Distributions and Functions A tempered distribution is an *abstract* object — a continuous linear functional — not a function. Expressions like $|\hat{u}(\xi)|^2$ or $\int (1+|\xi|^2)^s |\hat{f}(\xi)|^2 \, d\mathcal{L}^n(\xi)$ only make sense after establishing that the distribution is *represented by* a measurable function. This section explains when such identification is valid and where the boundary of temperedness lies. ### Embedding $L^p$ into $\mathcal{S}'$ Every measurable function $f$ with at most polynomial growth defines a tempered distribution $T_f$ via \begin{align*} T_f(\phi) &:= \int_{\mathbb{R}^n} f(x) \, \phi(x) \, d\mathcal{L}^n(x), \quad \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} The map $f \mapsto T_f$ is [injective](/theorems/450): $T_f = T_g$ implies $f = g$ a.e., so $L^p(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$ as a set-theoretic inclusion. [example: $L^p$ Functions As Tempered Distributions] For $f \in L^p(\mathbb{R}^n)$ with $1 \le p \le \infty$, let $q$ be the conjugate exponent with $1/p + 1/q = 1$ and choose $N > n/q$. Hölder's inequality gives \begin{align*} |T_f(\phi)| &\le \|f\|_{L^p}\|\phi\|_{L^q}. \end{align*} Since $\phi \in \mathcal{S}(\mathbb{R}^n)$ satisfies $|\phi(x)| \le \|\phi\|_{N,0}(1+|x|)^{-N}$ with $N = |\alpha_0|$, and $(1+|x|)^{-Nq}$ is integrable when $Nq > n$: \begin{align*} \|\phi\|_{L^q}^q &= \int_{\mathbb{R}^n} |\phi(x)|^q \, d\mathcal{L}^n(x) \le \|\phi\|_{N,0}^q \int_{\mathbb{R}^n} (1+|x|)^{-Nq} \, d\mathcal{L}^n(x) = C_N^q \|\phi\|_{N,0}^q. \end{align*} Therefore $|T_f(\phi)| \le \|f\|_{L^p} \cdot C_N \|\phi\|_{N,0}$, confirming the [semi-norm bound](/theorems/456) with a single Schwartz semi-norm. [/example] ### Functions of Polynomial Growth Many natural objects lie outside every $L^p$ space yet still define tempered distributions, provided they grow at most polynomially. [example: Functions Of Polynomial Growth] A measurable function $g: \mathbb{R}^n \to \mathbb{C}$ satisfying $|g(x)| \le C(1+|x|)^M$ defines a tempered distribution via $T_g(\phi) := \int g(x)\phi(x)\,d\mathcal{L}^n(x)$. Splitting the integrand: \begin{align*} |T_g(\phi)| &\le C \int_{\mathbb{R}^n} (1+|x|)^M |\phi(x)| \, d\mathcal{L}^n(x) \\ &= C \int_{\mathbb{R}^n} \frac{(1+|x|)^{M+n+1} |\phi(x)|}{(1+|x|)^{n+1}} \, d\mathcal{L}^n(x) \\ &\le C \cdot \|\phi\|_{M+n+1,\, 0} \int_{\mathbb{R}^n} (1+|x|)^{-(n+1)} \, d\mathcal{L}^n(x) \\ &= C \cdot K_n \cdot \|\phi\|_{M+n+1,\, 0}, \end{align*} where $K_n = \int(1+|x|)^{-(n+1)}\,d\mathcal{L}^n(x) < \infty$. This gives the [semi-norm bound](/theorems/456) with $N = M+n+1$ and $M_0 = 0$. [/example] [example: Plane Waves] For $k \in \mathbb{R}^n$, the plane wave $e^{ik \cdot x}$ is bounded ($M = 0$) and defines \begin{align*} T_{e^{ik \cdot (\cdot)}}(\phi) &= \int_{\mathbb{R}^n} e^{ik \cdot x}\phi(x)\,d\mathcal{L}^n(x) = \hat\phi(-k). \end{align*} This lies outside every $L^p$ space with $p < \infty$ (constant modulus $1$ over $\mathbb{R}^n$), yet it is a well-behaved tempered distribution. Its Fourier transform is a shifted Dirac delta: \begin{align*} \widehat{T_{e^{ik \cdot (\cdot)}}} &= (2\pi)^n \delta_k, \end{align*} formalising the idea that a plane wave has a single "frequency" at $k$. [/example] ### When a Tempered Distribution "Is" a Function Given $u \in \mathcal{S}'(\mathbb{R}^n)$, the question $u = T_f$ for some measurable $f$ does not always have a positive answer. The Dirac delta has no locally integrable representative (as proved on the [Distribution](/page/Distribution) page). When we write "$(1+|\xi|^2)^{s/2}\hat{u} \in L^2(\mathbb{R}^n)$," the left-hand side is first formed as a tempered distribution (the product defined in the multiplication section below), and "$\in L^2$" asserts that this distribution equals $T_g$ for some $g \in L^2(\mathbb{R}^n)$. Only *after* establishing this does $|\hat{u}(\xi)|^2$ acquire meaning — where $\hat{u}(\xi)$ denotes the $L^2$ representative, defined $\mathcal{L}^n$-a.e. Writing \begin{align*} \|u\|_{H^s} &= \left(\int_{\mathbb{R}^n} (1+|\xi|^2)^s |\hat{u}(\xi)|^2 \, d\mathcal{L}^n(\xi)\right)^{1/2} \end{align*} *without first establishing that $(1+|\xi|^2)^{s/2}\hat{u}$ is an $L^2$ function* conflates a tempered distribution with a pointwise-defined function. ### The Hierarchy of Function Spaces Inside $\mathcal{S}'$ There is a chain of continuous embeddings \begin{align*} \mathcal{D}(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n) \xrightarrow{\; f \,\mapsto\, T_f \;} \mathcal{S}'(\mathbb{R}^n) \hookrightarrow \mathcal{D}'(\mathbb{R}^n), \end{align*} where all arrows are injections. At the left end, every object is a classical function; moving rightward, objects become increasingly abstract. Each $\phi \in \mathcal{S}(\mathbb{R}^n)$ belongs to every $L^p$ space and has $\hat{\phi} \in \mathcal{S}$, so expressions like \begin{align*} \int_{\mathbb{R}^n} (1+|\xi|^2)^s |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi) \end{align*} are unambiguous for every $s$. This allows one to *define* the Sobolev norm on $\mathcal{S}$ first and take the completion to obtain $H^s$. ### The Boundary of Temperedness The spaces of [test functions](/page/Test%20Function) satisfy $\mathcal{D}(\mathbb{R}^n) \subseteq \mathcal{S}(\mathbb{R}^n)$ with a continuous inclusion. Dualising reverses the inclusion: \begin{align*} \mathcal{S}'(\mathbb{R}^n) \hookrightarrow \mathcal{D}'(\mathbb{R}^n), \end{align*} and the injection is proper. The boundary lies at polynomial growth: $|f(x)| \le C(1+|x|)^M$ implies $T_f \in \mathcal{S}'$, while super-polynomial growth may fail. [example: Failure Of Temperedness] Let $n = 1$ and $g(x) = e^{e^x}$. The functional $T_g(\phi) := \int_{\mathbb{R}} e^{e^x} \phi(x) \, d\mathcal{L}^1(x)$ is a distribution: for $\phi \in \mathcal{D}(\mathbb{R})$ supported in $[-R, R]$, \begin{align*} |T_g(\phi)| &\le e^{e^R}\|\phi\|_{L^1} \le 2R\,e^{e^R}\sup|\phi| < \infty. \end{align*} However, $T_g \notin \mathcal{S}'(\mathbb{R})$. If the [semi-norm bound](/theorems/456) held, testing against $\phi_R(x) = \eta(x - R)$ (a translated bump) would give \begin{align*} |T_g(\phi_R)| &\le C \sum_{j \le N, k \le M} \|\phi_R\|_{j,k} \le C'(R+1)^N \end{align*} (since $\|\phi_R\|_{j,k} \le (R+1)^j \|\eta^{(k)}\|_\infty$). But \begin{align*} T_g(\phi_R) &= \int e^{e^x}\eta(x-R)\,d\mathcal{L}^1(x) \ge c\, e^{e^{R-1/2}} \end{align*} for a constant $c > 0$. Since $e^{e^{R-1/2}}$ grows faster than any polynomial in $R$, this contradicts the bound for large $R$. [/example] The injection is injective: if $u \in \mathcal{S}'(\mathbb{R}^n)$ restricts to zero on $\mathcal{D}(\mathbb{R}^n)$, then $u = 0$ since $\mathcal{D}$ is [dense in $\mathcal{S}$](/theorems/455). The dual pairing reverses inclusion: $\mathcal{D} \subseteq \mathcal{S}$ implies $\mathcal{S}' \subseteq \mathcal{D}'$. The advantage of the smaller dual $\mathcal{S}'$ is that the Fourier transform [extends to it](/theorems/230). ## Multiplication by Slowly Increasing Functions The exchange identity $\widehat{\partial_j u} = i\xi_j \hat{u}$ requires multiplying the tempered distribution $\hat{u}$ by the polynomial $i\xi_j$. Since a tempered distribution is a functional, not a function, the pointwise product has no meaning. We define the product by transferring the multiplication to the test function: if $\psi$ is a function and $u$ is a tempered distribution, we set \begin{align*} (\psi u)(\phi) &:= u(\psi \phi). \end{align*} For this to define an element of $\mathcal{S}'$, we need (i) $\psi\phi \in \mathcal{S}$ whenever $\phi \in \mathcal{S}$, and (ii) the map $\phi \mapsto \psi\phi$ to be continuous $\mathcal{S} \to \mathcal{S}$. Both conditions restrict the growth of $\psi$. [definition: Slowly Increasing Function] A **slowly increasing function** (or **function of moderate growth**) is a smooth function $\psi \in C^\infty(\mathbb{R}^n)$ such that for every multi-index $\beta \in \mathbb{N}_0^n$, there exist constants $C_\beta > 0$ and $M_\beta \ge 0$ satisfying \begin{align*} |\partial^\beta \psi(x)| &\le C_\beta(1+|x|)^{M_\beta} \quad \text{for all } x \in \mathbb{R}^n. \end{align*} The space of all slowly increasing functions is denoted $\mathcal{O}_M(\mathbb{R}^n)$. [/definition] The class $\mathcal{O}_M(\mathbb{R}^n)$ includes all polynomials, all bounded smooth functions with bounded derivatives, and the Bessel potential $(1+|x|^2)^{s/2}$ for every $s \in \mathbb{R}$. What makes $\mathcal{O}_M$ the right class of multipliers is that the product $\psi\phi$ stays in $\mathcal{S}$ and depends continuously on $\phi$. For $\psi \in \mathcal{O}_M$ and $\phi \in \mathcal{S}$, the Leibniz rule gives \begin{align*} x^\alpha \partial^\beta(\psi\phi)(x) &= x^\alpha \sum_{\gamma \le \beta} \binom{\beta}{\gamma} (\partial^\gamma\psi)(x)(\partial^{\beta-\gamma}\phi)(x). \end{align*} Taking the supremum and applying the growth bound $|\partial^\gamma\psi(x)| \le C_\gamma(1+|x|)^{M_\gamma}$: \begin{align*} \|\psi\phi\|_{\alpha,\beta} &= \sup_x |x^\alpha \partial^\beta(\psi\phi)(x)| \\ &\le \sum_{\gamma \le \beta} \binom{\beta}{\gamma} C_\gamma \sup_x (1+|x|)^{M_\gamma} |x^\alpha| \, |\partial^{\beta-\gamma}\phi(x)| \\ &\le \sum_{\gamma \le \beta} \binom{\beta}{\gamma} C_\gamma \, \|\phi\|_{\alpha + M_\gamma e,\, \beta-\gamma} \end{align*} where $\|\phi\|_{\alpha + M_\gamma e,\, \beta-\gamma} = \sup_x (1+|x|)^{M_\gamma} |x^\alpha| \, |\partial^{\beta-\gamma}\phi(x)|$ is bounded by a finite combination of Schwartz semi-norms. Each $\|\psi\phi\|_{\alpha,\beta}$ is thus controlled by finitely many semi-norms of $\phi$, which gives continuity of $\phi \mapsto \psi\phi$ as a map $\mathcal{S} \to \mathcal{S}$. [definition: Product Of A Tempered Distribution With A Slowly Increasing Function] Let $u \in \mathcal{S}'(\mathbb{R}^n)$ and $\psi \in \mathcal{O}_M(\mathbb{R}^n)$. The **product** $\psi u$ is the tempered distribution defined by \begin{align*} (\psi u)(\phi) &:= u(\psi \phi) \quad \text{for every } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} This is well-defined because $\psi\phi \in \mathcal{S}$, linear because $u$ is linear, and continuous because the composition \begin{align*} \mathcal{S}(\mathbb{R}^n) \xrightarrow{\;\phi \,\mapsto\, \psi\phi\;} \mathcal{S}(\mathbb{R}^n) \xrightarrow{\;u\;} \mathbb{C} \end{align*} is a composite of continuous maps. [/definition] When $u = T_f$ is a [regular distribution](/page/Regular%20Distribution), this recovers the pointwise product: \begin{align*} (\psi T_f)(\phi) &= T_f(\psi\phi) = \int_{\mathbb{R}^n} f(x)\psi(x)\phi(x)\,d\mathcal{L}^n(x) = T_{\psi f}(\phi). \end{align*} ### Why Only Slowly Increasing Functions? If $\psi$ grows faster than polynomially, $\psi\phi$ may leave $\mathcal{S}$. Consider $\psi(x) = e^{|x|^2}$ and the Gaussian $\phi(x) = e^{-|x|^2} \in \mathcal{S}(\mathbb{R}^n)$. Then \begin{align*} (\psi\phi)(x) &= e^{|x|^2} \cdot e^{-|x|^2} = 1. \end{align*} The constant function $1$ is bounded but not rapidly decreasing, so $\psi\phi \notin \mathcal{S}(\mathbb{R}^n)$ even though $\phi \in \mathcal{S}(\mathbb{R}^n)$. In $\mathcal{D}'(\mathbb{R}^n)$, *any* smooth function is a valid multiplier because compact support of test functions kills all growth. The class $\mathcal{O}_M$ is the price of working with $\mathcal{S}'$ rather than $\mathcal{D}'$. [example: Multiplying The Dirac Delta By A Polynomial] The coordinate function $x_j \in \mathcal{O}_M(\mathbb{R}^n)$ is a polynomial. The product $x_j \delta_0$ acts by \begin{align*} (x_j \delta_0)(\phi) &:= \delta_0(x_j \phi) = (x_j\phi)(0) = 0 \cdot \phi(0) = 0. \end{align*} Thus $x_j \delta_0 = 0$ in $\mathcal{S}'(\mathbb{R}^n)$. For the product of $x_j$ with the derivative $\partial_k\delta_0$: \begin{align*} (x_j \cdot \partial_k\delta_0)(\phi) &= (\partial_k\delta_0)(x_j\phi) = -\partial_k(x_j\phi)(0) = -\delta_{jk}\phi(0) - 0 \cdot \partial_k\phi(0) = -\delta_{jk}\phi(0). \end{align*} So $x_j \cdot \partial_k\delta_0 = -\delta_{jk}\,\delta_0$. More generally, for a polynomial $p$ and multi-index $\alpha$: \begin{align*} (p \cdot \partial^\alpha \delta_0)(\phi) &= (-1)^{|\alpha|} \partial^\alpha(p\phi)(0) = (-1)^{|\alpha|} \sum_{\gamma \le \alpha} \binom{\alpha}{\gamma} (\partial^\gamma p)(0) \, (\partial^{\alpha-\gamma}\phi)(0), \end{align*} which is a finite linear combination of $\partial^{\alpha-\gamma}\delta_0$ weighted by $(\partial^\gamma p)(0)$. [/example] [example: The Sobolev Weight As A Multiplier] For $s \in \mathbb{R}$, define $\psi_s(\xi) := (1+|\xi|^2)^{s/2}$. This is smooth on $\mathbb{R}^n$ (the base is bounded below by $1$). The first derivative is \begin{align*} \partial_{\xi_j}\psi_s(\xi) &= s\,\xi_j\,(1+|\xi|^2)^{s/2 - 1}, \end{align*} and the general derivative has the form $\partial^\beta\psi_s(\xi) = P_\beta(\xi)(1+|\xi|^2)^{s/2 - |\beta|}$ where $P_\beta$ is a polynomial of degree $\le |\beta|$. Therefore \begin{align*} |\partial^\beta \psi_s(\xi)| &\le C_\beta(1+|\xi|^2)^{(s - |\beta|)/2} \cdot (1+|\xi|)^{|\beta|} \le C_\beta'(1+|\xi|)^{|s| + |\beta|}, \end{align*} confirming $\psi_s \in \mathcal{O}_M(\mathbb{R}^n)$. The product $(1+|\xi|^2)^{s/2}\hat{u}$ for $u \in \mathcal{S}'(\mathbb{R}^n)$ acts by \begin{align*} \bigl((1+|\xi|^2)^{s/2}\hat{u}\bigr)(\phi) &= \hat{u}\bigl((1+|\xi|^2)^{s/2}\phi\bigr) \quad \text{for every } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} This is the product appearing in the Fourier characterisation of [Sobolev spaces](/page/Inhomogeneous%20Sobolev%20Spaces). [/example] ## [Differentiation](/page/Derivative) For $u \in \mathcal{S}'(\mathbb{R}^n)$ and a multi-index $\alpha$, the [distributional derivative](/page/Distributional%20Derivative) is defined by \begin{align*} (\partial^\alpha u)(\phi) &:= (-1)^{|\alpha|} u(\partial^\alpha \phi). \end{align*} Since $\partial^\alpha: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ is [continuous](/theorems/454), $\partial^\alpha u \in \mathcal{S}'(\mathbb{R}^n)$. Every tempered distribution is infinitely differentiable. The derivative and the product interact via the Leibniz rule: for $\psi \in \mathcal{O}_M(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$, \begin{align*} \partial_j(\psi u) &= (\partial_j\psi) u + \psi (\partial_j u). \end{align*} This is verified by expanding both sides on a test function $\phi$. The left side gives \begin{align*} (\partial_j(\psi u))(\phi) &= -(\psi u)(\partial_j\phi) = -u(\psi \partial_j\phi). \end{align*} The right side gives \begin{align*} ((\partial_j\psi)u)(\phi) + (\psi \partial_j u)(\phi) &= u((\partial_j\psi)\phi) + (\partial_j u)(\psi\phi) \\ &= u((\partial_j\psi)\phi) - u(\partial_j(\psi\phi)) \\ &= u\bigl((\partial_j\psi)\phi - (\partial_j\psi)\phi - \psi\partial_j\phi\bigr) \\ &= -u(\psi\partial_j\phi), \end{align*} using the classical product rule $\partial_j(\psi\phi) = (\partial_j\psi)\phi + \psi(\partial_j\phi)$. ## The Fourier Transform on $\mathcal{S}'(\mathbb{R}^n)$ With multiplication and differentiation defined, we state the central result justifying the construction of $\mathcal{S}'(\mathbb{R}^n)$. [quotetheorem:230] The definition is \begin{align*} \hat{u}(\phi) &:= u(\hat{\phi}) \quad \text{for every } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} Since $\mathcal{F}: \mathcal{S} \to \mathcal{S}$ is a [topological automorphism](/theorems/228), $\hat{u} = u \circ \mathcal{F}$ is automatically continuous, and invertibility follows from $\mathcal{F}^{-1}$ being equally well-behaved. When $u = T_f$ for $f \in \mathcal{S}$, the transpose identity recovers the classical Fourier transform: \begin{align*} \widehat{T_f}(\phi) &= T_f(\hat{\phi}) = \int f \cdot \hat{\phi}\,d\mathcal{L}^n = \int \hat{f} \cdot \phi\,d\mathcal{L}^n = T_{\hat{f}}(\phi). \end{align*} More generally, the same compatibility $\widehat{T_f} = T_{\hat{f}}$ holds for any $f \in L^1(\mathbb{R}^n)$ — the proof requires only Fubini's theorem and the integrability of Schwartz functions. [quotetheorem:718] This contrasts with $\mathcal{D}'(\mathbb{R}^n)$: the Fourier transform does not map $\mathcal{D}$ into itself (by the Paley–Wiener theorem, a compactly supported function with compactly supported Fourier transform must be zero), so the transpose construction fails. ### Exchange Identities The Fourier transform interconverts differentiation and polynomial multiplication: \begin{align*} \widehat{\partial_j u} &= i\xi_j \hat{u}, \\ \widehat{x_j u} &= -i\partial_{\xi_j} \hat{u}. \end{align*} Both sides are tempered distributions. For the first identity, the left side acts on $\phi$ by \begin{align*} \widehat{\partial_j u}(\phi) &= (\partial_j u)(\hat{\phi}) = -u(\partial_j\hat{\phi}), \end{align*} and the right side by \begin{align*} (i\xi_j \hat{u})(\phi) &= \hat{u}(i\xi_j\phi) = u(\widehat{i\xi_j\phi}). \end{align*} These agree because the Schwartz-level exchange identity gives $\widehat{i\xi_j\phi} = -\partial_j\hat{\phi}$. Similarly, the second identity follows from $\widehat{\partial_{\xi_j}\phi} = -ix_j\hat{\phi}$. For a differential operator $P(D) = \sum_{|\alpha| \le m} a_\alpha \partial^\alpha$: \begin{align*} \widehat{P(D)u} &= P(i\xi)\hat{u}, \quad \text{where } P(i\xi) = \sum_{|\alpha| \le m} a_\alpha (i\xi)^\alpha. \end{align*} This converts constant-coefficient PDEs into algebraic equations. ### Fourier Transforms of Distinguished Distributions [example: Fourier Transforms Of Distinguished Tempered Distributions] **The constant function.** The constant $1 \in L^\infty(\mathbb{R}^n)$ defines $T_1$. Computing: \begin{align*} \hat{T}_1(\phi) &= T_1(\hat\phi) = \int_{\mathbb{R}^n} \hat\phi(x)\,d\mathcal{L}^n(x) = (2\pi)^n \phi(0) = (2\pi)^n \delta_0(\phi), \end{align*} where the third equality is the [Fourier inversion theorem](/theorems/364) ($\int \hat\phi\,d\mathcal{L}^n = (2\pi)^n\phi(0)$). Thus \begin{align*} \hat{1} &= (2\pi)^n\delta_0. \end{align*} The classical integral for $\hat{1}$ does not converge — the distributional framework is essential. **The Dirac delta.** Computing directly: \begin{align*} \hat\delta_0(\phi) &= \delta_0(\hat\phi) = \hat\phi(0) = \int_{\mathbb{R}^n} \phi(x)\,d\mathcal{L}^n(x) = T_1(\phi). \end{align*} Thus $\hat\delta_0 = T_1$, i.e., the Fourier transform of a point mass is the constant function $1$. Since $\hat{\delta}_0$ is represented by a function, $|\hat{\delta}_0(\xi)|^2 = 1$ is legitimate. **Polynomials.** The monomial $x^\alpha$ defines a tempered distribution. Applying the exchange identity $|\alpha|$ times: \begin{align*} \widehat{x^\alpha} &= (-i\partial_\xi)^\alpha \hat{1} = (-i)^{|\alpha|} \partial_\xi^\alpha \bigl[(2\pi)^n\delta_0\bigr] = (-i)^{|\alpha|} (2\pi)^n \partial^\alpha \delta_0. \end{align*} [/example] ### Fundamental Solutions via the Fourier Transform For a constant-coefficient operator $P(D)$, a **fundamental solution** $E$ satisfies $P(D)E = \delta_0$. The Fourier transform reduces this to \begin{align*} P(i\xi)\hat{E} &= \hat{\delta}_0 = 1, \end{align*} an algebraic division problem. [example: Fundamental Solution Of The Laplacian] For $-\Delta \Phi = \delta_0$ in $\mathcal{S}'(\mathbb{R}^n)$ with $n \ge 3$, the exchange identity gives \begin{align*} |\xi|^2 \hat\Phi(\xi) &= 1. \end{align*} Since $|\xi|^{-2}$ is locally integrable for $n \ge 3$ (the singularity $|\xi|^{-2}$ near the origin has $\int_{B(0,1)} |\xi|^{-2}\,d\mathcal{L}^n(\xi) = \omega_{n-1}\int_0^1 r^{n-3}\,dr < \infty$ when $n \ge 3$), we set \begin{align*} \hat\Phi(\xi) &= |\xi|^{-2}, \end{align*} which defines a regular tempered distribution. Inverting the Fourier transform yields the Newton potential $\Phi(x) = c_n|x|^{-(n-2)}$. [/example] ## Convolution and Regularisation The convolution of two arbitrary tempered distributions may not be well-defined, but the convolution with a Schwartz function always produces a smooth function of polynomial growth. [quotetheorem:458] The convolution is defined by duality: $u$ acts on the Schwartz function $y \mapsto \phi(x-y)$: \begin{align*} (u * \phi)(x) &:= u_y(\phi(x-y)). \end{align*} The result is a function (not a distribution), and the theorem guarantees smoothness and polynomial growth. The derivative can be placed on either factor: \begin{align*} \partial^\beta(u * \phi) &= (\partial^\beta u) * \phi = u * (\partial^\beta \phi). \end{align*} Even for $u = \partial^\alpha\delta_0$, the convolution is smooth: \begin{align*} (\partial^\alpha\delta_0 * \phi)(x) &= (\partial^\alpha\delta_0)_y(\phi(x-y)) = (-1)^{|\alpha|}\partial_y^\alpha\phi(x-y)\big|_{y=0} = (-1)^{|\alpha|}(-1)^{|\alpha|}\partial^\alpha\phi(x) = \partial^\alpha\phi(x). \end{align*} [example: Convolution With The Dirac Delta] For the Dirac delta itself ($\alpha = 0$): \begin{align*} (\delta_0 * \phi)(x) &= \delta_0(\phi(x - \cdot)) = \phi(x - 0) = \phi(x). \end{align*} The delta is the identity for convolution: $\delta_0 * \phi = \phi$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. [/example] ### Regularisation by Mollification Taking $\phi = \eta_\varepsilon$ (a [standard mollifier](/page/Standard%20Mollifier) with support in $B(0, \varepsilon)$), the [convolutions](/page/Convolution) $u * \eta_\varepsilon$ are smooth and converge in $\mathcal{S}'$: \begin{align*} u * \eta_\varepsilon &\to u \quad \text{in } \mathcal{S}'(\mathbb{R}^n) \text{ as } \varepsilon \to 0. \end{align*} This holds because $T_{u * \eta_\varepsilon}(\phi) = u(\check{\eta}_\varepsilon * \phi) \to u(\phi)$ as $\check{\eta}_\varepsilon * \phi \to \phi$ in $\mathcal{S}(\mathbb{R}^n)$. ### The Fourier Exchange Formula Convolution and the Fourier transform are intertwined: \begin{align*} \widehat{u * \phi} &= \hat{u} \cdot \hat{\phi}. \end{align*} On the left, $u * \phi$ has polynomial growth, so $\widehat{u * \phi} \in \mathcal{S}'$. On the right, $\hat{\phi} \in \mathcal{S} \subset \mathcal{O}_M$, so the product $\hat{u} \cdot \hat{\phi}$ is defined by \begin{align*} (\hat{u} \cdot \hat{\phi})(\psi) &= \hat{u}(\hat{\phi}\psi) \quad \text{for } \psi \in \mathcal{S}(\mathbb{R}^n). \end{align*} This is the engine behind fundamental solutions: if $P(D)E = \delta_0$, then $u = E * f$ solves $P(D)u = f$, and on the Fourier side \begin{align*} \hat{u} &= \hat{E} \cdot \hat{f}. \end{align*} ## Structure Theorem Every tempered distribution is a finite-order derivative of a continuous function of polynomial growth. This means $\mathcal{S}'(\mathbb{R}^n)$ is not as exotic as its abstract definition suggests. [quotetheorem:457] The proof constructs the continuous function $g$ by repeated integration of the semi-norm bound from the [characterisation theorem](/theorems/456): each integration reduces the order by one and introduces one order of polynomial growth. After finitely many integrations, one arrives at a continuous function. This is specific to $\mathcal{S}'$: in $\mathcal{D}'(\mathbb{R}^n)$, the distribution \begin{align*} u &= \sum_{k=0}^\infty \partial^k \delta_{x_k} \end{align*} for a sequence $x_k \to \infty$ requires derivatives of unbounded order. The temperedness condition forces the order to be finite. [example: Structure Theorem For Derivatives Of The Delta] The Dirac delta $\delta_0$ in dimension $n = 1$: the Heaviside function $H(x) = \mathbb{1}_{[0,\infty)}(x)$ is bounded and continuous away from $0$, and \begin{align*} H'(\phi) &= -\int_0^\infty \phi'(x)\,d\mathcal{L}^1(x) = \phi(0) = \delta_0(\phi), \end{align*} so $\delta_0 = H'$. For $\delta_0'$, define $g(x) = -\max(x,0)$, which is continuous with $|g(x)| \le |x|$. Then \begin{align*} g''(\phi) &= \int_{\mathbb{R}} g(x)\phi''(x)\,d\mathcal{L}^1(x) = -\int_0^\infty x\,\phi''(x)\,d\mathcal{L}^1(x). \end{align*} Integrating by parts twice (boundary terms vanish by rapid decay of $\phi$): \begin{align*} -\int_0^\infty x\,\phi''(x)\,d\mathcal{L}^1(x) &= -\bigl[x\phi'(x)\bigr]_0^\infty + \int_0^\infty \phi'(x)\,d\mathcal{L}^1(x) \\ &= 0 + \bigl[\phi(x)\bigr]_0^\infty - 0 = -\phi(0) = -\delta_0(\phi). \end{align*} So $g'' = -\delta_0$, hence $\delta_0' = -g''' = -(g'')' = (-(-\delta_0))' = \delta_0'$ — consistent. More directly, $\delta_0 = H' = (-g)''$, confirming the structure theorem with $\alpha = (2)$ and $g$ continuous with linear growth. [/example] ## Fourier Characterisation of Sobolev Spaces The principal payoff of tempered distributions for PDE theory is the ability to define [Sobolev spaces](/page/Sobolev%20Space) via the Fourier transform, extending the classical $W^{k,p}$ theory to all real orders. For $\phi \in \mathcal{S}(\mathbb{R}^n)$, Plancherel's theorem gives $\|\partial^\alpha \phi\|_{L^2} = \|\xi^\alpha \hat{\phi}\|_{L^2}$, so \begin{align*} \sum_{|\alpha| \le k} \|\partial^\alpha \phi\|_{L^2}^2 &= \int_{\mathbb{R}^n} \sum_{|\alpha| \le k} |\xi^\alpha|^2 \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi). \end{align*} The sum $\sum_{|\alpha| \le k} |\xi^\alpha|^2$ is comparable to $(1+|\xi|^2)^k$: both are degree-$k$ polynomials in $|\xi|^2$ that agree at leading order and are bounded below by $1$. Therefore \begin{align*} \sum_{|\alpha| \le k} \|\partial^\alpha \phi\|_{L^2}^2 &\sim \int_{\mathbb{R}^n} (1+|\xi|^2)^k \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi). \end{align*} **Having $k$ derivatives in $L^2$ is equivalent to the Fourier transform being square-integrable against the weight $(1+|\xi|^2)^k$.** The weight $(1+|\xi|^2)^{s/2}$ has a frequency interpretation: $\xi$ is the frequency, and the condition \begin{align*} \int_{\mathbb{R}^n} (1+|\xi|^2)^s |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi) &< \infty \end{align*} for $s > 0$ says that high-frequency components are suppressed; for $s < 0$, it allows rougher distributions. This motivates the definition: for $s \in \mathbb{R}$, the [inhomogeneous Sobolev space](/page/Inhomogeneous%20Sobolev%20Spaces) is \begin{align*} H^s(\mathbb{R}^n) &:= \left\{ u \in \mathcal{S}'(\mathbb{R}^n) \;\middle|\; (1+|\xi|^2)^{s/2}\hat{u} \in L^2(\mathbb{R}^n) \right\}, \end{align*} where $(1+|\xi|^2)^{s/2}\hat{u}$ is the product from the multiplication section, acting by \begin{align*} \bigl((1+|\xi|^2)^{s/2}\hat{u}\bigr)(\phi) &= \hat{u}\bigl((1+|\xi|^2)^{s/2}\phi\bigr). \end{align*} The condition "$\in L^2$" means this tempered distribution is represented by $g \in L^2(\mathbb{R}^n)$. The norm is \begin{align*} \|u\|_{H^s} &:= \|(1+|\xi|^2)^{s/2}\hat{u}\|_{L^2} = \left(\int_{\mathbb{R}^n} (1+|\xi|^2)^s \, |g(\xi)|^2 \, d\mathcal{L}^n(\xi)\right)^{1/2}. \end{align*} For $s = k \in \mathbb{N}_0$, this recovers $W^{k,2}(\mathbb{R}^n)$. For $s = 0$, $H^0 = L^2$. For $s < 0$, the space contains distributions: $\delta_0 \in H^s(\mathbb{R}^n)$ if and only if $s < -n/2$ (Problem 4). The Bessel potential operator $(1-\Delta)^{s/2}$, defined on the Fourier side as multiplication by $(1+|\xi|^2)^{s/2}$, gives an isomorphism \begin{align*} (1-\Delta)^{s/2}: H^t(\mathbb{R}^n) &\xrightarrow{\;\sim\;} H^{t-s}(\mathbb{R}^n) \end{align*} for every $t \in \mathbb{R}$, as made precise by the [Helmholtz isomorphism](/theorems/227). This gives the $H^s$ scale the structure of a "ruler" for regularity. ## References 1. L. Grafakos, *Classical Fourier Analysis*, 3rd ed. (2014). 2. L. Hörmander, *The Analysis of Linear Partial Differential Operators I* (1983). 3. M. Reed and B. Simon, *Methods of Modern Mathematical Physics I: Functional Analysis* (1980). 4. W. Rudin, *Functional Analysis* (1991). 5. E. M. Stein and R. Shakarchi, *Functional Analysis* (2011).