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 $\mathcal{S}'$