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The Schwartz space iswhereA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{S}$ exactly when every one of these seminorms of ${\phi }_j-\phi$ tends to zero.

The space of tempered distributions ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$. Its standard weak convergence is

Continuity immediately implies sequential continuity. Conversely, suppose a linear form $u$ is sequentially continuous but not continuous at zero. Enumerate an increasing family of seminorms that generates the Schwartz space topology, and letSince $u$ is unbounded on every neighborhood of zero, choose ${\phi }_k\in U_k$ with $|u({\phi }_k)|\ge 1$. For each fixed $j$, $p_j({\phi }_k)<1/k$ once $k\ge j$, so ${\phi }_k\to 0$ in $\mathcal{S}$. Sequential continuity would imply $u({\phi }_k)\to 0$, a contradiction. Hence $u$ is continuous and belongs to ${\mathcal{S}}^{\prime }$.

Each continuous $f_{\alpha }$ of polynomial growth defines a regular tempered distribution byChoosing an integer $L>M_{\alpha }+n$ giveswhich is bounded by finitely many Schwartz space seminorms. Its distributional derivative satisfiesand is therefore tempered. A finite sum of continuous linear forms is continuous, so

Although $|u(x)|=e^{x^2}$ has superpolynomial growth, the rapidly varying phase makes $u$ an oscillatory tempered distribution. Split the integral against $\phi \in \mathcal{S}(\mathbb{R})$ into $|x|\le 1$ and the two tails. On a tail, with $\Phi (x)=e^{x^4}$,Integration by parts transfers the derivative toThis function and its derivative are integrable because $e^{x^2-x^4}$ dominates every polynomial, and the boundary term at infinity vanishes. The result is bounded by finitely many Schwartz space seminorms. The compact part has the same property. Thus the cutoff integrals converge and define a continuous linear functional:

For an orthogonal matrix $R$, a unit-Jacobian change of variables gives $\widehat{\phi \circ R^t}=\hat{\phi }\circ R^t$. The dual definition of the Fourier transform of a tempered distribution consequently givesSince the Fourier transform is invertible on the Schwartz space and its dual, a tempered distribution is invariant under a rotation group exactly when its transform is invariant. This includes radial tempered distributions.

A Schwartz function is an element of the Schwartz space: it is a smooth function and every derivative decreases faster than every inverse polynomial. Equivalently, $x^{\alpha }{D}^{\beta }f(x)$ defines a bounded function for every pair of multi-indices $\alpha ,\beta$.

These seminorms measure all polynomially weighted derivatives and generate the Fréchet topology of the Schwartz space. Finitely many such bounds control each weighted derivative's integrable norm by adding a spatial decay factor stronger than $⟨x{⟩}^{-n}$. Continuity of the Fourier transform follows by applying those bounds after differentiation and integration by parts. Equivalent families use powers of $⟨x⟩$ instead of individual monomials.

For a locally convex space $E$, the strong dual topology on its continuous dual space $E^{\prime }$ is generated by seminorms $p_B(T)={\sup }_{\phi \in B}|⟨T,\phi ⟩|$, where $B$ runs through the bounded sets in a topological vector space $E$. Convergence therefore means uniform convergence of pairings on every such set. For the Schwartz space, the defining boundedness is boundedness of every Schwartz seminorm, rather than one norm or one common compact support.

A tempered distribution is a continuous linear functional on the Schwartz space. Tempered distributions include functions of at most polynomial growth and admit a Fourier transform by duality.

A sequence $u_j\in {\mathcal{S}}^{\prime }$ converges weakly to $u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function $\phi$. This is the weak-star topology of the continuous dual of the Schwartz space, and differs from choosing a strong dual topology defined by uniform convergence on bounded sets of test functions.