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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 angular-frequency Fourier transform and its inverse on $\mathcal{S}({\mathbb{R}}^n)$ areIt extends to a tempered distribution by duality:

Since ${\delta }_{1/t}\phi (x)=\phi (x/t)$ maps the Schwartz space continuously to itself, the formuladefines a continuous linear functional on $\mathcal{S}$, hence a tempered distribution. For a locally integrable function $u$, the change of variables formula $y=tx$ givesso this dilation of a distribution agrees with ordinary function dilation.

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.

Although $x\mapsto |x{|}^{-1}$ is not integrable on all of ${\mathbb{R}}^3$, it defines a tempered distribution. For $x\ne 0$, the Gamma function integral givesThe three-dimensional form of the Fourier transform of a Gaussian isTherefore, up to a nonzero constant depending on the Fourier-transform convention,For $\xi \ne 0$, the substitution $u=|\xi {|}^2/(4t)$ turns the last integral intoThus, as asserted by the Fourier transform of inverse distance in three dimensions,as a tempered distribution, where $C^{\prime }\ne 0$.

For $f\in L^1(\mathbb{R})$, its regular tempered distribution acts by integration, and the distributional transform agrees with the functionThe triangle inequality gives the uniform boundIf ${\lambda }_j\to \lambda$, then $f(x)e^{-i{\lambda }_{j}x}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the Lebesgue integrable function $|f|$. The Dominated convergence theorem therefore proves that $\hat{f}$ is continuous. In fact the Riemann-Lebesgue lemma also gives $\hat{f}(\lambda )\to 0$ as $|\lambda |\to \infty$.

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

With the angular-frequency convention, the Fourier transform of a Schwartz function isIt maps $\mathcal{S}$ continuously to itself. The transform of $u\in {\mathcal{S}}^{\prime }$ is defined through the dual pairing:up to the fixed reflection and $2\pi$ factor if the inverse-transform convention is used for the test function.

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus

For a compact convex set $K\subset {\mathbb{R}}^n$, letThe Paley–Wiener–Schwartz theorem says that if $u$ is a compactly supported distribution with support in $K$, its Fourier--Laplace transformis entire and, for some $C,N$,Conversely, every entire function satisfying such an estimate is the transform of a distribution supported in $K$.

For the forward direction, compact support lets $u$ act on the exponential after insertion of a cutoff equal to one near $K$. Differentiation in $z$ may be passed under the pairing, proving entire analyticity. The finite-order estimate for $u$ bounds derivatives of the exponential on $K$ by a polynomial in $|z|$ times $e^{H_K(\mathrm{Im}z)}$.

Conversely, restrict the entire function $F$ to ${\mathbb{R}}^n$. Its polynomial growth defines a tempered distribution $u$ by inverse Fourier transform. If a test function is supported outside $K$, separate its compact support from $K$ by a real vector $\eta$. Shifting the Fourier inversion contour from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n+it\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{t{H}_K(\eta )}$ as $t\to \infty$, so the pairing vanishes. Hence $\mathrm{supp}u\subset K$, completing the converse.

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

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

A function has polynomial growth when $|f(x)|\le C(1+|x|{)}^N$ for some constants $C,N$. Such growth is slow enough for a locally integrable function to define a tempered distribution.

A real simple pole is not locally integrable in the ordinary improper sense. Its symmetric Cauchy principal value defines a tempered distribution. For the $e^{-ikx}$ transform convention, contour indentation or sine integration gives the displayed nondecaying oscillatory contribution. Upper and lower boundary-value prescriptions instead change it by $\mp i\pi \delta (x-a)$ before transforming. A pole prescription must therefore be stated before claiming a Fourier asymptotic expansion.

For $n>1$, a tempered distribution $T$ on ${\mathbb{R}}^n$ is radial when $T\circ R=T$ for every $R\in SO(n)$, where $⟨T\circ R,\phi ⟩=⟨T,\phi \circ R^t⟩$. This agrees with the usual radial function definition for regular smooth distributions, because the special orthogonal group acts transitively on spheres.

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.

Generalized Gaussian white noise over a real Hilbert space $H=L^2$ is an isonormal Gaussian process $W(h)$ with covariance $⟨h,g{⟩}_H$. Its law $P_0$ is carried on a larger observation space; the identity is not the covariance of an infinite-dimensional $H$-valued Gaussian measure because it is not a trace-class operator. A deterministic signal $h\in H$ defines a translated observation law $P_h$ with Radon-Nikodym derivativeIn an orthonormal basis, $W_m(h)={\sum }_j{m}_j{h}_j$ is a stochastic series with total variance $\Vert h{\Vert }_H^2$, not an inner product of two $H$-valued observations. The formula follows from the Cameron-Martin theorem for a Gaussian measure in its generalized white-noise version, or from finite-dimensional Gaussian likelihood ratios. On ${\mathbb{R}}^2$, generalized noise can be realized on tempered distributions; an unweighted global negative Sobolev space is not automatically a suitable almost-sure support on an unbounded domain. The stochastic series avoids imposing that unsupported regularity.