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Given a bilinear pairing $⟨\cdot ,\cdot ⟩$ and a seminorm $p$, its dual seminorm isIt may be infinite when $y$ does not annihilate the kernel of $p$. Whenever it is finite, the defining inequality gives $|⟨x,y⟩|\le p(x)p^{\ast }(y)$.

A Fréchet space is a complete metrizable locally convex space. Its topology can be described by a countable family of seminorms.

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,

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 family $\mathcal{F}$ of real-valued functions on a finite set and normalized inner product $⟨\cdot ,\cdot ⟩$, the test-function seminorm isIt measures the largest correlation of $g$ with an allowed test. It is a seminorm, and becomes a norm when the tests separate points.