Editing Distribution (mathematics) (section)

====Schwartz space====

The [[Schwartz space]] <math>\mathcal{S}(\R^n)</math> is the space of all smooth functions that are [[rapidly decreasing]] at infinity along with all partial derivatives. Thus <math>\phi:\R^n\to\R</math> is in the Schwartz space provided that any derivative of <math>\phi,</math> multiplied with any power of <math>|x|,</math> converges to 0 as <math>|x| \to \infty.</math> These functions form a complete TVS with a suitably defined family of [[seminorm]]s. More precisely, for any [[multi-indices]] <math>\alpha</math> and <math>\beta</math> define
<math display=block>p_{\alpha, \beta}(\phi) = \sup_{x \in \R^n} \left|x^\alpha \partial^\beta \phi(x) \right|.</math>

Then <math>\phi</math> is in the Schwartz space if all the values satisfy
<math display=block>p_{\alpha, \beta}(\phi) < \infty.</math>

The family of seminorms <math>p_{\alpha,\beta}</math> defines a [[locally convex]] topology on the Schwartz space. For <math>n = 1,</math> the seminorms are, in fact, [[Norm (mathematics)|norms]] on the Schwartz space. One can also use the following family of seminorms to define the topology:{{sfn|Trèves|2006|pp=92-94}}
<math display=block>|f|_{m,k} = \sup_{|p|\le m} \left(\sup_{x \in \R^n} \left\{(1 + |x|)^k \left|(\partial^\alpha f)(x) \right|\right\}\right), \qquad k,m \in \N.</math>

Otherwise, one can define a norm on <math>\mathcal{S}(\R^n)</math> via
<math display=block>\|\phi\|_k = \max_{|\alpha| + |\beta| \leq k} \sup_{x \in \R^n} \left| x^\alpha \partial^\beta \phi(x)\right|, \qquad k \ge 1.</math>

The Schwartz space is a [[Fréchet space]] (that is, a [[Complete topological vector space|complete]] [[Metrizable topological vector space|metrizable]] locally convex space). Because the [[Fourier transform]] changes <math>\partial^\alpha</math> into multiplication by <math>x^\alpha</math> and vice versa, this symmetry implies that the Fourier transform of a Schwartz function is also a Schwartz function.

A sequence <math>\{f_i\}</math> in <math>\mathcal{S}(\R^n)</math> converges to 0 in <math>\mathcal{S}(\R^n)</math> if and only if the functions <math>(1 + |x|)^k (\partial^p f_i)(x)</math> converge to 0 uniformly in the whole of <math>\R^n,</math> which implies that such a sequence must converge to zero in <math>C^\infty(\R^n).</math>{{sfn|Trèves|2006|pp=92–94}}

<math>\mathcal{D}(\R^n)</math> is dense in <math>\mathcal{S}(\R^n).</math> The subset of all analytic Schwartz functions is dense in <math>\mathcal{S}(\R^n)</math> as well.{{sfn|Trèves|2006|p=160}}

The Schwartz space is [[Nuclear space|nuclear]], and the tensor product of two maps induces a canonical surjective TVS-isomorphisms
<math display=block>\mathcal{S}(\R^m)\ \widehat{\otimes}\ \mathcal{S}(\R^n) \to \mathcal{S}(\R^{m+n}),</math>
where <math>\widehat{\otimes}</math> represents the completion of the [[injective tensor product]] (which in this case is identical to the completion of the [[projective tensor product]]).{{sfn|Trèves|2006|p=531}}