Editing Distribution (mathematics) (section)

====Convolution of a smooth function with a distribution====

Let <math>f \in C^\infty(\R^n)</math> and <math>T \in \mathcal{D}'(\R^n)</math> and assume that at least one of <math>f</math> and <math>T</math> has compact support. The '''{{em|convolution}}''' of <math>f</math> and <math>T,</math> denoted by <math>f \ast T</math> or by <math>T \ast f,</math> is the smooth function:{{sfn|Trèves|2006|pp=284-297}}
<math display=block>\begin{alignat}{4}
f \ast T : \,& \R^n && \to    \,&& \Complex \\
             & x    && \mapsto\,&& \left\langle T, \tau_x \tilde{f} \right\rangle \\
\end{alignat}</math>
satisfying for all <math>p \in \N^n</math>:
<math display=block>\begin{align}
&\operatorname{supp}(f \ast T) \subseteq \operatorname{supp}(f)+ \operatorname{supp}(T) \\[6pt]
&\text{ for all } p \in \N^n: \quad
\begin{cases}\partial^p \left\langle T, \tau_x \tilde{f} \right\rangle = \left\langle T, \partial^p \tau_x \tilde{f} \right\rangle \\
\partial^p (T \ast f) = (\partial^p T) \ast f = T \ast (\partial^p f).
\end{cases}
\end{align}</math>

Let <math>M</math> be the map <math>f \mapsto T \ast f</math>. If <math>T</math> is a distribution, then <math>M</math> is continuous as a map <math>\mathcal{D}(\R^n) \to C^\infty(\R^n)</math>. If <math>T</math> also has compact support, then <math>M</math> is also continuous as the map <math>C^\infty(\R^n) \to C^\infty(\R^n)</math> and continuous as the map <math>\mathcal{D}(\R^n) \to \mathcal{D}(\R^n).</math>{{sfn|Trèves|2006|pp=284-297}}

If <math>L : \mathcal{D}(\R^n) \to C^\infty(\R^n)</math> is a continuous linear map such that <math>L \partial^\alpha \phi = \partial^\alpha L \phi</math> for all <math>\alpha</math> and all <math>\phi \in \mathcal{D}(\R^n)</math> then there exists a distribution <math>T \in \mathcal{D}'(\R^n)</math> such that <math>L \phi = T \circ \phi</math> for all <math>\phi \in \mathcal{D}(\R^n).</math>{{sfn|Rudin|1991|pp=149-181}}

'''Example.'''{{sfn|Rudin|1991|pp=149-181}} Let <math>H</math> be the [[Heaviside step function|Heaviside function]] on <math>\R.</math> For any <math>\phi \in \mathcal{D}(\R),</math>
<math display=block>(H \ast \phi)(x) = \int_{-\infty}^x \phi(t) \, dt.</math>

Let <math>\delta</math> be the Dirac measure at 0 and let <math>\delta'</math> be its derivative as a distribution. Then <math>\delta' \ast H = \delta</math> and <math>1 \ast \delta' = 0.</math> Importantly, the associative law fails to hold:
<math display=block>1 = 1 \ast \delta = 1 \ast (\delta' \ast H ) \neq (1 \ast \delta') \ast H = 0 \ast H = 0.</math>