Mathematics_for_Physicists/chapters/sections/tempered_distributions.tex
2021-07-30 00:32:15 +02:00

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\begin{document}
\section{Outlook: Tempered Distributions}
\begin{defi}
A tempered distribution $f$ is a continuous, linear mapping
\begin{align*}
f: \mathcal{S}(\realn^d) &\longrightarrow \cmpln \\
\phi &\longmapsto f(\phi) = (f, \phi) \left( = \int f(x) \phi(x) \dd{x} \right)
\end{align*}
\end{defi}
\begin{thm}
Tempered distributions are linear, continuous mappings.
\end{thm}
\begin{proof}
To prove linearity, let $\phi, \psi \in \mathcal{S}(\realn^d)$ and $\lambda \in \cmpln$. Then
\begin{equation}
(f, \phi + \lambda\psi) = (f, \phi) + \lambda(f, \psi)
\end{equation}
For the continuity, we want to consider any sequence $\anyseqdef[\phi]{\mathcal{S}(\realn^d)}$ that converges to $\phi \in \mathcal{S}(\realn^d)$. I.e.
\begin{equation}
\lim_{n \rightarrow \infty} \sup_{x \in \realn^d} \abs{x^{\beta} \partial^{\alpha} (\phi_n(x) - \phi(x))} = 0, \quad \forall \alpha, \beta \in \natn_0^d
\end{equation}
Then we can conclude that
\begin{equation}
\lim_{n \rightarrow \infty} \abs{(f, \phi_n) - (f, \phi)} = 0
\end{equation}
\end{proof}
\begin{rem}
The space of all tempered distributions is denoted as $\mathcal{S}'(\realn^d)$.
\end{rem}
\begin{eg}
One important example is the Dirac deltra distribution:
\[
\delta: \mathcal{S}(\realn^d) \longrightarrow \cmpln
\]
It maps a function to its value at $0$.
\[
(\delta, \phi) = \int \delta(x) \phi(x) \dd{x} = \phi(0) \in \cmpln
\]
\end{eg}
\end{document}