Locally Integrable Function defines Distribution

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Theorem

Let $f \in \map {L^1_{loc}} {\R^d}$ be a locally integrable function.

Let $\map \DD {\R^d}$ be the test function space.

Let $T_f : \map \DD {\R^d} \to \C$ be a mapping.


Then $T_f$ is a distribution.


Proof

Existence

Let $\phi \in \map \DD {\R^d}$ be a test function.

Let $T_f$ be defined as

$\ds T_f = \int_{\R^d} \map f {\mathbf x} \map \phi {\mathbf x} \rd \mathbf x$

By definition, $\phi$ has the compact support.

Together with the properties of $f$ we have that $T_f$ is bounded with respect to any compact range of integration.

$\Box$


Linearity

By Linear Combination of Integrals we have that $T_f$ is a linear mapping.

$\Box$


Continuity

By Convergent Sequence Minus Limit, we can shift the sequence to set its limit to zero.

Let $\mathbf 0 : \R^d \to 0$ be the zero mapping.

Let $\phi_n$ converge to $\mathbf 0$ in $\map \DD {\R^d}$:

$\phi_n \stackrel \DD {\longrightarrow} {\mathbf 0}$

Let $K \subseteq \R^d$ be the compact support of all $\phi_n$.

\(\ds \size {\map {T_f} {\phi_n} }\) \(=\) \(\ds \size {\int_{\R^d} \map f {\mathbf x} \map {\phi_n} {\mathbf x} \rd \mathbf x}\)
\(\ds \) \(\le\) \(\ds \int_K \size {\map f {\mathbf x} } \size {\map {\phi_n} {\mathbf x} } \rd \mathbf x\) Definition of Test Function, Definition of Locally Integrable Function
\(\ds \) \(\le\) \(\ds \norm {\phi_n}_\infty \int_K \size {\map f {\mathbf x} } \rd \mathbf x\) Definition of Supremum Norm
\(\ds \) \(\stackrel {n \mathop \to \infty} {\longrightarrow}\) \(\ds 0\) Convergence in test function space; Definition of Uniform Convergence

$\Box$


By definition, $T_f$ is a distribution.

$\blacksquare$


Sources