Multiplication of Distribution induced by Locally Integrable Function by Smooth Function

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Theorem

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

Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth function.

Let $T_f \in \map {\DD'} {\R^d}$ be a distribution induced by $f$.


Then in the distributional sense it holds that:

$\alpha T_f = T_{\alpha f}$


Proof

Let $\Omega \subseteq \R^d$ be a compact subset.

Then for all $\mathbf x \in \Omega$ we have that $\map \alpha {\mathbf x}$ is bounded.

Hence, $\alpha f$ is locally integrable.

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

Then:

\(\ds \alpha \map {T_f} \phi\) \(=\) \(\ds \map {T_f} {\alpha \phi}\) Definition of Multiplication of Distribution by Smooth Function
\(\ds \) \(=\) \(\ds \int_{\R^d} \map f {\mathbf x} \map \alpha {\mathbf x} \map \phi {\mathbf x} \rd \mathbf x\) Locally Integrable Function defines Distribution
\(\ds \) \(=\) \(\ds \map {T_{\alpha f} } \phi\) Locally Integrable Function defines Distribution

$\blacksquare$

Sources

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