Lebesgue Infinity-Space is Subset of Tempered Distribution Space

Theorem
Let $\map {L^\infty} \R$ be the Lebesgue infinity-space.

Let $\map {\SS'} \R$ be the tempered distribution space.

Then in the distributional sense it holds that:


 * $\map {L^\infty} \R \subseteq \map {\SS'} \R$

That is, every distribution defined by an element of $\map {L^\infty} \R$ is a tempered distribution.

Proof
Let $f \in \map {L^\infty} \R$.

Let $\phi \in \map \SS \R$ be a Schwartz test function.

Let $T_f : \map \SS \R \to \R$ be a mapping such that:


 * $\ds \map {T_f} \phi = \int_\R \map f x \map \phi x \rd x$

Then:

Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in $\map \SS \R$.

Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$ to $\mathbf 0$:


 * $\phi_n \stackrel \SS \longrightarrow \mathbf 0$

That is:


 * $\ds \forall l, m \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x} = 0$

Specifically:


 * $\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {\map {\phi_n} x} = 0$


 * $\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^2 \map {\phi_n} x} = 0$

Then:

On the other hand:


 * $\map {T_f} {\mathbf 0} = 0$

where $\mathbf 0$ denotes the zero Schwartz test function.

Hence:


 * $T_f \in \map {\SS'} \R$