Lebesgue 1-Space is Subset of Tempered Distribution Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map {L^1} \R$ be the Lebesgue $1$-space.

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


Then:

$\map {L^1} \R \subseteq \map {\SS'} \R$


Proof

Let $f \in \map {L^1} \R$.

By definition of the Lebesgue space:

$\ds \norm f_1 = \int_\R \size {\map f x} \rd x < \infty$

where $\norm {\, \cdot \,}_1$ denotes the 1-seminorm.

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

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

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

By Integral Operator is Linear, $\map {T_f} \phi$ is linear with respect to both $f$ and $\phi$.

Furthermore:

\(\ds \size {\map {T_f} \phi}\) \(=\) \(\ds \size {\int_\R \map f x \map \phi x \rd x}\)
\(\ds \) \(\le\) \(\ds \int_\R \size {\map f x} \size {\map \phi x} \rd x\)
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in \R} \size {\map \phi x} \int_\R \size {\map f x} \rd x\) Definition of Supremum of Real-Valued Function
\(\ds \) \(=\) \(\ds \norm f_1 \norm \phi_\infty\) Definition of Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions

Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in the Schwartz space.

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

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

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

That is:

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

Set $m = 0$ and $l = 0$.

Then:

\(\ds \lim_{n \mathop \to \infty} \norm {\phi_n}_\infty\) \(=\) \(\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {\map {\phi_n} x}\)
\(\ds \) \(=\) \(\ds 0\)

Therefore:

\(\ds \lim_{n \mathop \to \infty} \size {\map {T_f} {\phi_n} }\) \(\le\) \(\ds \lim_{n \mathop \to \infty} \norm f_1 \norm {\phi_n}_\infty\)
\(\ds \) \(=\) \(\ds 0\)

Thus:

$\ds \lim_{n \mathop \to \infty} \map {T_f} {\phi_n} = 0$

On the other hand:

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

Hence, from the definition of the tempered distribution it follows that:

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

$\blacksquare$



Sources