Tempered Distribution Space is Proper Subset of Distribution Space

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then $\map {\SS'} \R$ is a proper subset of $\map {\DD'} \R$:

$\map {\SS'} \R \subsetneqq \map {\DD'} \R$


Proof

By Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space we have that $\map {\SS'} \R \subseteq \map {\DD'} \R$.



Consider the real function $\map f x = e^{x^2}$.

We have that:

Real Power Function for Positive Integer Power is Continuous
Exponential Function is Continuous/Real Numbers
Composite of Continuous Mappings is Continuous

Thus, $f$ is a continuous real function.

Also:

$\forall x \in \R : e^{x^2} < \infty$

Hence, $f$ is locally integrable.

By Locally Integrable Function defines Distribution, $T_f \in \map {\DD'} \R$.

Aiming for a contradiction, suppose $T_f$ is a tempered distribution.

We have that $e^{-x^2}$ is a Schwartz test function.

Then:

\(\ds \map {T_f} {e^{-x^2} }\) \(=\) \(\ds \int_{- \infty}^\infty e^{x^2} e^{-x^2} \rd x\)
\(\ds \) \(=\) \(\ds \int_{- \infty}^\infty 1 \rd x\)
\(\ds \) \(=\) \(\ds \infty\)

Hence, $\map {T_f} {e^{-x^2} } \notin \R$.

This is a contradiction.

Therefore, $T_f \notin \map {\SS'} \R$ while at the same time $T_f \in \map {\DD'} \R$.

$\blacksquare$


Sources