Tempered Distribution Space is Proper Subset of Distribution Space
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 $\map {\SS'} \R \subseteq \map {\DD'} \R$.
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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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions