📂Distribution Theory

Proof That the Space of Test Functions is a Proper Subset of the Schwartz Space

Theorem¹

Let $\mathcal{D}$ be the space of test functions, and let $\mathcal{S}$ be the Schwartz space. Then the following equation holds.

$$ \mathcal{D} \subsetneq \mathcal{S} $$

Proof

Strategy: First, we show that all test functions belong to the Schwartz space, and then by providing an example of a Schwartz function that is not a test function, we prove the theorem.

Schwartz Functions

Define $\phi$ as a Schwartz function if it satisfies the following two conditions:

• (a) $\phi \in C^{\infty}$
• (b) For all multi-indices $\alpha$, $\beta$, $\left| x^{\beta}D^{\alpha}\phi (x) \right| <\infty$

Assume an arbitrary test function $\phi$ is given. Since $\phi \in C_{c}^{\infty}$, it satisfies condition (a). The support of $\phi$ is bounded, so there exists $r>0$ satisfying the following equation.

$$ \mathrm{supp}\phi \subset \overline{B}(r) $$

Here, $\overline{B}(r)$ is a closed ball centered at the origin with radius $r$. Moreover, due to the properties of test functions, for any multi-index $\alpha$, the following holds true.

$$ \mathrm{supp}D^{\alpha}\phi\subset \mathrm{supp}\phi \subset \overline{B}(r) $$

Now, let’s consider two different cases.

• Case 1. $x \notin \overline{B}(r)$

  Then, $D^{\alpha}\phi (x)=0$ and

  $$ \left| x^{\beta}D^{\alpha}\phi (x) \right|=0 <\infty $$

• Case 2. $x \in \overline{B}(r)$

  $D^{\alpha}\phi (x)$ is a continuous function and, since continuous functions have both maximum and minimum values on compact spaces, there exists some positive number $K>0$ for which the following equation holds.

  $$ \left| x^{\beta}D^{\alpha}\phi (x) \right|\le r^{\beta}K <\infty $$

  Therefore, because $\phi \in \mathcal{S}$, it follows that $\mathcal{D}\subseteq \mathcal{S}$ holds. Now, consider the following function.

  $$ \phi (x)=e^{-x^{2}} $$

  Then, it is evident that $\phi \in C^{\infty}$ holds. However, since $\phi (x)\ne0, \forall x\in \mathbb{R}$, $\phi$ does not have a compact support. Hence, $\phi$ is not a test function. Now, showing $\phi \in \mathcal{S}$ completes the proof. By the chain rule, $D^{\alpha}\phi$ can be expressed as the product of $\phi$ and any polynomial $P$.

  $$ D^{\alpha}\phi (x) = P(x)\phi (x) $$

  Therefore, the following equation holds.

  $$ \begin{equation} \left| x^{\beta}D^{\alpha}\phi (x) \right| =\frac{\left| x^{\beta}P(x) \right| }{e^{x^{2}}} \label{eq1} \end{equation} $$

  Since the numerator is a polynomial and the denominator is an exponential function, the following equation holds.

  $$ \lim \limits_{x\to \pm\infty} \frac{\left| x^{\beta}P(x) \right| }{e^{x^{2}}}=0 $$

  Therefore, for all $\varepsilon >0$, there exists $N>0$ such that the following equation holds.

  $$ \left| x \right| > N \implies \frac{\left| x^{\beta}P(x) \right| }{e^{x^{2}}} < \varepsilon $$

  Now, consider when $\left| x \right| \le N$. $[-N,N]$ is compact and $\eqref{eq1}$ is a continuous function, so it is bounded by some positive number $M>0$. Now, let $C_{\alpha,\beta}=\max \left\{ M,\varepsilon \right\}$, and then the following holds.

  $$ \left| x^{\beta}D^{\alpha}\phi (x) \right| \le C_{\alpha,\beta} $$

  Therefore, $\phi \in \mathcal{S}$ holds. Hence, $\phi \notin \mathcal{D}$ holds and because $\phi \in \mathcal{S}$ holds, the following statement is true.

  $$ \mathcal{D} \subsetneq \mathcal{S} $$

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