Test Function/Examples/Exponential of One over x Squared minus One
Example of Test Function
Let $\phi : \R \to \R$ be a real function with support on $x \in \closedint {-1} 1$ such that:
- $\map \phi x = \begin {cases} \map \exp {\dfrac 1 {x^2 - 1} } & : \size x < 1 \\ 0 & : \size x \ge 1 \end {cases}$
Then $\phi$ is a test function.
Proof
Consider a real function $f : \R \to \R$ such that:
- $\map f x = \begin {cases} \map \exp {-\dfrac 1 x} & : x > 0 \\ 0 & : x \le 0 \end {cases}$
From Nth Derivative of Exponential of Minus One over x:
- $\dfrac {\d^n} {\d x^n} \map \exp {-\dfrac 1 x} = \dfrac {\map {P_n} x} {x^{2 n} } \map \exp {-\dfrac 1 x}$
where $\map {P_n} x$ is a real polynomial of degree $n$.
Then the right hand side can be rewritten in terms of at most $n + 1$ terms of the form $\dfrac 1 {x^m} \map \exp {-\dfrac 1 x}$ where $m \in \N$.
Let us take the limit $x \to 0$ from the right:
\(\ds \lim_{x \mathop \to 0^+} \frac 1 {x^m} \map \exp {-\frac 1 x}\) | \(=\) | \(\ds \lim_{z \mathop \to \infty} \frac {z^m} {\map \exp z}\) | Substitution $\ds z = \frac 1 x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Limit at Infinity of Polynomial over Complex Exponential |
Therefore:
- $\ds \lim_{x \mathop \to 0^+} \frac {\map {P_n} x} {x^{2 n} } \map \exp {-\frac 1 x} = 0$
By construction:
- $\ds \map \phi x = \map f {1 - x^2} = \begin {cases} \map \exp {-\dfrac 1 {1 - x^2} } & : 1 - x^2 > 0 \\ 0 & : 1 - x^2 \le 0 \end {cases}$
where:
\(\ds 1 - x^2\) | \(>\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(>\) | \(\ds x^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size x\) | \(<\) | \(\ds 1\) |
Furthermore:
\(\ds \dfrac {\d \map \phi x} {\d x}\) | \(=\) | \(\ds \dfrac {\d \map f y} {\d y} \paren {-2 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map {P_1} y} {y^2} \map \exp {-\frac 1 y} \paren {-2 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map {M_3} x} {y^2} \map \exp {-\frac 1 y}\) |
where $y = 1 - x^2$ and $\map {M_3} x$ is a real polynomial of degree $3$.
Similarly, any higher derivative of $\map \phi x$ will have $\map {M_k} x$ instead of $\map {P_n} x$ with $k \ge n$.
Thus:
- $\dfrac {\d^n} {\d x^n} \map \phi x = \dfrac {\map {M_k} x} {y^{2 n} } \map \exp {-\dfrac 1 y}$
Consequently:
\(\ds \lim_{x \mathop \to -1^+} \frac {\map {M_k} x} {y^{2 n} } \map \exp {-\frac 1 y}\) | \(=\) | \(\ds \lim_{x \mathop \to -1^+} \map {M_k} x \lim_{x \mathop \to -1^+} \frac {\map \exp {- \frac 1 y} } {y^{2n} }\) | Product Rule for Limits of Real Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds C \lim_{y \mathop \to 0^+} \frac {\map \exp {- \frac 1 y} } {y^{2n} }\) | $C \in \R$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Analogously:
- $\ds \lim_{x \mathop \to 1^-} \frac {\map {M_k} x} {y^{2 n} } \map \exp {-\frac 1 y} = 0$
Since outside of the support we have that $\map \phi x = 0$, the limit coming from outside is also $0$.
Therefore, $\map \phi x$ is smooth at the boundaries of its support.
Also:
- $\forall x \in \openint {-1} 1 : \map \phi x \in \map {C^\infty} \R$
By definition, $\phi$ is a test function.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples