# Test Function/Examples/Exponential of One over x Squared minus One

## Example of Test Function The graph of the test function. It is smooth everywhere. Outside of its support denoted by dashed lines the function is identically zero. At boundary points it connects smoothly.

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:

$\map f x = \begin {cases} \map \exp {-\dfrac 1 x} & : x > 0 \\ 0 & : x \le 0 \end {cases}$
$\dfrac {\d^n} {\d x^n} \map \exp {-\dfrac 1 x} = \dfrac {\map {P_n} x} {x^{2 n} } \map \exp {-\dfrac 1 x}$

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:

$\forall x \in \openint {-1} 1 : \map \phi x \in \CC^\infty$

By definition, $\phi$ is a test function.

$\blacksquare$