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 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:

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:

Furthermore:

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:

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.