Are there test functions $\in \mathcal{C}_{\text{c}}^{\infty}$ which aren't in the algebra generated by $f_{a,b}(x) := e^{\frac{C}{(x-a)(x-b)}}$

Question

The standard test ($\in \mathcal{C}_{\text{c}}^{\infty}(\mathbb{R})$) function is the following \begin{equation*} f_{a,b}^{(c)}: \mathbb{R} \to \mathbb{R}_{\ge 0}, \ x \mapsto \begin{cases} \exp\left(\frac{C}{(x - a)(x - b)}\right), & \text{if } a < x < b, \\ 0, & \text{elsewhere.} \end{cases} \end{equation*} Where $C := \frac{c(a - b)^2}{4}$. Then, $f \in \mathcal{C}_{\text{c}}^{\infty}(\mathbb{R})$, supp$(f) \subset [a,b]$ and $\max\limits_{x \in \mathbb{R}} f(x) = e^{-c}$.

Are there fundamentally different test functions and if not, why?

I want "fundamentally different" to mean that you can't express it in the form exp of the reciprocal of a polynomial (mutiplied by a constant) for an interval $[a,b]$ and zero elsewhere.

Edit: (response to @md2perpe's comment about convoluting with an indicator function) We can construct a function which is very similar to the one you described by using the function from above (Note: this is not the exact function, but we had to this for a homework way back so I decided to post it here, it can surely be modified to be exactly your example).

For $x \in \mathbb{R}$ and $\alpha \in (0, \infty)$ define $$ h_{\alpha, x}: \mathbb{R} \to [0,1], t \mapsto \begin{cases} 1, & \text{if } t \in [x - \frac{\alpha}{2}, x + \frac{\alpha}{2}], \\ \exp\left(1 -\frac{1}{1-\left(\frac{2}{\alpha}(t + \frac{\alpha}{2} - x )\right)^2 } \right), & \text{if } t \in [x - \alpha, x - \frac{\alpha}{2}), \\ \exp\left(1 -\frac{1}{1-\left(\frac{2}{\alpha}(t - \frac{\alpha}{2} - x )\right)^2 } \right), & \text{if } t \in (x + \frac{\alpha}{2}, x + \alpha], \\ 0, & \text{else.} \end{cases} $$

Then, we have $$ h_{\alpha, x}(t) = \begin{cases} 1, & t \in [x - \frac{\alpha}{2}, x + \frac{\alpha}{2}] \\ 0, & t \in \mathbb{R} \setminus [x - \alpha, x + \alpha]. \end{cases}, $$ which is fundamentally different from the function above.

Answer 1

I found another way to construct such a test function here.

Define $$ f(x) = \begin{cases} \exp\left(- \frac{1}{x}\right), & x > 0, \\ 0, & \text{else.} \end{cases} \quad \text{and} \quad g(x) = \frac{f(x)}{f(x) + f(1-x)} = \begin{cases} 0, & x \le 0, \\ \frac{1}{\exp\left(\frac{2x - 1}{x(x-1)}\right) + 1}, & 0 < x < 1, \\ 1 & x \ge 0. \end{cases} $$ Then, the function $h(x) := g\left(\frac{x - a}{b - a}\right) g\left(\frac{d - x}{d - c}\right)$ equals one on $[b,c]$ and vanished outside $(a,d)$.

Answer 2

I want "fundamentally different" to mean that you can't express it in the form exp of the reciprocal of a even polynomial (mutiplied by a constant) for an interval $[a,b]$ and zero elsewhere.

With that criteria, it's easy to construct a fundamentally different example. Just square the polynomial: \begin{equation*} f_{a,b}^{(c)}: \mathbb{R} \to \mathbb{R}_{\ge 0}, \ x \mapsto \begin{cases} \exp\left(\frac{C}{(x - a)^2(x - b)^2}\right), & \text{if } a < x < b, \\ 0, & \text{elsewhere.} \end{cases} \end{equation*}