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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.

ViktorStein
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    I don't know why you got a downvote, maybe you could reformulate your question as: are there functions in $\mathcal{C}c^{\infty}(\mathbb{R})$ that are not in the algebra generated by the $f{a,b}: x \mapsto \begin{cases} \exp\left(\frac{C}{(x - a)(x - b)}\right), & \text{if } a < x < b, \ 0, & \text{elsewhere.} \end{cases}$? Can we find an explicit example? – charmd Apr 12 '19 at 11:43
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    @CharlesMadeline. Under what operations do you consider that algebra be generated? – md2perpe Apr 12 '19 at 20:57
  • @ViktorGlombik. For example $f_{-1,1}^{(1)} * \chi_{[-10, 10]}.$ – md2perpe Apr 12 '19 at 22:20
  • @md2perpe We have $$f_{-1,1}^{(1)} \ast \chi_{[-10,10]}(t) = \int_{\mathbb{R}} e^{\frac{1}{\tau^2 - 1}} \cdot 1_{[-10, 10]}(t - \tau) d \tau = \int_{t - 10}^{t + 10} e^{\frac{1}{\tau^2 - 1}} d \tau,$$ right? How does such a function look like? – ViktorStein Apr 12 '19 at 23:08
  • @md2perpe I was thinking about addition and multiplication (as a subalgebra of $\mathcal{C}^0$). But I don't really know what 'fundamentally different' meant for Viktor – charmd Apr 13 '19 at 05:46
  • @ViktorGlombik. The result will be a function that leaves zero at -11, reaches a level at -9, stays constant on that level until +9, and then becomes zero again at +11. – md2perpe Apr 13 '19 at 07:50

2 Answers2

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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*}

md2perpe
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  • And, even more general: $\exp\left(\frac{C \cdot (-1)^{n+1}}{(x - a)^{n} (x - b)^{n}}\right)$. – ViktorStein Apr 14 '19 at 19:32
  • Thanks to your example I realise I wan't precise enough in my comment you quoted. One should probably get rid of the even in that sentence. The example you gave would be different in this manner as well since it's constantly 1 on an interval, which isn't true for this function or the one I stated at the beginning of my question. – ViktorStein Apr 14 '19 at 19:35
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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 $(a,d)$.

ViktorStein
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    All explicit constructions of $C_c^\infty$-functions are (as far as I know) based on $f$. For example, $f_{a,b}^{(c)}(x) = f((x-a)(x-b)/C).$ So maybe the question should rather be if there are $C_c^\infty$-functions that do not belong to the algebra generated of functions of the form $f \circ u$ where $u$ is ... (what should it say here?). – md2perpe Apr 17 '19 at 13:21