Let $f(x)$ within $x\in[a,b]$ an absolute continuous function with
- $f(x)\geq0$
- $f(x_m)=0$ for all absolute minima $x_m$
- no other zeros than at $x_m$
I am trying to define a counting function for the $x_m$ within $x\in[a,b]$ which delivers at highest possible accuracy at any $a\lt x \lt b$, the number of zeros $x_m$ with $a \lt x_m \lt x$. For this purpose I try to construct a counter function, starting with the form:
$$\kappa=\int_a^x \exp\left(-\varepsilon f(t)\right)\mathrm dt$$
It seems that for sufficient large $\varepsilon$ we can achieve that $exp(-\varepsilon f(x))$ approximates zero for all $x\neq x_m$, and $exp(-\varepsilon f(x))=1$ for $x_m$.
However, I am still quite struggling, how to tailor $\kappa$ to its final form, which should deliver the number of $x_m$. Two major open questions are:
- How can I norm the integral, so it delivers at the end the correct numbers? - I guess, one must devide it by some norm.
- How shall $\varepsilon$ be mathematically defined to ensure accuracy of the result? - It looks to me a quite vague formulation to say for instance only $\varepsilon f(x)\gg 0$. I would much prefer to have a kind of big $\mathcal O$ formulation or stronger constraints for $\varepsilon$.
I would appreciate your support to find the final formulation of a mathemtically well formed counting function. Of course, any helpful references would be also welcome.
Many thanks in advance.
Doesn't any non-negative continuous function on $[a,b]$ satisfy the conditions given in the last two bullets? Also, using $\varepsilon$ for something a priori big is touche.
I also do not think that you can construct such a counting function using integral (w.r.t. Lebesgue measure): currently you are trying to approximate the length of $\{f = 0\}$, not the number of elements there. I'm pretty sure that it's not something you would like to do. The thing is that the number of zeros is a very unstable quantity: if you know $f$ precisely, why would you go for approximation of this number; if you do not know it precisely, that it may have $1000$ zeros or none at all, and you can't even conclude that from the approximation of $f$ that you have.