Consider $N \in \mathbb{N}$ real numbers $a_1 \leq a_2 \leq \ldots \leq a_N.$ Moreover, consider the following sequence of functions:
$$f_{k}(x) = \sum_{i=1}^N |x-a_i|^\frac{1}{k},$$
where $k \in \mathbb{N}$. Let
$$f(x) = \lim_{k \to \infty} f_k(x).$$
I would like to know which are the tools that I can use to find the value of $x \in [a_1, a_N]$ such that $f(x)$ attains its minimum (or its infimum).
The intuition led me the following road.
- Suppose that all the members of the set $A = \{a_1, a_2, \ldots a_N\}$ are distinct. If $x \not\in A$, then:
$$f(x) = \lim_{k \to \infty}\sum_{i=1}^N|x-a_i|^\frac{1}{k} =\\= \sum_{i=1}^N\lim_{k \to \infty}|x-a_i|^\frac{1}{k} = \sum_{i=1}^N (\text{some positive number})^0 = N.$$
Instead, if $x \in A$ (let say WLOG $x = a_N$) then:
$$f(x) = \lim_{k \to \infty}\sum_{i=1}^N|x-a_i|^\frac{1}{k} =\\= \sum_{i=1}^N\lim_{k \to \infty}|x-a_i|^\frac{1}{k} = \sum_{i=1}^{N-1} (\text{some positive number})^0 + 0^0 = N-1 + 0^0.$$
Numerically, I observe that "$0^0$" seems to be very close to $0$. This is the most controversial part of my efforts. How can I evaluate this indeterminate form resulting in a $0^0$?. Therefore, the minimum is attained for any $x \in A$.
- Suppose that all members of $A$ are distinct except for $a_r = a_{r+1} = \ldots = a_N.$ Therefore, if $x \not \in A$, we still get $N$. Anyway, if $x = a_r$, then:
$$f(x) = \lim_{k \to \infty}\sum_{i=1}^N|x-a_i|^\frac{1}{k} =\\= \sum_{i=1}^N\lim_{k \to \infty}|x-a_i|^\frac{1}{k} = \sum_{i=1}^{r-1} (\text{some positive number})^0 + (N-r+1)0^0 \\=\\ r-1 + (N-r+1)0^0.$$
Again, numerical experiments suggest my that in this case $0^0$ is close to $0$. Therefore the minimum is attained for $x = a_r$.
Do you have some hints for a more precise and rigorous approach? Thanks!
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