(That question was answered. I realized I asked a wrong question. So here is a modified question.)
I will call an average any continuous function $f(v_1,\dots,v_n)$ of $n$ arguments such that it lies in the closed interval $[\min(v_{1},\dots ,v_{n});\max(v_{1},\dots ,v_{n})]$, is symmetric for all permutations of arguments, and is homogeneous with the degree $1$.
Question: Can an average provided the below additional requirements (I ask for the case if all of them are required but also for milder cases of just a part of these requirements) not to tend to infinity when one of the arguments tends to infinity and the rest arguments are fixed nonnegative reals? (We can assume $n\geq 2$.)
- Our average does not tend to infinity when all arguments except one tend to infinity.
- Our average is $C^1$-smooth in every argument (maybe except the argument(s) value equal to zero).
- Our average is $C^\infty$-smooth in every argument (maybe except the argument(s) value equal to zero).
Take a weighted average where the weight of each argument decreases homogeneously as the argument tends to infinity. For example, if you choose the weight of an argument $v_i$ to be $w(v_i)=v_i^{-2}$, you get $$f(v_1,\dots,v_n)=\frac{v_1^{-1}+\dots+v_n^{-1}}{v_1^{-2}+\dots+v_n^{-2}}.$$
In fact, this formula remains bounded when up to $n-1$ values tend to infinity.
If you want something that behaves more like the traditional average, and in particular doesn't have $f(v_1,\dots,v_n)\to0$ when any $v_i\to0$, you can try $$f(v_1,\dots,v_n)=\frac{v_1^2(v_2+\dots+v_n)+\dots+v_n^2(v_1+\dots+v_{n-1})}{(v_1^2+\dots+v_n^2)(n-1)}.$$ This is the weighted average, with weights $v_i^2$, of each average where one argument $v_i$ excluded.
Compared to the previous formula, this one only remains bounded if at most one argument tends to infinity.