Does "finite average" exist?

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 not to tend to infinity when one of the arguments tends to infinity and the rest arguments are fixed nonnegative reals? (We can assume $n\geq 2$.)

My original question had a trivial solution $\min(v_{1},\dots ,v_{n})$, so let's add an additional requirement: our average not to tend to zero when one of the arguments tends to zero and the rest arguments are fixed positive reals? (We can assume $n\geq 2$.)

Answer 1

Not sure this matches your needs, but $f(v_1,...,v_n)=\min(v_1,...,v_n)$ might obey all your requirements. It is homogeneous of degree 1, symmetric in all permutation of arguments, and if only one argument goes to infinity, it will not change it's value. Also, it lies in the closed interval you want.

Answer 2

For $k=1,2,...,n$ define $m_k=\min_{i\neq k}\{v_i\}$ and $f(v_1,...,v_n)=\max_{k=1,...,n}\{m_k\}$. This can be generalized to exclude the first $l$ minima, which will alow $f$ to not go to zero, as the entries $v_1, ..., v_l\to 0$.