Continuity of generalized mean functions I'm studying generalized mean functions, and somewhere I found that a weighted mean function could be defined as $M: (0,\infty)^n \rightarrow (0,\infty)$ with the properties:


*

*Fixed Point : $M(1,1,\dotsc,1) = 1$

*Homogeneity : $M(\lambda x_1, \dotsc, \lambda x_n) = \lambda M(x_1, \dotsc, x_n)$

*Monotonicity : if $x_i \leq y_i$ for each $i$ then $M(x_1,\dotsc, x_n) \leq M(y_1, \dotsc, y_n)$


This imply some of the properties expected for mean functions, such boundedness: $\min\{x_i\} \leq M(x_i) \leq \max\{x_i\}$ and continuity. Though, I was unable to find a proof for continuity. I'm familiar with real analysis, but not in $\mathbb{R}^n$.
I tried to find a counterexample of such function, only to fail again. The non-homogeneous functions that would serve as counterexample fail to be monotonic or to be positive.
 A: By definition, $M$ is continuous if for every $(x_1,\dots,x_n) \in (0,\infty)^n$, and every $\epsilon > 0$, there exists $\delta > 0$ such that, if $(y_1,\dots,y_n) \in (0,\infty)^n$ is such that $|y_i-x_i| < \delta$ then $|M(y_1,\dots,y_n)-M(x_1,\dots,x_n)| < \epsilon$.
Let us show that this holds.
Let $(x_1,\dots,x_n) \in (0,\infty)^n$ and $\epsilon > 0$ be given.
We will figure out exactly what $\delta$ should be in a moment; for now, just let it be fixed.
If $|y_i-x_i| < \delta$ for all $i$, then this means that
$$ x_i - \delta < y_i < x_i + \delta, $$
so that
$$ y_i < (1+\frac\delta{x_i})x_i \leq (1+\frac\delta{\text{min}_i\, x_i})x_i, $$
and therefore, $(y_1,\dots,y_n) \leq (1+\frac\delta{\text{min}_i\,x_i})(x_1,\dots,x_n)$ so that
$$ M(y_1,\dots,y_n) \leq (1+\frac\delta{\text{min}_i\,x_i})M(x_1,\dots,x_n). $$
Likewise, we can show that
$$ M(y_1,\dots,y_n) \geq (1-\frac\delta{\text{min}_i\,x_i})M(x_1,\dots,x_n). $$
Thus, if we set $\delta < \frac{\epsilon\cdot\text{min}_i\,x_i}{M(x_1,\dots,x_n)}$ then $|y_i-x_i| < \delta$ for all $i$ implies $|M(y_1,\dots,y_n)-M(x_1,\dots,x_n)| < \epsilon$.
