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Is there a class of functions that fulfills or partly fulfills following conditions?

For any $n\in\mathbb{N}_{>1}$ there is a function $f_n$ with following properties:

  1. $f_n$ is a function of $n$ variables, i.e. $f_n=f_n(x_1,...,x_n)$ with $x_i\ge0 \;\forall i$
  2. if $x_i=a \;\forall i\;$ then $f_n=k_n a^{n-1}$ with $k_n>0$
  3. $f_n\ge0$
  4. $f_n=0$ iff $x_i=0 \;\forall i$

It is not known that 1-4 are sufficient conditions to define a unique function class but it would be good to know at least one class that fulfills all conditions.

What is known?

Following function scheme fulfills conditions 1,2 but not conditions 3,4 and might be a starting point:

$$f_n=\frac{\prod_{i=1}^{n} x_i}{\frac{1}{n}\sum_{i=1}^{n} x_i}$$ As an expample $n=4$ is used:

condition 1 fulfilled: $$f_4=\frac{x_1x_2x_3x_4}{\frac{1}{4}\left(x_1+x_2+x_3+x_4\right)}$$

condition 2 fulfilled: $$f_4=\frac{a\cdot a\cdot a\cdot a}{\frac{1}{4}\left(a+a+a+a\right)}=a^3 \;\;\;\text{with}\;\;\; k_4=1$$

condition 3 only fulfilled if $\neg(x_i=0\; \forall i)$:

$$\sum_{i=1}^n x_i\ge0 \land \prod_{i=1}^n x_i\ge0 \to f_4\ge0$$

condition 4 not fulfilled: $$f_4=\frac{0\cdot 1\cdot 1\cdot 1}{\frac{1}{4}\left(0+1+1+1\right)}=0 \nleftrightarrow x_i=0\; \forall i$$

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granular_bastard
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1 Answers1

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Accepting the comment from user Rainer Glüge following function class fulfills all conditions. $$f_n=\left(\sum_{i=1}^n x_i\right)^{n-1}$$

It can easily be shown that the 4 conditions are fulfilled and it is exemplary shown for $n=4$.

condition 1 fulfilled: $$f_n=(x_1+x_2+x_3+x_4)^3$$

condition 2 fulfilled: $$f_n=(a+a+a+a)^3=k_4a^3\;\;\;\text{with}\;\;\;k_4=64$$

condition 3 fulfilled:

follows from definition of $x_i\ge0\;\forall i$

condition 4 fulfilled:

follows from definition of $x_i\ge0\;\forall i$, $$x_1+x_2+x_3+x_4=0 \leftrightarrow x_i=0\;\forall i$$ $$0^3=0 $$

There might be other more complicated expressions that also fulfill all conditions.

granular_bastard
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