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I am seeking a general mathematical proof & a reference for the proof for something I know intuitively to be true, and can demonstrate by example, but would like to prove. Assume a function with 6 inputs [ f(z,y,x,w,v,u) ] that is non-linear with respect to all six variables. I do not get the same answer averaging the outputs vs using the average input and and then averaging the outputs. I need to know if there is a way to prove that this will always be the case, or to identify the minimum set of assumptions required to assert that it will not be true, or alternatively to identify the set of assumptions that MUST be true in order to get the same result (such that I might infer that if those assumptions are not true, then the results will not be equal).

This is not for school, it is for work. I am at a loss for where to find a proof for something so simple, but I need it to convince a co-worker who does not seem to believe my examples.

thanks

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I think mathstackexchange is the appropriate forum for your question. This one is for research-oriented math. Please try there and good luck. –  Benjamin Steinberg Apr 15 '13 at 16:38
    
typically, you will not have much luck finding a proof that two complex expressions differ for all possible inputs; I imagine typically you will be able to find exceptional inputs that equate the two expressions. –  Carlo Beenakker Apr 15 '13 at 16:44
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@Douglas, does not seem to be posted at math.stackexchange.com/questions?sort=newest where it would get more basic responses. Looking at it, I am not sure how the word average is being used. Some example calculations, here or at MSE, would probably settle the question. –  Will Jagy Apr 15 '13 at 19:38
    
PleaseHelpMe, when you say "average," do you mean a few dozen sextuplets $(z,y,x,w,v,u)$ as inputs, with the corresponding few dozen outputs? How many numbers are in a single output? –  Will Jagy Apr 15 '13 at 19:57
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closed as off topic by Benjamin Steinberg, Carlo Beenakker, Noah Stein, Henry Cohn, Nik Weaver Apr 15 '13 at 17:04

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1 Answer

Counterexample:

Let $\lbrace e_i \rbrace_{i\in I}$ be a Hamel basis, a basis for $\mathbb R$ over $\mathbb Q$, which means every real can be written uniquely as a finite linear combination of the basis elements with rational coefficients. The set of indices $I$ is uncountable. Choose a particular basis element $e_0$ (which might be taken to be $1$). Let $f$ assign to $(z,y,x,w,v,u)$ the coefficient of $e_0$ in the expression of $z$ as a rational linear combination of the Hamel basis. Then $f$ is discontinuous (it only takes rational values) and therefore not linear over the reals, but it satisfies $\operatorname{Avg}(f) = f(\operatorname{Avg})$ because it is $\mathbb Q$-linear.

There are foundational issues. I can't write down a Hamel basis, but one exists if you assume the Axiom of Choice. See this question.

If you assume that $f$ is continuous, or just measurable then it must actually be linear.

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truly the answer of a MathGod –  Carlo Beenakker Apr 15 '13 at 19:38
    
I suppose then that the $s$ is at the end of "Douglas" ? –  stankewicz Apr 15 '13 at 20:52
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