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If I have an arbitrary scalar function f(x,y,z) that is non-zero and positive within a volume $V$ of $\Bbb R^3$, and I have a constraint that requires

$$\int_{V} f(x,y,z)\, dV =\int_{V} f(x,y,z) \cdot g(x,y,z)\, dV$$

where g(x,y,z) is a unique given function and f(x,y,z)>0 inside $V$ and =0 outside $V$ .

It would appear to me that this requires g(x,y,z)=1 given that f(x,y,z) is arbitrary.

What would be the best way to strictly prove this mathematically?

Thanks in advance.

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2 Answers 2

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Since $g$ is fixed it can come before the integral(as any number would).

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  • $\begingroup$ g is not a number but a function of x,y.x as well, only that it is a unique, given function (I changed the formulation in the OP in this respect) but the equation should hold for arbitrary f(x,y,z) $\endgroup$
    – Thomas
    Apr 28, 2018 at 15:19
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Is there not a uniqueness theorem for integration? This would reduce the problem to identifying the g function as a multiplicative identity by inspection which is 1.

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