If for n-dimensions f>=0 , prove the integral of f >=0

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The discussion centers on proving that if a function f(x1, x2, ..., xn) is non-negative (f >= 0) in n-dimensional space, then the integral of f over that space is also non-negative. Participants emphasize the importance of demonstrating that the supremum of f is greater than or equal to zero. They suggest that without a specific theorem linking inequalities of functions to their integrals, one must rely on the definition of the integral to establish this proof.

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  • Familiarity with the concept of supremum in mathematical analysis
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  • Basic understanding of inequalities in calculus
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Homework Statement



1j9DZU8.png

Homework Equations

The Attempt at a Solution


a) The furthest I have got to understanding the solution is that I need to find a way to show that sup(f(x1,x2,...,xn) >= 0. Intuitively and graphically I can see why the statement is obvious, I'm just having a hard time starting to write the proof...
 
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gradivcurl said:

Homework Statement



1j9DZU8.png

Homework Equations

The Attempt at a Solution


a) The furthest I have got to understanding the solution is that I need to find a way to show that sup(f(x1,x2,...,xn) >= 0. Intuitively and graphically I can see why the statement is obvious, I'm just having a hard time starting to write the proof...

If ##f(x_1,..x_n) \ge 0## then it's pretty obvious ##sup(f(x_1,..x_n)) \ge 0##. If you don't have a convenient theorem like if ##f \ge g## then ##\int f \ge \int g##, then you might have to use the definition of the integral.
 

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