Show f(x) >= g(x) given integral of f is >= g on any subinterval of [a,b]

[afkguy]

Homework Statement

f,g are cont. fcns on [a,b] and $$\int f$$ $$\geq$$ $$\int g$$ for any subinterval [i,j] of [a,b]. Show f(x) $$\geq$$ g(x) on [a,b].

Homework Equations

Don't know if I should have used something like the fundamental theorem or mean value of integrals or something. I was using theorems like those for other problems in this section but wasn't sure about this one so I took a different direction.

The Attempt at a Solution

Assume that this isn't true, i.e., there exists some c in [a,b] s.t. g(c) > f(c).

f and g are continuous so let's choose our $$\epsilon$$ = g(c) - f(c), so there should exist some $$\delta$$ > 0 where:

|f(x) - f(c)| < $$\epsilon$$ whenever |x - c| < $$\delta$$

Thus 2f(c) - g(c) < f(x) < g(c) when x is in (c - $$\delta$$, c + $$\delta$$)

So f(x) < g(c) whenever x is in (c - $$\delta$$, c + $$\delta$$).

So consider any partition p of any subinterval [i,j] of [a,b] where the one of the subintervals of [i,j] is (c - $$\delta$$, c + $$\delta$$).

I stopped here because I'm pretty sure I chose the wrong epsilon and I wasn't really sure if I was even in the right direction.

 

[Asphyxiated]

well i am not sure how you would go about 'proving' this but knowing that the area of f is always greater than or equal to g on any sub interval [i,j] in the interval [a,b] implies that f never intersects g, they may touch and be equal but f will never be less than g because the area would be less on the sub interval [i,j] where f<g. The fact that this is true with any sub interval (implying infinitesimal intervals as well) means that f>=g.

 

[Gib Z]

Proof by contradiction is a good strategy here and you had good ideas, but you are muddling yourself up by dealing with details that we don't need, such as finding epsilons.

1) Assume g(c) > f(c) for some c in [a,b]
2) Use the continuity property to deduce there is some neighborhood (c - e, c + e) ( e>0 ) where g > f in that neighborhood.
3) Consider the integral over this interval to find a contradiction.

 

[Delta2]

To prove 2) stated above use the function g(x)-f(x) which is also continuous and you know it has a positive limit at x=c.