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

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Homework Statement



f,g are cont. fcns on [a,b] and [tex]\int f[/tex] [tex]\geq[/tex] [tex]\int g[/tex] for any subinterval [i,j] of [a,b]. Show f(x) [tex]\geq[/tex] 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 [tex]\epsilon[/tex] = g(c) - f(c), so there should exist some [tex]\delta[/tex] > 0 where:

|f(x) - f(c)| < [tex]\epsilon[/tex] whenever |x - c| < [tex]\delta[/tex]

Thus 2f(c) - g(c) < f(x) < g(c) when x is in (c - [tex]\delta[/tex], c + [tex]\delta[/tex])

So f(x) < g(c) whenever x is in (c - [tex]\delta[/tex], c + [tex]\delta[/tex]).

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

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
Gib Z
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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.
 
  • #4
Delta2
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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.
 

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