# Riemmann/Darboux-Integral Question

1. Apr 4, 2008

### end3r7

1. The problem statement, all variables and given/known data

(i) Suppose that F is continuous on [a,b] and $$\int_a^b FG = 0$$ for all continuous functions G on [a,b]. Prove that F = 0.

(ii) Suppose now that G(a) = G(b) = 0. Does it again follow that F must be identically zero?

2. Relevant equations
Let P be any partition of [a,b]

Upper Darboux
$$U(F,P) = \sum_{k=0}^\n sSupF(x_k)(t_k - t_{k-1})$$
Lower Darboux
$$L(F,P) = \sum_{k=0}^\n iInfF(x_k)(t_k - t_{k-1})$$

F is integrable iff
inf U(F,P) = U(F) = L(F) = sup L(F,P)

3. The attempt at a solution

For (i) I argued the following.

Let G = F.
Clear H = F*F >= 0 and is continuous.
Let's say H(y) > 0.
So I can find an interval (y - c, y + c) because H is continuous, where H(x) > H(y)/2.
Clearly U(H,P) > H(y)/2*c
Which means U(H) >= H(y)/2*c > 0.

Thus F must be identically zero.

For (ii), I said it still needed to be identically zero, and I'm not certain how the argument needs to be changed. Perhaps if F was only nonzero very near a and b, but because of the continuity of F, wouldn't that still bee a problem?

2. Apr 4, 2008

### Dick

How about G(x)=F(x)*(x-a)*(b-x)? There's nothing special about (x-a)*(b-x), it's just a positive continuous function on [a,b] that vanishes only at the endpoints.

3. Apr 4, 2008

### end3r7

Thanks Dick, works perfectly. =)