Riemmann/Darboux-Integral Question

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SUMMARY

The discussion addresses two problems related to the Riemann/Darboux integral. In the first problem, it is established that if F is continuous on [a,b] and the integral of F multiplied by any continuous function G equals zero, then F must be identically zero. The second problem explores whether the same conclusion holds if G is constrained to vanish at the endpoints a and b. The participant suggests that F still needs to be identically zero, supported by the continuity of F and the properties of the chosen function G.

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Homework Statement



(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?

Homework 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)

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?
 
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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.
 
Thanks Dick, works perfectly. =)
 

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