Recent content by Misswfish

ahhhhhh that doesn't sound right. I thought about using the definition of uniformaly continuous. I picked my eplision to be f(p)/2. I then used the theorem "Suppose S is a set of numbers and M is the least upper bound of S. If p is a number such that p< M , then there is an x in S such that p <...

Ahhh my teacher told me to pick a "clever" g(x) so that we can use this theorem and therefore f(x) = 0. I was thinking contradiction but my teacher shot that down

Suppose f is continuous function on [a,b] such that for each continuous function g, \int(fg)dj = 0 (Note: integral is from a to b) , then f(x) = 0 for each x in [a,b]. I know that I should use the theorem If is continuous on [a,b], f(x)\geq0 for each x in [a,b] and theree is a number p i n...

but how does the def of continuous show that the integral is greater than zero

well I thought about putting a "box" around it and picking variable to the right and left of p. Is that it?

If f is continuous on [a,b], f(x) \geq 0 for each x in [a,b] and there is a number p in [a,b] such that f(p) > 0, then \int f dj > 0 ( Note : the integral is from a to b)