The discussion revolves around the integrability of the product of two functions, f and g, defined on the interval [a, b]. Participants are exploring the implications of integrability in the context of Riemann integration.
The discussion is ongoing, with participants providing different perspectives on the integrability of functions. Some have raised counterexamples that challenge the original assertion, indicating a lack of consensus on the validity of the claim.
Participants are discussing the definitions of integrability and the implications of specific examples that may violate the assumptions of the problem. There is a focus on the conditions under which the original statement holds true.
You can't prove it, it's not true. For example, if f(x)= 1 if x is rational, -1 if x is irrational, the f2(x)= 1 so f2(x) is integrable on, say, [0, 1] but f is not. Also, if g(x)= 2, then g2(x)= 4 so is integrable on [0,1] but fg(x)= 2 if x is rational, -1 if x is irrational is not itegrable.a1010711 said:The Attempt at a Solution
It suffices to show that f^2 is integrable, since
fg= [(f+g)^2-(f-g)^2]/4
The function x --> x^2 is uniformly continuous on the range of f,
im not sure how to turn this into a formal proof, I am lost
Riemann integration
What if you can find a function f that is integrable on [a,b] but not square integrable?a1010711 said:The Attempt at a Solution
It suffices to show that f^2 is integrable
That much is true.HallsofIvy said:You can't prove it, it's not true.
That's fine, but the title of the thread is "Prove that if f and g are integrable on [a, b], then so is fg", so you are picking some f that violates the given conditions.For example, if f(x)= 1 if x is rational, -1 if x is irrational ...