Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].
Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.
I came up with the following solution, I hope it is not so wrong.
Suppose that is not for all , then there exists such that (If the is no such , there must be such that .In that case we will look at ).
f is continuous function therefor exists for which for all .
Now we define -'staircase'-characteristic function of open interval .
Hence, we will get:
.
So, we got a contradiction to the fact that if is continuous on and h is staircase function .
But we also know that if is integrable function on , so for every exists staircase function so that
Combining these two results we will conclude the needed.