Determining f

[e(ho0n3]

The problem statement, all variables and given/known data
Suppose f: [0,1] --> R is continuous and

\int_0^1 f(x) x^n \, dx = 0

for all n = 0, 1, ... Prove that f(x) = 0.

The attempt at a solution
There's a hint that says: Prove that

\int_0^1 (f(x))^2 \, dx = 0.

I don't know how to prove this hint and I don't know how that would help in determining what f is. Any tips?

[Dick]

HAH! This is cosmic. Look at this thread.
http://www.physicsforums.com/showthread.php?p=2064400#post2064400
I think the course I was outlining for Hitman2-2 will work perfectly for you, and even more simply since you have the condition for all n>=0. Use Stone-Weierstrass. You have integral f(x)=0 since it's true for n=0. Hitman2-2 only gave me n an even natural number. Is zero an even natural number? I would have said, no. And that's the roadblock for that thread.

[Dick]

The problem statement, all variables and given/known data
Suppose f: [0,1] --> R is continuous and

\int_0^1 f(x) x^n \, dx = 0

for all n = 0, 1, ... Prove that f(x) = 0.

The attempt at a solution
There's a hint that says: Prove that

\int_0^1 (f(x))^2 \, dx = 0.

I don't know how to prove this hint and I don't know how that would help in determining what f is. Any tips?

Are you SURE you don't know why the integral of f^2 equals zero wouldn't solve the problem?

[Brian_C]

The integral of f^2 can only be zero if f is identically zero. If f(x) were greater than zero at some point in the interval, there would be a positive contribution to the integral from that point with no negative contribution to cancel it out (since the square of a real function is non-negative). Think about it.

I don't know how to prove the assertion. I will have to think about it.

[e(ho0n3]

That thread was really helpful. Now I know why the hint is true. However...

I don't know why the integral of f^2 equals zero wouldn't solve the problem. You asked Hitman2-2 the same question as well, but he/she didn't respond.

[Dick]

Brian_C just answered why the integral of f^2 equal zero solves the problem.

[Brian_C]

I have an idea how to prove that the integral of f^2 is equal to zero. Try expanding one of the f(x)'s in the integral of f^2 as a taylor series in x. You should end up with an infinite series of integrals of the form f(x) * x^n, which should all vanish (by assumption), thus proving that the integral of f^2 vanishes. The only problem is, I don't know how to prove that a Taylor series will converge on some interval for this particular function.

[Dick]

The taylor series doesn't have to converge. The function is only continuous, it doesn't have to be differentiable, much less analytic. Check Hitman2-2's thread. You have to use the Weierstrass approximation theorem.