# Integration and sequences of functions

1. Feb 5, 2009

### Hitman2-2

1. The problem statement, all variables and given/known data

Let f be a continuous function on [0,1]. Prove that if

$$\int_{0}^{1} x^n f(x) dx = 0$$

for all even natural numbers n, then f(x) = 0 for all $$x \in [0,1]$$.

2. Relevant equations

3. The attempt at a solution

I'm pretty much stuck on this problem. All I know is that by the Weierstrass Approximation Theorem, there exists a sequence of polynomials Pn that converges uniformly to f, so

$$\lim_{n\rightarrow 0} \int_{0}^{1} P_n f(x) dx = \int_{0}^{1} f(x) \lim_{n\rightarrow 0} P_n dx = \int_{0}^{1} f^2(x) dx$$

but that's about as far as I can get.

2. Feb 5, 2009

### Dick

How about if you extend f(x) to an even function g(x) on the interval [-1,1]? And while you are at it subtract a constant so the integral of g(x) over [-1,1] is zero. Now the integral of x^k times g(x) equals zero for all k greater than or equal zero.

3. Feb 5, 2009

### Hitman2-2

What would be the relationship between

$$\int_{0}^{1} x^n f(x) dx$$

and

$$\int_{-1}^{1} x^n g(x) dx$$

4. Feb 5, 2009

### Dick

If n is even then the second one is twice the first one, right? Or zero. If n is odd then the second one vanishes because its the integral of an odd function.

5. Feb 5, 2009

### Dick

Hmmmm. I've been trying to show integral of g^2 over [-1,1] is zero. But I can't figure out what to do with the 'constant part' of g(x). My suggestion earlier that you just 'subtract it off' doesn't work. Maybe there's an easier way. Any new ideas?

6. Feb 5, 2009

### Hitman2-2

No, unfortunately. But I'll let you know if I figure anything out. Thanks again.

7. Feb 5, 2009

### Dick

Do that, thanks. This is annoying me.

8. Feb 5, 2009

9. Feb 5, 2009

### Hitman2-2

For the course I'm currently in, I think so. We know 0 is an even number. In my class, {0,1,2,...} is taken to be the set of natural numbers (though other professors I've had take {1,2,3,...} to be the natural numbers).

10. Feb 5, 2009

### Dick

Ok, that fixes it. The 'constant part' of g(x) must be zero. Now can you show that the integral of g(x)^2 must be zero? Just as you first started to do with Weierstrass. Do you see how that proves g(x)=0, hence f(x)=0?

11. Feb 5, 2009

### Hitman2-2

Well, g(x) would be continuous on [-1,1] so we have a sequence of polynomials pn converging uniformly to g, hence

$$0 = \int_{-1}^{1} g(x) \lim_{n\rightarrow 0} p_n dx = \int_{-1}^{1} g^2(x) dx$$

right?

12. Feb 5, 2009

### Dick

Yes, if you have followed everything that went before. All the powers in p_n integrated against g are zero.

13. Feb 5, 2009

### Hitman2-2

all right, I think I see what's going on. Thanks again for all your assistance.