# Prove that f = 0 integral

1. Mar 24, 2006

"Suppose f is continuous on [a,b] and for all k = 0,1,2,3,4,5,... $$\int_a^b x^kf(x) = 0$$. Prove that f = 0."

What I know so far:

f(c)=0 for some c in [a,b]

$$\int_a^b f=0$$

$$\int_a^b x^k \int_a^b f=0$$

Any hint on how to proceed would be appreciated.

2. Mar 24, 2006

### AKG

Use Weierstrass approximation, and consider $\int f^2$.

3. Mar 25, 2006

I can't see the solution. Can you elaborate a little more?

Last edited: Mar 25, 2006
4. Mar 25, 2006

### AKG

First, do you know the Weierstrass approximation theorem? Next, if p is any polynomial, what is $\int pf$? Putting those two facts together, what can you conclude about $\int f^2$?

5. Mar 25, 2006

Weierstrass approximation theorem tells me that f(x) can be uniformly approximated by polynomials. So give e>0, for some p(x), |p(x)-f(x)|<e for all x. x^k * f(x) is apprximately x^k * p(x). But I don't understand how this has to do with f^2.

6. Mar 25, 2006

### AKG

f is bounded, so there exists M such that M > |f(x)| for all x in [a,b]. For all e > 0, there exists a polynomial p(x) such that |p(x) - f(x)| < e/M for all x. Thus, for such a polynomial, |f(x)p(x) - f²(x)| < e

$$|\int f^2| = |\int f^2 + 0| = |\int f^2 - \int fp| = |\int f^2 - fp| \leq \int |fp - f^2| \leq (b-a)e$$

Since we can make e arbitarily small, $|\int f^2| = 0$, but f² is non-negative continuous, so f² is zero, hence so is f.

7. Mar 26, 2006

At no point did you use the hypothesis of 'x^k for all k=0,1,2,...' . This would imply that if f is continuous on [a,b] and $$\int_a^b f=0$$, then f=0.

8. Mar 26, 2006

### shmoe

Try to justify each of AKG's steps carefully.

9. Mar 26, 2006