Integration and sequences of functions

[Hitman2-2]

Homework Statement

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

$$<br /> \int_{0}^{1} x^n f(x) dx = 0<br />$$

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

Homework Equations

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 P_n that converges uniformly to f, so

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

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

 

[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.

 

[Hitman2-2]

What would be the relationship between

$$<br /> \int_{0}^{1} x^n f(x) dx<br />$$

and

$$<br /> \int_{-1}^{1} x^n g(x) dx<br />$$

 

[Dick]

What would be the relationship between

$$<br /> \int_{0}^{1} x^n f(x) dx<br />$$

and

$$<br /> \int_{-1}^{1} x^n g(x) dx<br />$$

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.

 

[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?

 

[Hitman2-2]

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

 

[Dick]

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

Do that, thanks. This is annoying me.

 

[Dick]

Um, you wouldn't consider n=0 an even natural number, would you? See https://www.physicsforums.com/showthread.php?t=290229

 

[Hitman2-2]

Um, you wouldn't consider n=0 an even natural number, would you? See https://www.physicsforums.com/showthread.php?t=290229

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).

 

[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?

 

[Hitman2-2]

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?

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

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

right?

 

[Dick]

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

 

[Hitman2-2]

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