Showing that $\int_0^1 x^{2n}f(x) dx = 0$ implies $f = 0$

This is my question:

Show that if $f \in C[0,1]$ satisfies $\int_0^1 x^{2n}f(x) dx = 0$, then $f$ is the zero function.

Note: I am aware that a similar question to this has been asked on maths SE at least once, but those questions deal with $\int_0^1 x^{n}f(x) dx = 0$, ie $n \in \Bbb N_0$ - I am interested in $2n \in 2\Bbb N_0$.

The first part of the question was to show that it is true when any $n$ is allowed. To show this, it asked me to use the Weierstrass Approximation Theorem (see this question).

The question then continues on to ask whether or not it is true for $x^{2n+1}$. Hopefully, if I can determine for $2n$, then I should be able to adapt that to $2n+1$, or if I can't then come up with a counter-example.

Oh, and please note: I'm interested in learning from this question, so please don't just put the answer without an explanation! If you can, I'd most appreciate a hint as to how to do it, then I'll work it out myself.

• The Muntz-Szasz theorem is a generalization of the Weierstrass theorem which applies to the second situation. This might be overkill, though. – user38355 Nov 6 '14 at 14:58
• Yeah, haven't come across that theorem in lectures... – Sam T Nov 6 '14 at 15:01
• You haven't stated that $f$ is continuous. It's necessary though, as you could pick the characteristic of $\Bbb Q$ as a counterexample (using Lebesgue integral so that the function is actually integrable). – Jean-Claude Arbaut Nov 6 '14 at 15:27
• Yeah, sorry, of course. I mean, just take $f(x) = 0$ for all $x \neq 1/2$ and $f(1/2) = 1$. Obviously satisfies the condition. Thanks for pointing that out - edited now. – Sam T Nov 6 '14 at 15:29

Extending $f$ to a continuous, even function on $[-1, 1]$, we then have $\int_{-1}^1 x^n f = 0$ for all $n$, as $\int_{-1}^1 x^{2n} f = 2\int_0^1 x^{2n}f = 0$. Thus $f = 0$. For the case of odd $n$, consider $g(x) = x f(x)$.

• Sorry, why do we know that $\int_0^1 x^n f(x) dx = 0$ just by extending to $[-1,1]$? – Sam T Nov 6 '14 at 15:03
• Use the fact that $f$ is even. – anomaly Nov 6 '14 at 15:05
• You don't even need to perturb $f$ at all, you can extend it to an even function directly. – Daniel Fischer Nov 6 '14 at 15:13
• Yes, the even extension of $f\colon [0,1]\to\mathbb{R}$ is automatically continuous. You only need $f(0)=0$ for an odd extension. – Daniel Fischer Nov 6 '14 at 15:16
• Ignore that: $$x^{2n+1}f(x) = x^{2n} (x(f(x)) = x^{2n}g(x)$$ not $$x^{2n+1}f(x) = x^{2(n+1)}g(x).$$ Sorry about that! – Sam T Nov 6 '14 at 15:40

The Stone-Weierstrass theorem (a natural generalization of the Weierstrass approximation theorem) says that polynomials with all terms of even degree are uniformly dense in continuous functions on $[0,1]$, since they form a subalgebra which separates points and contains the constants. Using this you may proceed as in the proof of the first part. Note that the Stone-Weierstrass theorem does not apply to polynomials with all terms of odd degree as these don't even form an algebra.

• I did wonder if this might be the case - I didn't realise that that was Stone-Weierstrass. In fact, I went through the proof but then wasn't sure how to justify the claim. (Further, I also required $[0,1]$, not $[-1,1]$ - as I see you have edited your answer to contain.) – Sam T Nov 6 '14 at 15:17
• Yeah, I was noticing points are not separated on $[-1,1]$ – Seth Nov 6 '14 at 15:18
• I was just thinking that surely $f(x) = x$ can't be approximated by an even polynomial? – Sam T Nov 6 '14 at 15:30
• Why not? Certainly it can't on [-1,1] but according to the theorem it can on [0,1] – Seth Nov 6 '14 at 17:17
• Sorry, I was meaning that it couldn't be on $[-1,1]$. :) – Sam T Nov 6 '14 at 20:57