# If $\int_0^1 f(x)x^{2n+1}dx=0$ for all but finitely many $n\in\mathbb{N}$, then $f=0$ on $[0,1]$. [closed]

Let $f$ be continuous real function. Assume $\int_0^1 f(x)x^{2n+1}dx=0$ for all but finitely many $n\in\mathbb{N}$, then $f=0$ on $[0,1]$.

Use Stone-Weierstrass theorem (not change of variables.)

*If we loosen up the problem: say $\int_0^1 f(x)x^ndt=0$ for all but finitely many $n$, without changing the variable how would stone weierstrass work here?

*What about $x^2n$ for all $n\geq 0$, infinity many terms are 0 and infinity many terms may not be 0?

Thank you!

## closed as off-topic by Sahiba Arora, Daniel W. Farlow, Leucippus, JonMark Perry, Claude LeiboviciAug 8 '17 at 6:59

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Let $g(x) = xf(x).$ Then $\int_0^1g(x)x^{2n}\, dx=0$ for $n\ge$ some $N.$ In particular

$$\tag 1 \int_0^1g(x)(x^{2N})^m\, dx=0,\,\, m=1,2,\dots.$$

Now by Stone-Weierstrass, polynomials in $x^{2N}$ are dense in $C[0,1].$ Hence there is a sequence of polynomials $p_k$ such that $p_k(x^{2N}) \to g(x)$ uniformly on $[0,1].$ But note $g(0)=0.$ It follows that $p_k(x^{2N})-p_n(0) \to g(x)$ uniformly on $[0,1].$ Since each $p_k(x^{2N})-p_n(0)$ is a finite linear combination of the monomials $(x^{2N})^m,\, m=1,2,\dots,$ we have by $(1)$

$$\int_0^1 g(x)[p_k(x^{2N})-p_n(0)]\,dx = 0$$

for all $k.$ Therefore $\int_0^1 g(x)^2\,dx =0,$ which implies $g\equiv 0,$ hence $f\equiv 0.$

• Can this method be generalized to show the following: if there is a strictly increasing sequence $a_k:\mathbb{N} \to \mathbb{N}$ such that $\int_{0}^{1}f(x)x^{n} dx = 0$ whenever $n=a_i$ for some $i$, then $f \equiv 0$? – MathematicsStudent1122 Aug 7 '17 at 20:27
• @MathematicsStudent1122 Good question. The answer is certainly no. Look up the Muntz-Szasz theorem in all its glory, a most beautiful theorem indeed. I wonder if there is an easy answer to your question. – zhw. Aug 7 '17 at 22:07

Müntz–Szász theorem provides an interesting overkill. The series $\sum_{d\geq 0}\frac{1}{2d+1}$ is clearly divergent and it stays so if we remove from it a finite number of terms. It follows that the span of $x^{2d+1}$ is dense in $C^0=[0,1]$, which is dense in $L^2(0,1)$. So if the original identity holds, $f$ has to be $0$ almost everywhere on $(0,1)$. Since $f$ is a continuous function, $f\equiv 0$.

As an alternative, let us consider the polynomials of the form $$P_{a,b}(x)=C_{a,b}\, x^{2a+1}(1-x^2)^b$$ with $a,b\in\mathbb{N}$, with $C_{a,b}$ chosen in such a way that $\int_{0}^{1}P_{a,b}(x)\,dx=1$.
We may chose $a$ and $b$ in such a way that $P_{a,b}(x)\geq 0$ is concentrated in a arbitrarily small neighbourhood of any $x_0\in(0,1)$, since $P_{a,b}$ attains its maximum at $\sqrt{\frac{2a+1}{2a+2b+1}}$.
Assume that $f\neq 0$ in a neighbourhood of $x_0\approx\sqrt{\frac{2a+1}{2a+2b+1}}$. Then $$\int_{0}^{1}P_{a,b}(x)\,f(x)\,dx$$ is arbitrarily close to both $f(x_0)\neq 0$ and $0$, contradiction.

• This is a little hard for me to digest... Can we simply use Stone weierstrass theorem with some not specified set of polynomials? Thank you – 2ndaccount Aug 7 '17 at 18:53
• What if $N=k+1?$ – zhw. Aug 7 '17 at 18:55
• my idea: (probably false) Since $f$ is continuous on a closed set, we have a set of polynomials approaching $f$ uniformly. Let this set of polynomials be in $t^\lambda$. It would be nice if the polynomials are $0$ through out all the terms. But I have problem with this"all but finitely many..." – 2ndaccount Aug 7 '17 at 18:56
• Life is hard. In order to tackle the problem through Stone-Weierstrass, you have to show that your algebra separates points. How do you plan to do it without invoking polynomials? – Jack D'Aurizio Aug 7 '17 at 18:56
• Do I really need to show what kind of polynomials are in the set? Seems like you listed a specific kind of polynomials. – 2ndaccount Aug 7 '17 at 18:57