1
$\begingroup$

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!

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Sahiba Arora, Daniel W. Farlow, Leucippus, JonMark Perry, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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 $f\equiv 0.$

$\endgroup$
  • $\begingroup$ 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 $f \equiv 0$? $\endgroup$ – MathematicsStudent1122 Aug 7 '17 at 20:27
  • $\begingroup$ @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. $\endgroup$ – zhw. Aug 7 '17 at 22:07
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – 2ndaccount Aug 7 '17 at 18:53
  • $\begingroup$ What if $N=k+1?$ $\endgroup$ – zhw. Aug 7 '17 at 18:55
  • $\begingroup$ 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..." $\endgroup$ – 2ndaccount Aug 7 '17 at 18:56
  • $\begingroup$ 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? $\endgroup$ – Jack D'Aurizio Aug 7 '17 at 18:56
  • $\begingroup$ Do I really need to show what kind of polynomials are in the set? Seems like you listed a specific kind of polynomials. $\endgroup$ – 2ndaccount Aug 7 '17 at 18:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.