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Suppose that $f \colon [0,1] \rightarrow \mathbb{R}$ is a continuous function on $[0,1]$ with $$\int_0^1 f(x)\ dx = \int_0^1 f(x)(x^n+x^{n+2})\ dx$$ for all $n=0,1,2, \dots$. Show that $f\equiv 0$. Can someone help me with this question? Is this one of the questions where we apply the Stone-Weierstrass Theorem? Thanks

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To write $\LaTeX$, just enclose the code in dollar signs. No need for the backticks—they'll just escape everything. – Dylan Moreland May 20 '12 at 14:37
Thanks a lot, my first time here, been searching how to get the typesetting right. Thanks again. – KWO May 20 '12 at 14:39
Hint: $f(x)(x^n+x^{n+2})=[f(x)(1+x^2)]x^n$. – user31373 May 20 '12 at 14:47
@KWO No trouble at all. Hope you enjoy it here! – Dylan Moreland May 20 '12 at 15:46
up vote 2 down vote accepted

Using the relation, we get $\int_0^1f(x)dx$, letting $n\to +\infty$. Indeed $$\left|\int_0^1f(x)dx\right|\leq \max_{0\leq t\leq 1}|f(t)|\int_0^1(x^n+x^{n+2})dx=\max_{0\leq t\leq 1}|f(t)|\left(\frac 1{n+1}+\frac 1{n+3}\right).$$

By induction, we deduce that $\int_0^1f(x)x^{2k}dx=0$ for $k\geq 0$ and that $\int_0^1f(x)x^{2k+1}dx=(-1)^k\int_0^1f(x)xdx$. We get $\int_0^1xf(x)dx=0$ letting $k$ going to $+\infty$, then that $\int_0^1f(x)dx=0$.

So what we got is that $\int_0^1f(x)x^k=0$ for all $k\geq 0$, and Stone-Weierstrass applies.

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Got it, thanks a lot. – KWO May 20 '12 at 15:03
I've added the details. – Davide Giraudo May 20 '12 at 15:07
A bit of comment, actually when we have $\int_0^1 f(x)x^{2k} \ dx =0$, then the problem is already solved, as the polynomial of the form $a_0+a_2 x^2 +\cdots +a_{2n}x^{2n}$ is dense in $C([0,1])$. – KWO May 20 '12 at 15:12
If you know this version of Stone-Weierstrass or if you write how you deduced it (for example using composition) it's an alternative way. – Davide Giraudo May 20 '12 at 15:14
I think we just need to check that the set of polynomial in this form form an algebra, it contains the function $x\rightarrow 1$ and that it separate points, then the Stone Weierstrass Theorem will apply and the results followed. Important thing is that in [0,1], $x^{2n}$ is injective. Thanks again for your help, greatly appreciate it. – KWO May 20 '12 at 15:20

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