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
Application of Stone-Weierstrass Theorem
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real-analysis
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0@KWO No trouble at all. Hope you enjoy it here! – 2012-05-20
1 Answers
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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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0It works and this approach is indeed better. I confused Stone-Weierstrass theorem with Weierstrass (density of the polynomial, which is a particular case). – 2012-05-20