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Let $f$ be a continuous function over $[0,1]$ and $\displaystyle \int_0^1 f(x)\,dx=0$. If for any positive integer $n$, $\displaystyle \int_0^1 x^{12+3n}f(x)\,dx=0$ then using Stone-Weiersterass Theorem prove that $f\equiv 0$ in $[0,1]$.

Since $f$ is contnuous on $[0,1]$ so by Stone-Weierstrass Theorem there exists a sequence of polynomials $\{p_n(x)\}_n$ which converges uniformly to $f$ on $[0,1]$. That is , $p_n\to f$ , implies $p_n(x)f(x)\to f^2(x)$ as $n\to \infty$, which implies that $\displaystyle \int_0^1p_n(x)f(x)\,dx \to \int_0^1 f^2(x)\,dx$. But from the given condition I can't say that $\displaystyle \int_0^1 f^2(x)\,dx =0$. So how to proceed to solve this problem ? Any hint. please.

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    $\begingroup$ Try to apply Stone-Weierstrass with not all the polynomials, but just the polynomials of a certain form. $\endgroup$ Commented Nov 26, 2020 at 14:33
  • $\begingroup$ @TheSilverDoe Not getting the point. Can I get hint a little more ? $\endgroup$
    – Empty
    Commented Nov 26, 2020 at 14:46
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    $\begingroup$ Stone-Weierstrass theorem gives that under some conditions, a subset of $\mathcal{C}([0,1])$ is dense. What you know (and what you wanted to use) is that Stone-Weierstrass can be applied with the subset of all polynomial functions. But you can also try to apply it with another subset, maybe smaller than all the polynomials, which would also satisfy the conditions of Stone-Weierstrass, and hence, would be dense... $\endgroup$ Commented Nov 26, 2020 at 14:48

2 Answers 2

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The span of the monomials $\{1\} \cup \{x^{3n+12} : n\in\Bbb{N}\}$ is dense in $C[0,1]$ because $$\sum_{n=1}^\infty \frac1{3n+12} = \frac13 \sum_{n=1}^\infty \frac1{n+4} = \frac13 \sum_{n=5}^\infty \frac1n = +\infty$$

so the Müntz–Szász theorem gives the conclusion.

Now approximate $f$ by polynomials from $\operatorname{span}(\{1\} \cup \{x^{3n+12} : n\in\Bbb{N}\})$.

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Setting $g(x)=x^{12}f(x)$, we know that $$ \int_0^1 x^{3n}g(x)\,dx=0, $$ for every $n>0$, and hence also $$ \int_0^1 p(x^3)g(x)\,dx=0, $$ for every polynomial $p$ with vanishing constant term. Since the set of functions of the form $p(x^3)$, with $p(0)=0$, is a dense subalgebra of $$ C_0((0,1]) = \{h∈ C([0,1]): h(0)=0\} $$ by Stone-Weierstrass, we deduce that $\int_0^1 h(x)g(x)\,dx=0$, for all $h$ in that subalgebra, so $g=0$ on $(0,1]$.

Consequently $f=0$ on $(0,1]$ as well and, since $f$ is continuous, we conclude that $f$ vanishes everywhere.

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