2
$\begingroup$

Problem: Let $\varphi_{1}, \varphi_{2}, \cdots, \varphi_{n},\cdots$ be non-negative continuous functions on $[0,1]$ such that the limit $$\lim_{n \to \infty} \int\limits_{0}^{1} x^{k} \varphi_{n}(x) \ \text{dx}$$ exists for every $k \in \mathbb{Z}_{+}$. Does this imply the limit $$\lim_{n \to \infty} \int\limits_{0}^{1} f(x) \varphi_{n}(x) \ \text{dx}$$ exists for every continuous function $f(x)$ on $[0,1]$.

No idea on how to proceed. Any help would be useful.

Added. As everybody suggested the idea is to use the Stone-Weierstrass theorem in the space $\mathcal{C}[0,1]$ equipped with the $\text{Sup-norm}$. So let $\{p_{n}\}$ be a sequence of polynomials which converge to $f$ in the $\text{sup-norm}$. With our hypothesis, we can conclude that there is some positive $C$ such that $$\Biggl|\int\limits_{0}^{1} \varphi_{n}(x) \ \text{dx}\Biggr| \leq C$$ for all $n$. Since $p_{n} \to f$, we have for given $\epsilon >0$, there exists an integer $N$ such that $||f-p_{N}||< \epsilon$ for all $k \geq N$. I could only reach till here by applying the definitions.

  • 0
    Stone-Weierstrass suggests itself...2011-05-04
  • 0
    Maybe you can try approximating $f(x)$ by polynomials using the [Stone-Weierstrass Theorem](http://en.wikipedia.org/wiki/Stone-Weierstrass_theorem)?2011-05-04
  • 0
    Enlighten by this Stone-Weierstrass Theorem based problem, I wonder if the following is true: if $\displaystyle \lim_{n\to\infty} \int^1_0 \sin(kx)\phi_n(x) \,dx = 0$ for any $k$, does this implies $\displaystyle \lim_{n \to \infty} \int^1_0 f(x) \phi_{n}(x) = 0$ for any $f\in C([0,1])$?2011-05-04
  • 0
    Yes, i tried using Ston-Weierstrass and ended up somewhere.2011-05-04
  • 0
    I couldn't reach the conclusion.2011-05-04
  • 0
    @Chandru1: Well, you have uniform convergence of the integrand if you replace $f$ by $p_k$ (not $p_n$!), then by uniform convergence you can switch limits for example and the integrals also converge uniformly to the integral of the limit function.2011-05-04
  • 1
    You shouldn't just apply the definitions blindly but apply what the definitions imply!2011-05-04
  • 0
    @Jonas T: Well, thanks for the hint. I did what i could, this is what i came up with.2011-05-04

1 Answers 1

5

We can quickly see that for all polynomials $p$ there holds that

$$\lim_{n \to \infty} \int_0^1 p(x) \phi_n(x) \, dx \text{ exists}$$

So now, approximate your $C[0,1]$ function $f$ uniformly by polynomials $p_k$.

We would now like to show that

$$\lim_{n \to \infty} \int_0^1 \lim_{k \to \infty} p_k(x) \phi_n(x) \, dx \text{ exists}.$$

By uniform continuity we know that $\int f_n \to \int f$ if $f_n \to f$ uniformly. Now note that $p_k \phi_n \to f \phi_n$ uniformly as well as $k \to \infty$.