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This is from an old qualifying examination question.

The premise of the question is as follows: Let $f:\mathbb R \rightarrow \mathbb R$ be continuous. Show that

$$\lim_{n\rightarrow \infty} \int_0^1nx^{n-1}f(x)dx=f(1) $$

I know how to show this without using Weierstrass approximation theorem. But I'd still like to see someone else's take on this.

My questions are

1) Is there a way to show this using just the Weierstrass approximation theorem (i.e., without using Stone's generalization). Weierstrass says that a continuous function defined on a closed bounded interval can be uniformly approximated by polynomials.

2) a sequence of functions is uniformly convergent iff it is convergent in the sup norm. Is there a analogous characterization for uniform continuity interms of suprema?. What about continuity?

Any comments hints are appreciated.


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up vote 2 down vote accepted

I don't really understand your second question so I'll only adress the first.

  1. Assume $f\in C^1$ then an integration by parts gives $$ \int_0^1 nx^{n-1}f(x)dx = f(1) -\int_0^1 x^nf'(x)dx $$ the second integral can easily be seen to converge to $0$ (for example split the integral $\int_0^1 = \int_0^{1-\varepsilon} + \int_{1-\varepsilon}^1$. In the first interval the integrand converges unifromly to $0$ and is uniformly bounded in the second as $\varepsilon \to 0$).

  2. Now assume $f$ is only continuous and take a sequence $f_m\to f$ unfiormly such that $f_m\in C^1$ (for example polynomials via Wierstrass' theorem), and take $\varepsilon>0$, we estimate $$ \left| f(1)-\int_0^1nx^{n-1}f(x)dx\right| \leq |f(1)-f_m(1)| + \left| f_m(1)-\int_0^1nx^{n-1}f_m(x)dx\right| + \left| \int_0^1nx^{n-1}(f_m(x)-f(x))dx\right| $$ If we pick $m$ such that $\sup|f_m-f|<\varepsilon/3$. Then the first term is clearly $\leq \varepsilon/3$, as is the third term since $\int_0^1nx^{n-1} =1$. Now pick $n$ big enough so that the second term is below $\varepsilon/3$ (we can because $f_m\in C^1$) and the convergence follows.

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Thanks for your answer. But if $f\in C^1$ then $||f'||_\infty <\infty$ and You don't even need Weierstrass. :). f is only continuous not $C^1$. – Jack Dawkins Jan 7 '13 at 5:39
@user54755: I've edited to make the approximation step more explicit. – Jose27 Jan 7 '13 at 5:45

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