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If $f \in L(0,1)$ show that $x^{n} f(x) \in L(0,1)$ for $n \in \mathbb{N}$ and that $\displaystyle \int_{0}^{1} x^{n} f(x) dx \rightarrow 0$ as $n \to \infty$.

Is the following attempt correct?

Since $0<x<1$ then $0<x^{n}<1$ also $x^{n}$ is measurable (being cts) and $f$ is measurable because it is in $L(0,1)$ so $x^{n} f(x)$ is measurable. Now certainly $|x^{n}f(x)| \leq |f(x)|$ so this implies that $x^{n} f(x)$ is integrable so in $L(0,1)$.

Now to finish the second part can we simply say that:

$x^{n} f(x) \rightarrow 0$ as $n \rightarrow \infty$ pointwise so we may apply convergence theorem.

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@user10: What is $L(0,1)$ –  anonymous Nov 19 '10 at 23:47
    
I think $L(0,1)$ stands for set of all Lebesgue integrable functions on $(0,1)$. –  anonymous Nov 19 '10 at 23:50
    
@Chandru, yes. The vector space of all measurable and integrable functions with respect the seminorm $||f||_{1} = \int_{X} |f| d\mu$, i.e $L^{1}(0,1)$. –  student Nov 19 '10 at 23:50
    
@user10: Ok. Understood –  anonymous Nov 19 '10 at 23:51

1 Answer 1

I am assuming that by $L(0,1)$, you mean $\mathcal{L}^1 (0,1)$.

You have $x^nf(x)$ dominated by $f(x)$ in the interval $(0,1)$ and $f \in \mathcal{L}^1 (0,1)$. Hence, you can use dominated convergence theorem to swap the limit and the integral.

And hence you will get the desired result since $\displaystyle \lim_{n \rightarrow \infty} x^n f(x) = 0$ almost surely.

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