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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 then $0 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: Ok. Understood2010-11-19

1 Answers 1

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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.