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I am struggling with this question!

For each $n > 1$, let $g_n(x)=\frac{1}{x} \chi_{(1,n)} (x) : \mathbb{R} \rightarrow \mathbb{R}$, where $\chi_(1,n)$ is the characteristic function of $(1,n)$. Show that $g_n \in L_1(\lambda)$. Show that for every $n>1$, $\exists$ a m>1 such that $||g_n - g_m||_1>1$. Here $\lambda$ is the Lebesgue measure on $\mathbb{R}$.

Now it appears to me that $||g_n - g_m||_1 = \int^m_1 \frac{1}{x} dx - \int^n_1\frac{1}{x}dx = \ln(\frac{m}{n}) $. However, I am not too sure how to show that this is greater than 1.

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  • $\begingroup$ If $m = en$ then $\ln(m/n) = 1$. So if $m > en$ then... $\endgroup$
    – Umberto P.
    May 12, 2017 at 13:37
  • $\begingroup$ take $n=m/3$ if $m$ odd and $m/4$ if $m$ even. $\endgroup$
    – user386627
    May 12, 2017 at 13:37
  • $\begingroup$ @UmbertoP.: I guess that $m,n\in\mathbb N$ $\endgroup$
    – user386627
    May 12, 2017 at 13:38
  • $\begingroup$ so all I have to do is define what m to take and then I have answered the question? $\endgroup$ May 12, 2017 at 13:38
  • $\begingroup$ @user386627 I didn't check but I'm fairly certain there is a natural number larger than $en$. $\endgroup$
    – Umberto P.
    May 12, 2017 at 13:39

1 Answer 1

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After your computation, the problem reduces to the following: prove that for every $n\gt 1$, there exists a integer $m\gt 1$ such that $\ln\left(m/n\right)\gt 1$.

So let $n\gt 1$ be fixed. Using equality $\ln\left(m/n\right)=\ln m-\ln n$, we have to find $m$ such that $\ln m\gt 1+\ln n$. We know that $\lim_{m\to +\infty}\ln m=+\infty$, so we are sure that such an $m$ exists. But we can give an explicit choice: take $m=3n$. Then $$\ln m=\ln 3+\ln n\gt \ln e+\ln n=1+\ln n.$$

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