Lebesgue measure of $\{ u \ge \epsilon\}$

Question

Let $$u(x) := \sum_{n=1}^\infty \frac{1}{n^2(n+1)} 1_{(n,2n]}(x),$$ and show that for any $\epsilon > 0$, $\lambda(\{x \ | \ u(x) \ge \epsilon \}) \le 1/\epsilon$.

Any suggestions?

Answer 1

Assuming that $x$ is positive and large, $$ u(x) = \sum_{\substack{n\geq\frac{x}{2}\\ n<x}}\frac{1}{n^2(n+1)}\approx \int_{x/2}^{x}\frac{dz}{z^2(z+1)}=\frac{1}{x}+\log(1+x)-\log(2+x)\approx\frac{3}{2x^2}. $$ Moreover, $u(x)$ is non-negative and Lebesgue integrable over $\mathbb{R}^+$. Since: $$ \int_{0}^{+\infty}u(x)\,dx = \sum_{n\geq 1}\frac{1}{n(n+1)} = \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+1}\right)= 1 $$ your bound follows from Markov's inequality. We may also compute:

$$ \|u\|_{L^2(\mathbb{R}^+)}^2=\sum_{n\geq 1}\frac{1}{n^3(n+1)^2}+2\sum_{n=1}^{+\infty}\sum_{\substack{m<n\\ m\geq\frac{n}{2}}}\frac{2m-n}{n^2 m^2(n+1)(m+1)}$$ and get a slightly better bound from Chebyshev's inequality.