I was going through Measure theory book by Dorogovtsev and found this:
Let $p_k, k \geq 1$ be such numbers that $$ \sup_{0<s<1} \sum_{k=1}^\infty \frac{\sin^2(ks)}{s^2} p_k < +\infty $$ Proof that $$ \sum_{k=1}^\infty k^2 p_k < +\infty $$
Do you know any other (than the one I'm submitting) approach to proof this?
proof:
Let $(N_+, \lambda({k}):=p_k)$ be a measure space (equipped with natural sigma-algebra of course). Also we have: $$\lim_{s\to 0} \frac{\sin^2(ks)}{s^2} = k^2$$ Now applying Fatou's lemma ($ \int_X \lim \inf f_k d \lambda \leq \lim \sup \int_X f_k d\lambda $) $$ \int_{N_+} k^2 d\lambda = \int_{N_+} \lim_{s\to 0} \frac{\sin^2(ks)}{s^2} d\lambda = \int_{N_+} \lim_{s_0 \to 0} \inf_{s \leq s_0} \frac{\sin^2(ks)}{s^2} d\lambda \leq \lim_{s_0 \to 0} \sup_{s \leq s_0} \int_{N_+} \frac{\sin^2(ks)}{s^2} d\lambda < \infty $$
Since $(N_+,\lambda(k):=p_k)$ is a measure space, I will assume that $p_k\ge0$. Then, for any $n$, $$ \begin{align} \sum_{k=1}^nk^2p_k &=\lim_{s\to0}\sum_{k=1}^n\frac{\sin^2(ks)}{s^2}\,p_k\\ &\le\limsup_{s\to0}\sum_{k=1}^\infty\frac{\sin^2(ks)}{s^2}\,p_k\\ &\le\sup_{0\lt s\lt1}\sum_{k=1}^\infty\frac{\sin^2(ks)}{s^2}\,p_k\\[9pt] &=L\lt\infty \end{align} $$ Therefore, the the sequence of partial sums is increasing and bounded above. Thus, $$ \sum_{k=1}^\infty k^2p_k\le L $$