Inequality of double sum with max operator and squared

Question

I am trying to understand a proof, however one inequality remains unclear:

$\mathbb{E} \big( \frac{1}{k_1 k_0} \sum_{i=1}^{k_1} \sum_{j=1}^{k_0} a_{j,i} -b_{j,i} \big)^2 \leq \max_i \frac{1}{k_m^2}\sum_{j=1}^{k_0}\mathbb{E}\big( a_{j,i} -b_{j,i}\big)^2 $, where $a_{j,i}$ and $b_{j,i}$ are random variables over which the expectation is taken. Is this a consequence of Cauchy Inequality?

Thanks a lot!

Answer 1

This looks very elementary if $k$ could depend on $k_0$ and $k_1$. By Cauchy Schwarz inequality LHS does not exceed $C \sum\limits_{i=1}^{k_1}\sum\limits_{j=1}^{k_0}E(a_{ji}-b_{ji})^{2}$ for some $C=C(k_0,k_1)$ and hence LHS $ \leq \max_i Ck_1 \sum\limits_{j=1}^{k_0} E(a_{ji}-b_{ji})^{2}$.