Let $a_{ij}>0$ for $\forall i\in[m],\forall j\in[n]$. Here $[m]$ denotes $\{1,2,\ldots,m\}$. ($\forall i) \sum_{j=1}^na_{ij}\leq d$. If $a_{ij}<\epsilon$ for all $i\in[m],j\in[n]$, then $\forall i$, $O(\lambda^2+\frac{\sum_{j=1}^n a_{ij}^{1.5}}{\lambda})\leq O(\lambda^2+\frac{\sqrt{\epsilon}d}{\lambda})$, if we choose $\lambda=(\sqrt{\epsilon} d)^{1/3}$, then the above term $\leq O((\sqrt{\epsilon} d)^{2/3})$.
Now we have $\{p_i\}$ $i\in[m]$ such that $\forall i,p_i>0$ and $\sum_{i=1}^mp_i=1$. Now we change the condition "$\forall i\in[m],\forall j\in[n],a_{ij}<\epsilon$" into "$\forall j\in[n],\sum_{i=1}^mp_ia_{ij}<\epsilon$" with other conditions unchanged. Then for the term $O(\sum_{i=1}^mp_i(\lambda_i^2+\frac{\sum_{j=1}^n a_{ij}^{1.5}}{\lambda_i}))$, can we choose a set of $\{\lambda_i>0\}$, such that the term above can be upperbounded by another term $A(\epsilon,d)$, such that $\lim_{\epsilon\to0}A(\epsilon,d)=0$, and $A$ is independent of $m$ or $n$ or {$p_i$}?
Here is my attempt which doesn't meet the final requirement. $O(\sum_{i=1}^mp_i(\lambda_i^2+\frac{\sum_{j=1}^n a_{ij}^{1.5}}{\lambda_i}))\leq O(\sum_{i=1}^mp_i(\lambda_i^2+\frac{(\sum_{j=1}^n a_{ij})^{1.5}}{\lambda_i}))$ and choose $\lambda_i=(\sum\limits_{j=1}^n a_{ij})^{0.5}$, then it is upperbounded by $O(n\epsilon)$, but it depends on $n$.