Does $\xi_{ni}\overset{p}\to0$ imply $\frac1n\sum_{i=1}^n\xi_{ni}\overset{p}\to0?$ Suppose for each $i=1,\cdots,n$, we have $$\xi_{ni}\overset{p}\to0.$$ Can we claim that $$\frac1n\sum_{i=1}^n\xi

Does $\xi_{ni}\overset{p}\to0$ imply $\frac1n\sum_{i=1}^n\xi_{ni}\overset{p}\to0?$ Suppose for each $i=1,\cdots,n$, we have $$\xi_{ni}\overset{p}\to0.$$ Can we claim that $$\frac1n\sum_{i=1}^n\xi_{ni}\overset{p}\to0?$$ Here $\overset{p}\to$ means convergence in probability. That is, for any $\varepsilon>0$, we have $\lim_{n\to\infty}P(|\xi_{ni}|>\varepsilon)=0$

You deleted a similar question where there was only one subscript. But anyway, the result is false.

Let $\xi_{n,i}=n 1_{[\frac{i-1}n,\frac{i}n)}$ be a sequence of random variables on the usual $[0,1)$. Then for each $i$ (we can even allow it to depend on $n$), $\xi_{n,i}\xrightarrow{\mathbb{P}}_n0$ because $\mathbb{P}(\xi_{n,i}>\varepsilon)=\frac1n$ as long as $\varepsilon$ is small. However, $n^{-1}\sum_i\xi_{n,i}=1$. You can give more extreme examples.

This also gives a counterexample to your previous question --- if $\xi_n\xrightarrow{\mathbb{P}}_n0$, the Cesaro sums need not converge to 0 in probability --- just lay these $\xi_{n,i}$ out like $$\xi_{1,1},\xi_{2,1},\xi_{2,2},\xi_{3,1},\xi_{3,2},\xi_{3,3},\dots,\xi_{n,1},\dots,\xi_{n,n},\dots$$ and note the Cesaro limit has to be the constant function 1.