Proving the limit of a linear transformation Assume that 
$$
\lim_{J\rightarrow \infty}| J f(j,J)-g(j,J)|=0,\quad\forall j \in \{1,\cdots,J\}
$$
where


*

*$f: \mathbb{N}\times \mathbb{N}\to [0,1]$,

*$g: \mathbb{N}\times \mathbb{N}\to [0,\infty)$.


I want to show that
$$
(*) \hspace{1cm}\lim_{J\to\infty}\left| \left(1-\sum_{j=1}^J f(j,J)\right)-\left(1-\frac{1}{J}\sum_{j=1}^Jg(j,J)\right)\right|=0.
$$
Below is how I proceed. 


*

*Could you tell me whether my steps are correct (My big doubt is about step (A))? 

*In case $(*)$ is wrong, could you specify which additional conditions would help to claim it? 

*Lastly, is Step A applicable to the setting in this other question of mine? I guess not (because for that different case the book suggests to go through local uniformity) but I do not understand why.

$$
\lim_{J\to\infty}\left| J f(j,J)-g(j,J)\right|=0,\quad\forall j \in \{1,\cdots,J\}\\
\Updownarrow\\
\lim_{J\rightarrow \infty}\Big\{ J f(j,J)-g(j,J)\Big\}=0 \text{ }\forall j \in \{1,...,J\}\\
\Downarrow \text{Step (A): following a comment below - still, it is unclear to me how to show this}\\
\lim_{J\to\infty}\left\{ \frac{1}{J} \left(J \sum_{j=1}^Jf(j,J)-\sum_{j=1}^J g(j,J)\right)\right\}=0\\
\Updownarrow\\
\lim_{J\to\infty}\left\{ -\sum_{j=1}^Jf(j,J)+\frac{1}{J}\sum_{j=1}^J g(j,J)\right\}=0\\
\Updownarrow\\
\lim_{J\to\infty}\left\{ 1-\sum_{j=1}^Jf(j,J)+\frac{1}{J}\sum_{j=1}^J g(j,J)\right\}=1\\
\Updownarrow\\
\lim_{J\to\infty}\left\{ \left(1-\sum_{j=1}^Jf(j,J)\right)-\left(1-\frac{1}{J}\sum_{j=1}^J g(j,J)\right)\right\}=0\\
\Updownarrow\\
\lim_{J\to\infty}\left| \left(1-\sum_{j=1}^Jf(j,J)\right)-\left(1-\frac{1}{J}\sum_{j=1}^J g(j,J)\right)\right|=0 
$$
 A: First, the proposition is not necessarily true.
Counter-example: Take$$
n f(m, n) - g(m, n) = \frac{m^a}{n^b}, \quad \forall m, n \in \mathbb{N}_+
$$
where $a > b > 0$ are constants, then$$
\lim_{n \to \infty} |n f(m, n) - g(m, n)| = 0. \quad \forall m \in \mathbb{N}_+
$$
However,\begin{align*}
\frac{1}{n} \sum_{m = 1}^n (n f(m, n) - g(m, n)) = \frac{1}{n^{b + 1}} \sum_{m = 1}^n m^a \sim \frac{1}{a + 1} · \frac{n^{a + 1}}{n^{b + 1}}, \quad n \to \infty
\end{align*}
which implies$$
\lim_{n \to \infty} \left| \sum_{m = 1}^n f(m, n) - \frac{1}{n} \sum_{m = 1}^n g(m, n) \right| = +\infty.
$$

Second, a useful but rather strong condition that guarantees the proposition is $h(m, n) := nf(m, n) - g(m, n)$ uniformly tends to $0$ as $n \to \infty$ for $m \in \mathbb{N}_+$, i.e.$$
\lim_{n \to \infty} \sup_{m \in \mathbb{N}_+} |h(m, n)| = 0.
$$

Last, due to the same reason as that in the counter-example given above, the convergence of Cesaro sum involving two variables is not guaranteed without further condition such as uniformity.
