Let $f_1,f_2,\ldots,g\colon\mathbb{Z}\rightarrow\mathbb{R}$ be functions such that $|f_N(n)|\leq g(n)$, $\sum_{n=-\infty}^\infty g(n)<\infty$, and $\lim_{N\rightarrow\infty}f_N(n)=f(n)$. Then show that $$\lim_{N\rightarrow\infty}\sum_{n=-\infty}^\infty f_N(n)=\sum_{n=-\infty}^\infty f(n).$$

This looks like the dominated convergence theorem, but how can we prove it directly?

Edit: As T. Bongers helpfully pointed out, this can be shown using the dominated convergence theorem. Is there a direct way to do it without the theorem?

  • $\begingroup$ This is exactly the dominated convergence theorem applied to the measure which is the sum of the point-mass measures at each integer. $\endgroup$ – user61527 Nov 29 '13 at 4:51
  • $\begingroup$ @T.Bongers Ah right. I was also wondering if there's a way to prove it directly (more easily) than proving the dominated convergence theorem? $\endgroup$ – Mika H. Nov 29 '13 at 4:55
  • $\begingroup$ Could you name the book/source you got this from? $\endgroup$ – Brofessor Apr 5 '18 at 10:45

The direct proof would still follow the proof of DCT (with simplifications). It suffices to show that $$\liminf_{N\to\infty} \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{n\in\mathbb Z} f(n) \tag{1}$$ because by applying (1) to $-f_N$ and $-f$ you get the reverse inequality which completes the proof. In order to show (1), add $\sum_{n\in\mathbb Z} g(n)$ to both sides. This reduces (1) to the case when all functions are nonnegative. Given $\epsilon>0$, pick $K$ such that
$$ \sum_{|n|\le K} f(n)> \sum_{n\in\mathbb Z} f(n)-\epsilon$$ Due to pointwise convergence, for all sufficiently large $N$ you have $f_N(n) > f(n)-\epsilon/K$, for every $n=-K,\dots,K$. Hence, for such $N$ $$ \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{|n|\le K} f_N(n) >\sum_{|n|\le K} f(n)-3\epsilon > \sum_{n\in\mathbb Z} f(n)-4\epsilon$$ which proves (1).


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