If $(s(n,m))$ is a double sequence such that

(i) the iterated limit $\lim_{m \to \infty} (\lim_{n \to \infty }s(n,m))= a$, and

(ii) the limit $\lim _{n \to \infty}s(n,m)$ exists uniformly for every $m \in \mathbb N$,

then the double limit $\lim_{n,m \to \infty}s(n,m)=a$.

Can anyone please tell me how to prove this theorem?

  • $\begingroup$ What is your problem? Why are you untagging again and again?@YuiTo Cheng $\endgroup$
    – cmi
    May 4, 2019 at 4:33
  • $\begingroup$ Please DO NOT DO THIS... This is my question ... I can create a tag.. I can do whatever I want to ....Please stay away from my account... If you do this one more time , I will take step to shut down your account..@YuiTo Cheng $\endgroup$
    – cmi
    May 4, 2019 at 4:44
  • 1
    $\begingroup$ I don’t know what is going on. However, if you have a grievance then it is best to flag for the attention of a moderator. Best not to make any threats. Conflicts here can be resolved. $\endgroup$
    – RRL
    May 4, 2019 at 6:23
  • $\begingroup$ But where to flag? Should I flag my own comment ? @RRL $\endgroup$
    – cmi
    May 4, 2019 at 6:55

1 Answer 1


Your hypothesis (i) -- that $\lim_{m \to \infty} (\lim_{n \to \infty }s(n,m))= a$ -- requires that the inner limit must exist for every $n$.

Even if we only assume that the single limit $\lim_{n \to \infty}s(n,m) = a_m$ exists for every $m$, and the limit $\lim_{m \to \infty} s(n,m) = b_n$ exists uniformly for every $n$. It then follows that

$$\lim_{n,m \to \infty} s(n,m) = \lim_{m \to \infty} a_m = \lim_{n \to \infty} b_n$$

By the uniform convergence of $s(n,m) \to b_n$, for any $\epsilon > 0$ there exists a positive integer $Q(\epsilon)$ such that if $m \geqslant Q(\epsilon)$, then for all n,

$$\tag{*}|s(n,m) - b_n| < \epsilon $$

Taking any fixed $q \geqslant Q(\epsilon)$, we have

$$|b_j - b_k|\leqslant |b_j - s(q,j)| + |s(q,j) - s(q,k)| + |s(q,k) - b_k|$$

By (*) it follows that $|b_j - s(q,j)| < \epsilon$ and $|s(q,k) - b_k| < \epsilon$ for any values of $j$ and $k$. Since $\lim_{m \to \infty} s(q,m) = a_q$ exists, it follows that with $q$ fixed $s(q,m)$ is a Cauchy sequence and there exists $M(\epsilon,q)$ such that if $j,k \geqslant M(\epsilon,q)$, then $|s(q,j) - s(q,k)|< \epsilon$.

Altogether, we have for all $j,k \geqslant M(\epsilon,q)$, $|b_j - b_k| < 3\epsilon$. Therefore, the sequence $(b_n)$ is a Cauchy sequence and there exists a number $b$ such that

$$\lim_{n \to \infty} \lim_{m \to \infty} s(n,m) = \lim_{n \to \infty}b_n = b$$

Again, for any $\epsilon > 0$, there exists a positive integer $N(\epsilon)$ such that if $n \geqslant N(\epsilon)$, then $|b_n -b | < \epsilon$.

Thus, if $n,m \geqslant \max(N(\epsilon),Q(\epsilon))$, then

$$|s(n,m) - a| \leqslant |s(n,m) - b_n| + |b_n - b| < 2\epsilon,$$

which proves that

$$\lim_{n,m \to \infty} s(n,m) = \lim_{n \to \infty} b_n = b$$

Given hypothesis (i) we must have $b = a$.

Finally, we can prove the existence of the other iterated limit and that

$$\tag{**}a = \lim_{m \to \infty} (\lim_{n \to \infty }s(n,m))= \lim_{n,m \to \infty} s(n,m) = b$$

For every $\epsilon > 0$, we have for all sufficiently large $n,m $,

$$|s(n,m) - b| < \epsilon$$

Thus, for sufficiently large $m$,

$$| \lim_{n \to \infty}s(n,m) - b | = \lim_{n \to \infty}|s(n,m) - b| \leqslant \epsilon,$$

which proves (**).

  • 1
    $\begingroup$ @cmi: Is this clear now? $\endgroup$
    – RRL
    May 4, 2019 at 17:39
  • $\begingroup$ You seem to have proven different thing...@RRL $\endgroup$
    – cmi
    May 7, 2019 at 7:31
  • $\begingroup$ It seems what you have proved is not the more general case of my statement..@RRL $\endgroup$
    – cmi
    May 7, 2019 at 7:45
  • 1
    $\begingroup$ It is not really less general. You assume that $\lim_{m \to \infty} (\lim_{n \to \infty }s(n,m))= a$. For this to be true we must have the inner limit defined, i.e. $(\lim_{n \to \infty }s(n,m)) = a_m$ for all $m$. This part along with (ii) is enough to prove that the double limit and all the iterated limits exist and are equal. So by calling the iterated limit in (I) "a" we must have all the limits equal to "a". I'll edit above to make this more clear. $\endgroup$
    – RRL
    May 7, 2019 at 17:06
  • $\begingroup$ @RRL If inner limit is uniform and outer limit exist then can we say double limit exist? $\endgroup$
    – Meet Patel
    Dec 5, 2023 at 14:02

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