Does $x_{n, m_n} \overset{n \to \infty}{\to} x$ hold, if $m_n = o(n)$ and $x_{n,m} \overset{n \to \infty}{\to} x_m \overset{m \to \infty}{\to} x$?

Question

Suppose we have $(x_{n,m}) \subset \mathbb R$ and $(x_m)\subset \mathbb R$ and $x \in \mathbb R$ such that

• For all $ m \in \mathbb N$ we have $x_{n,m} \overset{n \to \infty}{\longrightarrow} x_m \,$; and
• $x_m \overset{m \to \infty}{\longrightarrow} x$.

Let $(m_n)_n \subseteq \mathbb N$ be an non-decreasing sequence of indices such that $m_n \overset{n \to \infty}{\longrightarrow} \infty$ and $m_n = o(n)$.

Does it then hold that $x_{n, m_n} \overset{n \to \infty}{\longrightarrow} x$?

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My intuition (not rigorous):

I think this could be correct, since $m_n = o(n)$ means that $m_n$ is kind of "standing still" compared to $n$, so $\lim_{n \to \infty} x_{n, m_n} = \lim_{m \to \infty} \lim_{n \to \infty} x_{n, m}$.

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Thoughts/Attempt:

Using the triangle inequality, we get $$ |x_{n,m_n} - x| \leq |x_{n, m_n} - x_{m_n}| + |x_{m_n} - x|. $$ We know that $|x_{m_n} - x| \overset{n \to \infty}{\longrightarrow} 0$. Therefore, it would suffice to show $|x_{n, m_n} - x_{m_n}| \overset{n \to \infty}{\longrightarrow} 0$.

I'm hoping that $m_n = o(n)$ will guarantee than this converges to zero. But I don't know if that's the case.

Possible problem: What if $(|x_{n,m} - x_m|)_n$ converges increasingly slowly as $m$ gets larger? Could the $|x_{n,m_n} - x_{m_n}|$ then not converge to zero?

Note: For my purposes at the moment, I just need this for numbers in $[0,1]$. So I could replace $\mathbb R$ above with $[0,1]$. (Maybe this works, but only on compact sets?)

Answer 1

The little-o assumption doesn't change much. The convergence can be even slower than the growth of $m_n$ (whatever $m_n$ is!).

To see this, first consider the sequence $$ y_{n,m} = \begin{cases} 0 & \text{if } n \le m, \\ 1 & \text{if } n > m. \end{cases} $$ Clearly, $y_{n,m} \xrightarrow{n \to \infty} 1$ for each $m$, but the subsequence $y_{n,n}$ is constantly $0$ and doesn't converge to $1$. This corresponds to $m_n = n$, which is excluded in your question.

So let $m_n$ be any non-decreasing sequence such that $m_n \xrightarrow{n \to \infty} \infty$ and $\frac{m_n}{n} \xrightarrow{n \to \infty} 0$ (although the latter won't play any role). We will construct a counterexample $x_{n,m}$ as follows. First, define the sequence $$ t_m := \max \{ n : m_n \le m \}. $$ One should check from the definition that

• for each $m$, $\{ n : m_n \le m \}$ is a bounded subset $\mathbb{N}$ and thus has a maximum,
• $t_{m_n} \ge n$ for each $n$,
• $t_m \xrightarrow{m \to \infty} \infty$.

Now we define: $$ x_{n,m} := y_{n,t_m}. $$ Finally, it's easy to convince yourself that $$ x_{n,m_n} = y_{n,t_{m_n}} = 0 \not\to 1, $$ thanks to $n \le t_{m_n}$.

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The construction is probably clearer if we distinguish between $n$ and $m$. Let's use the notation $$ y^{(m)} = (y_{1,m},y_{2,m},y_{3,m},\ldots), $$ so that we have a sequence $y^{(m)}$ (and each term is a sequence itself). Then the sequence $x^{(m)}$ is chosen from $y^{(m)}$ as a subsequence: $x^{(m)} = y^{(t_m)}$. It's not exactly a subsequence, as $t_m$ doesn't have to be strictly increasing, but in principle $t_m$ grows fast enough to compensate for the slowness of $m$. This last fact is encapsulated in the condition $t_{m_n} \ge n$.

Answer 2

Let $$x_{n,m}:=\left(1-\frac1m\right)^{\log n}.$$ Then $x_{n,m}\xrightarrow[n\to\infty]{}0=:x_m$, and $x_m\xrightarrow[m\to\infty]{}0=:x.$ However, for $m_n:=\lceil\sqrt n\rceil=o(n)$, $$x_{n,m_n}=\exp\left(\log n\cdot\log\left(1-\frac1{\lceil\sqrt n\rceil}\right)\!\right)=\exp\left(-\frac{\log n}{\sqrt n}+o\!\left(\frac{\log n}{\sqrt n}\right)\!\right)\xrightarrow[n\to\infty]{}1.$$