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Suppose $(x_n)$ is bounded and $\lim_{n\rightarrow \infty} y_n = 0. $ It is well known that $\lim_{n\rightarrow \infty}(x_ny_n) = 0.$ My question is the following: if for all sequences $(y_n)$ for which $\lim_{n\rightarrow \infty} y_n = 0$ we have $\lim_{n\rightarrow \infty}(x_ny_n) = 0,$ is $(x_n)$ necessarily bounded?

  • Do you really intend to require the statement to hold for all $(y_n)_n$ converging to $0$ simultaneously, or for each such sequence? Perhaps more directly: what order of quantifiers did you intend? – Eric Towers Oct 14 '19 at 02:06
  • Yes, that was quite deliberately for every sequence $(y_n)$ simultaneously. The counterexample you posted would work otherwise. Sorry if this was unclear. –  Oct 14 '19 at 02:10
  • After looking at your other questions, you are too advanced to not be able to do this on your own, or to at least show an attempt. – Carsten S Oct 14 '19 at 16:31

3 Answers3

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Yes.

Suppose that $(x_n)$ is not bounded. We construct a sequence $(y_n)$ that converges to $0$ but for which $(x_ny_n)$ does not converge to $0$.

Since $(x_n)$ is not bounded, given any $N\gt 0$ and $m\gt 0$, there exists $n\gt m$ such that $|x_n|\gt N$ (there must be infinitely many terms with absolute value greater than $N$).

So, first, let $n_1$ be such that $|x_{n_1}|\gt 1$. Let $n_2\gt n_1$ be such that $|x_{n_2}|\gt 2$. Assuming $n_k$ has been defined such that $|x_{n_k}|\gt k$, let $n_{k+1}\gt n_k$ be such that $|x_{n_{k+1}}|\gt k+1$.

Now define $y_n$ to be $0$ if $n\notin \{n_k\mid k=1,2,3\ldots\}$, and $y_{n_k}=\frac{1}{x_{n_k}}$.

Note that since $|x_{n_k}|\to\infty$, then $y_n\to 0$. However, the sequence $(x_ny_n)$ consists of $0$ in all indices different from the $n_k$, and $1$s in the $n_k$ terms; this sequence either does not converge at all, let alone to $0$, or converges to $1$ (as it has a subsequence that is constant $1$).

Arturo Magidin
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Yes. Suppose $(x_n)_n$ is an unbounded sequence. Let $s_n = \max(1,\max_{i \in [1,n]} |x_n|)$. Then $y_n = 1/s_n$ gives a $(y_n)_n$ where $\lim_{n \rightarrow \infty}y_n = 0$, so is in the set of sequences converging to zero. For this $(y_n)_n$, $\limsup_{n \rightarrow \infty} |x_n y_n| = 1$, so $\lim_{n \rightarrow \infty} x_n y_n$, if it exists, is not $0$. By contraposition, we must have that $(x_n)_n$ is a bounded sequence.

Eric Towers
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Here's an alternative to a rephrasing of Arturo's proof.

Suppose $(x_n)$ is not bounded. Then $(x_n)$ contains a monotone subsequence $(x_{n_i})$ diverging to $\pm\infty$. Let $y_n={1\over x_n}$ when $n=n_i$ for some $i$, and $y_n=0$ otherwise. Then $(y_n)$ converges to zero (you can fill in the details), but $(x_ny_n)$ does not converge to $0$ because it has a subsequence $(x_{n_i}y_{n_i})$ where each term equals 1.

Steve Kass
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  • Do we not need to add something saying only finitely many 0's occur in $(x_n)$? Otherwise you can't always define $(y_n)$ or replace terms without affecting the limit. –  Oct 14 '19 at 02:31
  • @KLC: Huh? First, it is not necessarily true that only finitely many terms of $x_n$ are zero; the sequence that is $n$ when $n$ is even and $0$ when $n$ is odd is not bounded. The only detail needed is to request that the subsequence not be $0$. – Arturo Magidin Oct 14 '19 at 02:40
  • You should also require the subsequence to be bounded away form $0$, lest you have a bunch of $0$s at the beginning. – Arturo Magidin Oct 14 '19 at 02:41
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    I would call this a rephrasing of Arturo's proof instead of an alternative to it. But I suppose with just a 15 minute gap in posting time that you didn't want to abandon the work you had already done just because he beat you to it. – Paul Sinclair Oct 14 '19 at 16:57