Let $\{x_n\}$ be convergent to $L > 0$ and $\{y_n\}$ be bounded. Show that $$\limsup_{n \to \infty} (x_ny_n) = (\lim_{n \to \infty} x_n)(\limsup_{n \to \infty} y_n).$$

For my first attempt at this problem, I falsely claimed $\sup(x_ny_n) = \sup(x_n)\sup(y_n)$, which is clearly not true, but from which the statement follows easily.

Next, working from the right hand side, I tried $$(\lim_{n \to \infty} x_n)(\limsup_{n \to \infty} y_n) = L(\limsup_{n \to \infty} y_n) = L(\lim_{m \to \infty} \sup(y_n)_{n \geq m}) = (\lim_{m \to \infty} L \sup(y_n)_{n \geq m}).$$

From there, can we say $$(\lim_{m \to \infty} L \sup(y_n)_{n \geq m}) = (\lim_{m \to \infty} \sup(Ly_n)_{n \geq m})?$$

If so, since for all $\epsilon > 0$ there exists some $m$ such $|x_n - L| < \epsilon$ when $n \geq m$ we get $$\limsup_{n \to \infty}(x_n - \epsilon)(y_n) = (\lim_{m \to \infty} \sup((x_n-\epsilon)y_n)_{n \geq m}) \leq (\lim_{m \to \infty} \sup(Ly_n)_{n \geq m}) \leq (\lim_{m \to \infty} \sup((x_n+\epsilon)y_n)_{n \geq m}) = \limsup_{n \to \infty}(x_n + \epsilon)(y_n).$$

So... since $\epsilon$ was arbitrary...? Am I doing this right?

  • $\begingroup$ It seems that at the end you tend to vary $\epsilon>0$ inside the $\limsup$, which I think your lecturer will not be very happy with it. Strictly speaking, that is prohibited but hardly to say wrong at all. $\endgroup$
    – user284331
    Nov 30, 2016 at 1:37

1 Answer 1


For arbitrary $\epsilon>0$ such that $\epsilon<L$, find some $N\in{\bf{N}}$ such that $|x_{k}-L|<\epsilon$ for all $k\geq N$. Then $\limsup_{n}x_{n}y_{n}=\inf_{n\geq 1}\sup_{k\geq n}x_{k}y_{k}=\inf_{n\geq N}\sup_{k\geq n}x_{k}y_{k}\leq\sup_{k\geq n}x_{k}y_{k}\leq\sup_{k\geq n}(L+\epsilon)y_{k}=(L+\epsilon)\cdot\sup_{k\geq n}y_{k}$, so $\limsup_{n}x_{n}y_{n}\leq(L+\epsilon)\cdot\limsup_{n}y_{n}$.

On the other hand, for all $n\geq N$, $\sup_{k\geq n}x_{k}y_{k}\geq\sup_{k\geq n}(L-\epsilon)y_{k}=(L-\epsilon)\sup_{k\geq n}y_{k}\geq(L-\epsilon)\inf_{n\geq N}\sup_{k\geq n}y_{k}=(L-\epsilon)\cdot\limsup_{n}y_{n}$ so $(L-\epsilon)\limsup_{n}y_{n}\leq\limsup_{n}x_{n}y_{n}\leq(L+\epsilon)\limsup_{n}y_{n}$. If $\limsup_{n}y_{n}=\infty$ or $-\infty$, the equality holds, for otherwise, varying $\epsilon>0$ gives the result.


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