2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '16 at 1:37
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.