This is from a practice exam for my quals.

Let $(x_n)_{n=0}^\infty$ be an arbitrary sequence in $\mathbb R$ satisfying:

$x_n \ge 0$

$x_n + 2x_{n+1} \le 6$ for all $n \ge 0$

Part $b)$ is the part I need help with.


$a)\ \limsup x_n \le 3$ and $\liminf x_n \le 2$. Show that equality can happen.

Let $\varepsilon \ge 0$ and set $x_{n+1} = 3 + \varepsilon$. Then

$x_n + 2x_{n+1} \le x_n + 6 + 2\varepsilon \le 6$ with $x_n \ge 0$.

So $\varepsilon = 0$ if $\limsup x_n = 3 + \varepsilon$.

For $\liminf$, use superadditivity:

$6 \ge \liminf (x_n + 2x_{n+1}) \ge \liminf x_n + 2 \liminf x_{n+1} = 3 \liminf x_n$

So $3 \liminf x_n \le 6 \implies \liminf x_n \le 2$.

It isn't hard to construct examples to show that equality can happen.

$b)$ Show that we cannot have $\limsup x_n = 3$ and $\liminf x_n = 2$ simutaneously. Describe the region of admissible values of the pair $(\limsup x_n, \liminf x_n)$

This is what I need help with. I'd normally take a subsequence converging to the $\limsup x_n$ and $\liminf x_n$ for exercises like these but that wouldn't work if I want to consider $\limsup x_n$ and $\liminf x_n$ at the same time. I'm not sure how to analyze it.

I think "the region of admissible values" would be a union of two intervals. It will be the bound on $\limsup x_n$ given $\liminf x_n = 2$ union with the bound on $\liminf x_n$ given $\limsup x_n =3$

  • $\begingroup$ @copper.hat sorry it was a typo, it was supposed to be a $2$. Corrected. $\endgroup$ – GFauxPas Jul 1 at 17:23


If $\limsup_nx_n>3$ then there are infinitely many $x_n$'s that are larger that 3, say $\{x_k:k\in\mathbb{N}\}$. Then $$ x_{n_k-1}+2x_{{n_k}}>0+ 2*3=6 $$ a contradiction.

If $\limsup_nx_n>2$, then all but finitely may $x_n$'s are larger than $2$. Then for all sufficiently large $n$ $$ x_n+2x_{n+1}>2+2*2=6 $$ contradiction.


Suppose $3=\limsup_nx_n$. Let $x_{n_k}$ a a subsequence along which that limit is attained. If $\liminf_nx_n=2$, then all but finitely many $x_n$'s are greater that $2-\varepsilon$ for some fixed but small $\varepsilon<\frac13$. Then for all $k$ large enough $$ x_{n_k-1}+2x_{n_k}>2-\varepsilon + 2x_{n_k}>2-3\varepsilon+6=8-3\varepsilon$$ contradiction.

For the last question, consider sequences that alternate between values $x$ and $y$ where

  • $0\leq x\leq 3$
  • $0\leq y\leq 2$
  • $ x+2y\leq 6$
  • $ y + 2x\leq 6$
  • $y\leq x$

This type of sequences will have $y=\liminf_nx_n\leq\limsup_nx_n=x$.

  • $\begingroup$ Thanks but I needed help with part $b)$, sorry that wasn't clear.I amended my question to be explicit that I have the first part done.. $\endgroup$ – GFauxPas Jul 1 at 17:31
  • $\begingroup$ I also fixed a major typo, sorry. $\endgroup$ – GFauxPas Jul 1 at 17:39
  • $\begingroup$ This is very helpful, thank you. Do you know what the question is asking for about admissible values of the pair? I copied the question verbatim. $\endgroup$ – GFauxPas Jul 1 at 19:00
  • $\begingroup$ Why are we considering sequences that alternate between $x$ and $y$? $\endgroup$ – GFauxPas Jul 1 at 20:46
  • 1
    $\begingroup$ because you can generate this way simple sequences for which $\liminf_nx_n=y$ and $\limsup_nx_n=x$ satisfying the conditions of your problem. $\endgroup$ – Oliver Diaz Jul 1 at 22:19

Since $x_n \ge0$ we see that $x_{n+1} \le 3 -{1 \over 2} x_n \le 3$ and so $\limsup_n x_n \le 3$.

Suppose $c=\liminf_n x_n$ and let $c'<c$. Then $x_{n+1} \le 3 - {1 \over 2} c'$ for $n $ sufficiently large. Then $c' < c=\liminf_n x_n \le 3 - {1 \over 2} c'$ and hence $c' <2$. Hence $c \le 2$.

Note that since $x_{n+1} \le 3 -{1 \over 2} x_n$, we must have $\liminf_n x_n \le 3 - { 1\over 2} \limsup_n x_n$. Hence it is not possible to have $\liminf_n x_n =2 $ and $\limsup_n x_n = 3$ with the same sequence.

The limits are separately attained with the sequences $2,2,2,...$ and $0,3,0,3,0,...$.

  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Jul 1 at 23:12
  • $\begingroup$ Maybe because you didn't address the second half of part $b)$? $\endgroup$ – GFauxPas Jul 2 at 16:45
  • $\begingroup$ Maybe, but neither did the other answer. It just stated the result. It is not difficult to guess or obtain from the formula, but presumably the method not the result is of interest. In any event, I think it incumbent on down voters to explain their concerns. $\endgroup$ – copper.hat Jul 2 at 20:15

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