$\limsup$ and $\liminf$ of family of sequences with given properties

limsup-and-liminfreal-analysissequences-and-series

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.

Show:

$a)\ \limsup x_n \le 3$ 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 $\limsup x_n =3$

Best Answer

Part(a)

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.

Part(b)

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 $y=\liminf_nx_n\leq\limsup_nx_n=x$.

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