I wanted to prove something here about $\liminf$ and $\limsup$, but it is not coming out, I need some help.

Let $\{x_n\}$ and $\{y_n\}$ two sequences of real numbers such that $0\leq y_n\leq x_n$ for all $n$ and $\lim_{n\to\infty}x_n$ exists. I need to prove that

$$ \lim_ {n \to \infty} x_n - \liminf y_n = \limsup (x_n - y_n). $$

Ok, I did the following:

The left side can be writen as $$ \lim_ {n \to \infty} x_n - \liminf y_n =\lim_ {n \to \infty} x_n - \lim_ {n \to \infty}\inf\{ y_n, y_{n+1},y_{n+2},\ldots\} = $$ $$= \lim_ {n \to \infty}(x_n- \inf\{ y_n, y_{n+1},y_{n+2},\ldots\}) = \lim_ {n \to \infty}\sup \{ x_n-y_n, x_n-y_{n+1},x_n-y_{n+2},\ldots\}$$

while the right side is $$\limsup (x_n - y_n) = \lim_{n\to\infty}\sup\{x_n-y_n, x_{n+1}-y_{n+1}, x_{n+2}-y_{n+2},\ldots\}.$$

I can't see a good way to compare this sets.

Any help is welcome, thanks!

  • $\begingroup$ Let $A_n = \{ x_n - y_k : k \geqslant n\}$ and $B_n = \{ x_k - y_k : k \geqslant n\}$. Let $\varepsilon > 0$. Show that there is an $n(\varepsilon)$ such that $\sup A_n \leqslant \varepsilon + \sup B_n$, and $\sup B_n \leqslant \varepsilon + \sup A_n$ for $n \geqslant n(\varepsilon)$. $\endgroup$ Jan 18, 2015 at 22:19

1 Answer 1


Let's define $A_n = \{x_n - y_n, x_n-y_{n+1}, x_n - y_{n+2}, \ldots\}$, $B_n = \{x_n - y_n, x_{n+1} - y_{n+1}, x_{n+2} - y_{n+2}, \ldots\}$. Then, what we wish to prove is that $\lim\sup A_n = \lim\sup B_n$.

Let $z_n$ be an element of $A_n$ with $z_n \ge \sup A_n - \frac{1}{n}$. We'll show that $\lim_{n \to \infty} z_n \le \lim\sup B_n$. Suppose $z_n = x_n - y_m$. Then, there exists a corresponding element $w_n = x_m - y_m$ in $B_n$. We have that $m \ge n$, so for every $\epsilon > 0$, by making $n$ sufficiently large we can guarantee

$$z_n < w_n + \epsilon \le \sup B_n + \epsilon$$

based on the fact that $x_n$ is Cauchy. Since this is true for all $\epsilon > 0$ by making $n$ large enough, it follows that $\lim\limits_{n \to \infty} z_n \le \lim\limits_{n \to \infty} \sup B_n$. But, $\lim_{n\to\infty}z_n = \lim_{n\to\infty}\sup A_n$, so $$\lim_{n \to \infty} \sup A_n \le \lim_{n \to \infty} \sup B_n.$$

Using the same technique, we can similarly show that $\lim\sup B_n \le \lim\sup A_n$. It thus follows that $\lim\sup A_n = \lim\sup B_n$, which is exactly what you wanted to show.

Here's a proof I think is slightly easier. Let's let $x = \lim_{x \to \infty} x_n$. Then, for any $\epsilon > 0$, for $n$ sufficiently large, we have $$x - \epsilon - y_n < x_n - y_n < x + \epsilon - y_n$$ So, taking the $\lim\sup$, $$x - \epsilon + \lim\sup(-y_n) \le \lim\sup (x_n - y_n) \le x + \epsilon + \lim\sup(-y_n)$$

since $\epsilon > 0$ was arbitrary, we have $$x + \lim\sup (-y_n) \le \lim\sup (x_n - y_n) \le x + \lim\sup(-y_n)$$

So, using the well-known identity $\lim\sup (-y_n) = -\lim\inf y_n$ and observing that $x = \lim_{n\to\infty}x_n$, $$\lim_{n\to\infty}x_n - \lim\inf y_n = \lim\sup(x_n-y_n)$$

  • 1
    $\begingroup$ Your first prove I'm not sure because you don't know if $z_n$ is in the set, so it is not written as $z_n = x_n - y_m$. The second proof I really appreciate. looks fine and clean. $\endgroup$
    – Integral
    Jan 18, 2015 at 23:33
  • $\begingroup$ @Integral Good point about the first proof. Thankfully, it's not too hard to correct the error you found; I've made the necessary changes so $z_n \in A_n$. $\endgroup$
    – user88319
    Jan 19, 2015 at 0:08

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