Suppose $a_n$ is bounded and let $c_0$ be the set of all sequences $x_n$ such that $\lim\limits_{n\to\infty} x_n =0$. I need to show $$\inf\limits_{\{x_n\} \in c_0} \sup \{a_n + x_n\} = \limsup\limits_{n\to\infty} a_n.$$

Since $x_n$ converges to $0$, I know that $\limsup\limits_{n\to\infty} a_n = \limsup\limits_{n\to\infty}(a_n + x_n)$. Also, I thought I had one inequality $\inf\limits_{\{x_n\} \in c_0} \sup \{a_n + x_n\} \ge \limsup\limits_{n\to\infty} a_n$ since it first holds for any arbitrary $x_n \in c_0$, but now I'm doubting it because it's the $\inf$. Any help with the whole shebang would be great.


Take $x_i=-a_i$ for $1 \leq i \leq N$ and $x_i=0$ for $i >N$ to see that the left side does not exceed $\sup_{n>N} a_n$. Since $N$ is arbitrary this gives LHS $ \leq $RHS. Now let $x_n \to 0$. Choose $N$ such that $|x_i|<\epsilon$ for $i >N$. Then $sup (x_n+a_n) \geq \sup_{n>N} (a_n-\epsilon) =\sup_{n>N} a_n-\epsilon$. Can you take it from here?

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  • $\begingroup$ I do not understand how LHS $\le$ RHS since you picked a specific $x_n$. $\endgroup$ – user439126 Oct 14 '19 at 3:27
  • $\begingroup$ @user439126 Infimum of any set is less than or equal to any element of the set. Since there is an infimum on LHS we can pick any $\{x_n\}$ with limit $0$ and conclude that LHS $\leq \sup \{a_n+x_n\}$. $\endgroup$ – Kavi Rama Murthy Oct 14 '19 at 4:46
  • $\begingroup$ Great, that helps a ton. For the last part, are you suggesting I show that $\sup\{a_n+x_n\}$ is the greatest lowerbound of $\sup\limits_{n>N}a_n$? $\endgroup$ – user439126 Oct 14 '19 at 18:48
  • $\begingroup$ @ Kavi Rama Murphy I don't quite see where to go next $\endgroup$ – user439126 Oct 15 '19 at 4:37

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