I am currently looking at a result that states, for some nonnegative function $g: \mathbb{R} \times \mathbb{R} \to [0,1]$, there exists a constant $c> 0$ such that \begin{align} \lim_{k \to \infty} \limsup_{n \to \infty} |g(n,k) - c | = 0. ~(*) \end{align} It seems to me that the $\limsup$ is unnecessary, and could have been replaced by a $\lim$. This is because I believe that the above equation states that $\forall \epsilon >0, \exists K_{\epsilon}$ such that $\forall k > K_{\epsilon}, \exists N_{k, \epsilon}$ such that \begin{align} \sup_{n > N_{k,\epsilon}} |g(n,k) - c| < \epsilon. \end{align} (I found a nice explanation of the meaning of a double limit in quantifiers here Doubt about double limit definition. ).
Doesn't the $\sup$ imply that $|g(n,k) - c| < \epsilon|$ for all $n > N_{k,\epsilon}$, and vice versa? Also, is there any difference between $(*)$ and the following: \begin{align} \lim_{k \to\infty} \liminf_{n \to \infty} \frac{g(n,k)}{c} = \lim_{k \to\infty} \limsup_{n \to \infty} \frac{g(n,k)}{c} = 1 \end{align} ? Thanks so much.
Consider $g(n,k)=\frac12+\frac1{k^2+5} \cos(\pi n)$ and $c=\frac12$. It satisfies the condition, but $\lim_{n\to\infty}\lvert g(n,k)-c\rvert$ does not exist for any $k$.