Is $\limsup_{z\to z_0}f(z)=\limsup_{k\to\infty}f(z_k)$?

Question

Let $\Omega\in\Bbb C$ open, $f:\Omega\to\Bbb R$ a generic function. Let $(z_k)_k\subset\Omega$ s.t. $\lim_{k}z_k=:z_0\in\Omega$.

The question is the following: is true that $$ \limsup_{z\to z_0}f(z)=\limsup_{k\to\infty}f(z_k)\;\;\;\;\;? $$

I recall that $\limsup_{z\to z_0}f(z):=\lim_{r\to0^+}\left(\sup_{B(z_0,r[}f(z)\right)$ and $\limsup_{k\to\infty}f(z_k):=\lim_{k\to\infty}\left(\sup_{n\ge k}f(z_n)\right)$.

My feeling is the answer is yes, but I wasn't able to prove this. Can someone help me? Many thanks!

Answer 1

You have $\limsup_{z\to z_0} f(z) \ge \limsup_{k\to\infty} f(z_k)$. You do not always get equality. If you always do, then the limit actually exists.

Answer 2

Let $f(x) = \sin(\pi/x)$.

Then $\limsup_{x\to 0} f(x) = 1$, since we can find a sequence converging to $0$ for which $f(x_n) \to 1$. In particular if $x_n = (2n + 1/2)^{-1}$ we have $f(x_n) = \sin(\pi/2 + 2\pi n)= 1$ for all $n$.

However, when considering the sequence $x_n = 1/n$ we have $\limsup_{n\to\infty} f(x_n) = 0$, since $f(x_n)= \sin(\pi n) = 0$ for all $n$.

Thus we see that $$\limsup_{x\to 0} f(x) \neq \limsup_{n\to \infty} f(x_n)$$ for some sequences for which $x_n \to 0$.