# Interchange of Limsup and sup

Let $\phi_n: U(\subset\mathbb R^2\sim\mathbb C)\to \mathbb R$, be sequence of continuous functions.

Then for any compact set $K\subset U$, what are the necessary and sufficient condition on $\phi_ns$ or on domain of $\phi_ns$ to hold: $$\sup_{z\in K}[\limsup_n\phi_n(z)]= \limsup_n[\sup_{z\in K}\phi_n(z)]$$

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The equality holds if and only if there is a point $y\in K$ such that for all $\epsilon\gt0$ every neighbourhood of $y$ contains a point $x$ such that $\limsup_n\phi_n(x)$ is within $\epsilon$ of $a:=\limsup_n[\sup_{z\in K}\phi_n(z)]$.
For the "only if" direction, assume that the equality holds. Then for every $k\in\mathbb N$ there is a point $x_k\in K$ such that $\limsup_n\phi_n(x_k)$ is within $1/k$ of $a$. Since $K$ is compact, the sequence $x_k$ has an accumulation point $y$, and every neighbourhood of $y\in K$ contains points $x_k$ with $\limsup_n\phi_n(x_k)$ arbitrarily close to $a$.
For the "if" direction, if there is such a point $y$, then there are points $x$ with $\limsup_n\phi_n(x)$ arbitrarily close to $a$, so $\sup_{z\in K}[\limsup_n\phi_n(z)]\ge \limsup_n[\sup_{z\in K}\phi_n(z)]$. Since also $\sup_{z\in K}[\limsup_n\phi_n(z)]\le \limsup_n[\sup_{z\in K}\phi_n(z)]$ (independent of any conditions), it follows that $\sup_{z\in K}[\limsup_n\phi_n(z)]=\limsup_n[\sup_{z\in K}\phi_n(z)]$.