# 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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## 1 Answer

I doubt that you'll find a nice answer; the answer can't involve global properties of the functions since the equality depends only on the function values near the suprema and the functions are free to do whatever they want away from the suprema whether the equality holds or not.

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 $\sup_{z\in K}[\limsup_n\phi_n(z)]=\limsup_n[\sup_{z\in K}\phi_n(z)]$.

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