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Let $U \subset\mathbb{R}^n$ and $u:U\to\mathbb{R}$ be an arbitrary function.

Is $\lim_{r\to 0} \left(\sup_{y\in B_r(x)}u(y)\right) = \limsup\limits_{y\to x}u(y)$ always correct? "$\geq$" should always hold as long as I am not mistaken.

amWhy
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HelloEveryone
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  • What does $\limsup\limits_{y\to x}u(y)$ mean? I have only seen a definition of $\limsup$ when it comes to sequences... – Adam Rubinson May 31 '22 at 13:28
  • This is not the definition of $\limsup_{y\to x} u(y)$. You cannot just take any sequence. The definition is usually exactly what you wrote in your question, the limit as $r\to 0$ of supremums over $B_r(x)$, so the identity is trivial. In particular, the answer below is wrong. – Jeff Jun 07 '22 at 16:21

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I'll assume $\lim_{r\to 0} \left(\sup_{y\in B_r(x)}u(y)\right)=c\in\mathbb{R}. $

Then it is not true that every sequence with $y_n\to x$ also has $u(y_n)\to c.$ And since $\displaystyle\lim_{n\to\infty}u(y_n) \not\to c,$ we also have that $\displaystyle\limsup_{n\to\infty}u(y_n) \not\to c.$

An example of such a function is as follows:

$f:\mathbb{R}\to\mathbb{R};\ f\left( \frac{1}{n} \right) = 3\ \text{ if }\ n\in\mathbb{N},\quad f(x) = 7\ \text{ if }\ x\in\mathbb{R}\setminus\mathbb{N}. $

In this example, we see that $\lim_{r\to 0} \left(\sup_{y\in B_r(0)}f(y)\right)=7,\ $ whereas $\ \limsup\limits_{y\to 0}f(y),$ if it exists, must be equal to both $3$ and $7,$ and therefore this limit doesn't exist.

Adam Rubinson
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