# Question about sequential depiction of upper limit of a function

Let $X$ be a metric space, $D$ be a nonempty subset of $X$ and $x_0$ be a limit point of $D.$ Assume $f: D\to \mathbb{R}.$ We define as usual the upper limit $\limsup_{x\to x_0} f(x)$ of $f$ as $x$ tends to $x_0$ as $$\limsup_{x\to x_0}:=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},$$ where $\hat{\mathbb{B}}(x_0,r)$ denote the punctured ball of $x_0$ with radian $r.$ Write \begin{gather*} L:=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},\\ \begin{aligned} L^*:=\sup\{ L'\mid \exists&\text{ sequence } (x_n) \text{ of points in } D\backslash\{x_0\}, \text{ such that } x_n\to x_0, f(x_n)\to L', \text{ as } n\to \infty\}. \end{aligned} \end{gather*} I want to prove that $L=L^*.$ It is easy for me to prove $L\geq L^*,$ and I have given my proof of this below. But I have trouble in proving the converse part, that is, $L\leq L^*.$

Proof of $L\geq L^*"$ $\quad$ It suffices to prove for the case that $L\in\mathbb{R}.$ Since $L=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},$ by the definition of inf, we have:

$(\alpha)$ $\forall r>0, \sup\{f(x)\mid x\in D\cap \hat{\mathbb{B}}(x_0,r)\}\geq L,$

and

$(\beta)$ $\forall \epsilon>0~\exists r_{\epsilon}>0 : \sup\{f(x)\mid x\in D\cap \hat{\mathbb{B}}(x_0,r_{\epsilon})\}<L+\epsilon.$

By $(\beta),$ we have: $\forall \epsilon>0~\exists r_{\epsilon}>0 : \sup\{f(x)\mid x\in D\cap \hat{B}(x_0,r_{\epsilon})\}<L+\epsilon.$ Assume $(x_n)$ is a sequence of points in $D\backslash\{x_0\}$ with $x_n\to x_0$ such that $f(x_n)\to L'.$ Now that $x_n\to x_0,$ for the $r_{\epsilon},$ there exists $N_1\in\mathbb{N}$ such that for all $n\in\mathbb{N}:$ $n>N_1\implies x_n\in D\cap \hat{\mathbb{B}}(x_0,r_{\epsilon}).$ Thus
$f(x_n)\leq \sup \{f(x)\mid x\in D\cap \hat{B}(x_0,r_{\epsilon})\}<L+\epsilon,$ if $n>N_1.$ Since $f(x_n)\to L',$ letting $n\to \infty,$ we have $L'=\lim_{n\to\infty} f(x_n)\leq L+\epsilon.$ Since our choice of $(x_n)$ is arbitrary, the last inequality implies that $L^*\leq L+\epsilon.$ Because $\epsilon$ is arbitrary, we conclude that $L^*\leq L.$

My question is: How to prove that $L^*\geq L?$

• Sketch of proof: For each positive integer $n$, we can find some $r_n$ with $0 < r_n < 1/n$, such that $$L \leq \sup\{f(x) : x \in D \cap \hat{B}(x, r_n)\} \leq L + 1/n$$ This means that there is some $x_n \in D \cap \hat{B}(x, r_n)$ such that $$L - 1/n \leq x_n \leq L + 1/n$$ and in particular, $x_n \to L$. This shows that $L$ is an element of the set of which $L^*$ is the supremum, and so $L^* \geq L$. (Notice that we need the axiom of countable choice in order to construct the sequence $(x_n)$.) – Bungo Mar 9 '18 at 1:57
• @Bungo: I have actually do this, by showing that $L$ is indeed the supremum of the set of partial limits of $f$ as $x\to x_0.$ My question is: Is it possible to prove $L^*\geq L$ directly, without showing that $L$ is the" least upper bound? – azc Mar 9 '18 at 2:01