I recently learned of the following extension to the definition of a limit: Let $S \subset \mathbb{R}$, let $f: S \rightarrow \mathbb{R}$, let $a$ be a limit point of $S$, and let $L \in \mathbb{R}$. Then
$$\lim_{x \to a \\ x \in S } f(x) = L$$
means the following: For each $\epsilon > 0$ there exists some $\delta > 0$ such that
$$x \in S \text{ and } |x-a| < \delta \implies |f(x) – L| < \epsilon.$$
(I assume this definition extends similarly for general metric spaces.) I am now trying to understand the analogous extension for limit sup's and lim inf's. To quote Wikipedia:
There is a notion of lim sup and lim inf for functions defined on a metric space whose relationship to limits of real-valued functions mirrors that of the relation between the lim sup, lim inf, and the limit of a real sequence. Take a metric space $X$, a subspace $E$ contained in $X$, and a function $f : E \rightarrow R$. Define, for any limit point a $a$ of $E$,
\begin{align} \limsup_{x \to a} f(x) := \lim_{\epsilon \to 0} \left(\sup \big\{f(x):x \in E \cap B (a;\epsilon) \setminus \{a\} \big\}\right) \hspace{3cm} (1) \end{align}and
$$ \liminf_{x \to a} f(x) := \lim_{\epsilon \to 0} \left(\inf \big\{f(x):x \in E \cap B (a;\epsilon) \setminus \{a\} \big\}\right) \hspace{3.3cm} (2) $$
where $B(a;\epsilon)$ denotes the metric ball of radius $\epsilon$ about $a$.
My first question is: In the context of the setting described above, do $\displaystyle \lim\sup_{x \to a \\ x \in E} f(x) = L$ and $\displaystyle \lim\sup_{x \to a} f(x) = L$ mean the same thing? (And similarly for $\lim\inf$?). My question comes from my reading of Dini numbers in Stein and Shakarchi's Real Analysis. They define
$$D^{+}(F)(x) := \lim \sup_{h \to 0, \\ h > 0} \Delta_h(F)(x), \hspace{3cm} (3) $$
where $\Delta_h(F)(x) := \frac{F(x + h) – F(x)}{h}$ (where $F:[a,b] \rightarrow \mathbb{R}$), and similarly for the other three Dini numbers. I just want to ensure that I fully understand what this means. Does $(3)$ translate to the following?
\begin{align*}
\lim \sup_{h \to 0, \\ h > 0} \Delta_h(F)(x) &:= \lim_{\epsilon \to 0^{+}} \left(\sup \left\{\Delta_h(F)(x): x \in (0,\infty) \cap (-\epsilon, \epsilon) \setminus \{0\} \right\}\right) \\[4pt]
&= \lim_{\epsilon \to 0^{+}} \left(\sup \left\{\Delta_h(F)(x): x \in (0,\epsilon) \right\}\right) \quad ?
\end{align*}
(I take it that $h > 0$ means that $E = (0,\infty)$ in this case?)
Best Answer
For your first question, $\displaystyle\limsup_{x\to a;x\in E}f(x)$ is usually defined as (1) in your post. So $\displaystyle\limsup_{x\to a}f(x)$ and $\displaystyle\limsup_{x\to a;x\in E}f(x)$ are trivially equivalent in your context.
Note that if $a$ is a limit point of $S$, then for small enough $\epsilon>0$, you have[Incorrected remark deleted.]For your second question, both of your interpretations are incorrect. Be careful that $x$ is a fixed number(!) in the definition of $D^+(F)(x)$.
What you should have is: $$ D^+(F)(x)=\lim_{\epsilon\to 0+}\big(\sup\{\Delta_h(F)(x): h\in(0,\epsilon)\}\big) $$ which is the same as $$ D^+(F)(x)=\lim_{\epsilon\to 0}\big(\sup\{\Delta_h(F)(x): h\in(0,+\infty)\cap(-\epsilon,\epsilon)\}\big) $$