limit superior of real value function (Exercise 9.3.4 in Tao Analysis I)

Question

Here is the exercise:

Propose a definition for limit superior $\lim\sup_{x \to x_0 ; x \in E} f(x)$ and limit inferior $\lim\inf_{x \to x_0 ; x \in E} f(x)$, and then propose an analogue of Proposition 9.3.9 for your definition. (For an additional challenge: prove that analogue.)

Here is Proposition 9.3.9:

Let $X$ be a subset of $\mathbf{R}$, let $f : X \to \mathbf{R}$ be a function, let $E$ be a subset of $X$, let $x_0$ be an adherent point of $E$, and let $L$ be a real number. Then the following two statements are logically equivalent:
(a). $f$ converges to $L$ at $x_0$ in $E$.

(b). For every sequence $(a_n)_{n = 0}^\infty$ which consists entirely of elements of $E$ and converges to $x_0$, the sequence $(f(a_n))_{n = 0}^\infty$ converges to $L$.

Here is my attemped:

Let $X$ be a subset of $\mathbf{R}$, let $f : X \to \mathbf{R}$ be a function, let $E$ be a subset of $X$, and let $x_0$ be an adherent point of $E$. We define limit superior at $x_0$ in $E$ as $$ \limsup_{x \to x_0 ; x \in E} f(x) = \inf\Big\{\sup\big\{f(x) : x \in E \land |x - x_0| < \delta\big\} : \delta \in \mathbf{R}^+\Big\} $$ Let $L \in \mathbf{R} \cup \{-\infty, +\infty\}$. We claim that the following statements are equivalent:

(a). $\limsup_{x \to x_0 ; x \in E} f(x) = L$

(b). For every sequence $(a_n)_{n = 1}^\infty$ which consists entirely of elements of $E$ and converges to $x_0$, the sequence $(f(a_n))_{n = 1}^\infty$ has limit superior $\limsup_{n \to \infty} f(a_n) \leq L$ There exists a sequence $(b_n)_{n = 1}^\infty$ which consists entirely of elements of $E$ and converges to $x_0$, and $\limsup_{n \to \infty} f(b_n) = L$

I managed to prove (a) implies (b). But I don't know how to prove (b) implies (a). (Note that $L$ can be $-\infty$ or $+\infty$) Any helps are appreciated.

Answer 1

\begin{equation} \limsup _{x\rightarrow x_{0} ;x\in E} f(x)=\inf\Bigl\{\sup \bigl\{f(x):x\in E\land |x-x_{0} |< \delta \bigr\} :\delta \in \mathbf{R}^{+}\Bigr\} \tag{A} \end{equation}

We'll prove that the following are equivalent (for $L\in \mathbb R$):

1. $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =L$

2. For every $\displaystyle \epsilon >0$ and every sequence $\displaystyle ( x_{n}) \subset E-\{x_{0} \}$ converging to $\displaystyle x_{0}$, there exists $\displaystyle N\in \mathbb{N}$ such that $\displaystyle n\geq N\Longrightarrow $$\displaystyle f( x_{n}) < L+\epsilon $ and $\displaystyle f( x_{m}) >L -\epsilon $ for infinite $\displaystyle m$.

First let's consider the case when $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x)$$\displaystyle \in \mathbb{R}$. .

$\displaystyle ( 1) \Longrightarrow ( 2) :$Fix an $\displaystyle \epsilon >0$ and let $\displaystyle ( x_{n}) \subset E-\{x_{0} \}$ be any sequence converging to $\displaystyle x_{0}$.

By $\displaystyle (A) :\ \exists \ \delta >0$ such that $\displaystyle \sup \{f( x) :\ x\in E\land |x-x_{0} |< \delta \}< L+\epsilon $

It follows that $\displaystyle x\in E\land |x-x_{0} |< \delta \Longrightarrow f( x) \leq \sup \{f( x) :\ x\in E\land |x-x_{0} |< \delta \}< L+\epsilon $

Now $\displaystyle \left( x_{n}\rightarrow x_{0}\right) \ \Longrightarrow \exists N\ ( n\geq N\Longrightarrow |x_{n} -x_{0} |< \delta $)

It follows that $\displaystyle f( x_{n}) < L+\epsilon $

I leave the other part to you.

For the converse,

$\displaystyle ( 2) \Longrightarrow ( 1) :$As per our condition, $\displaystyle \limsup $ exists. Suppose it is equal to $\displaystyle r\neq L$.

If $\displaystyle r >L$, then we must have by $\displaystyle ( 1) \Longrightarrow ( 2)$ that $\displaystyle f( x_{n}) >r-( r-L) =L$ for infinitely many $\displaystyle n$, which contradicts $\displaystyle ( 2)$.

Similarly if $\displaystyle r< L$, then $\displaystyle f( x_{n}) >L-( L-r) =r$ for infinitely many $\displaystyle n$ but this contradicts the assumption that $\displaystyle r$ is a limsup.

So $\displaystyle r=L$ and we are done!

What if $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =+\infty $?

In that event, $\displaystyle f$ is not bounded from above in any neighborhood of $\displaystyle x_{0}$. So for every sequence $\displaystyle x_{n}\rightarrow x_{0} ,\ x_{n} \neq x_{0}$ for any $\displaystyle n$, we can find a subsequence $\displaystyle x_{n_{k}}$ such that $\displaystyle f( x_{n_{k}})\rightarrow +\infty $

Similarly, you may handle the case when $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =-\infty $