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We have the following proposition in Real Analysis

Let $X\subseteq\mathbb{R}$, let $f: X\rightarrow\mathbb{R}$ be a function, let $E\subseteq X$, let $x_{0}\in\overline{E}$ and let $L\in\mathbb{R}$. Then the following are equivalent: $(a)$ $\lim_{x\rightarrow x_{0};x\in E} f(x) = L$. $(b)$ For every sequence $(a_{n})_{n = 0}^{\infty}$ of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$ we have $\lim_{n\rightarrow\infty}f(a_{n}) = L$.

The limit superior of a real-valued function is defined as $\limsup_{x\rightarrow x_{0};x\in E}f(x) = \inf_{\varepsilon >0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$. Is the following generalization of the above proposition true?

Let $X\subseteq\mathbb{R}$, let $f: X\rightarrow\mathbb{R}$ be a function, let $E\subseteq X$, let $x_{0}\in\overline{E}$ and let $L\in\mathbb{R}$. Then the following are equivalent: $(a)$ $\limsup_{x\rightarrow x_{0};x\in E} f(x) = L$. $(b)$ For every sequence $(a_{n})_{n = 0}^{\infty}$ of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$ we have $\limsup_{n\rightarrow\infty}f(a_{n}) = L$.

Thoughts: This is what I have got so far: Suppose $(a)$ holds and let $(a_{n})_{n = 0}^{\infty}$ be a sequence of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$. Then for every $\varepsilon > 0$ there exists $N_{\varepsilon}\in\mathbb{N}$ such that for all $n\geq N_{\varepsilon}$ we have $a_{n}\in\{x\in E: |x-x_{0}| < \varepsilon\}$. Therefore, for all $n\geq N_{\varepsilon}$ we have $f(a_{n})\leq \sup\{f(x)\in E: |x-x_{0}| < \varepsilon\}$. $$\implies \limsup_{n\rightarrow\infty}f(a_{n})\leq \sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$$ Since this inequality is true for every $\varepsilon > 0$ we get $$\limsup_{n\rightarrow\infty}f(a_{n})\leq \inf_{\varepsilon > 0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x) = \limsup_{x\rightarrow x_{0};x\in E}f(x) = L$$ If the proposition is true, how would one go about showing the reverse inequality and also the converse (i.e. $(b)\implies (a)$)?

Karthik Kannan
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    No, for every sequence you have $\limsup_n f(a_n) \le L$ and there exists a sequence such that $L$ is attained. Take $f=1_\mathbb{Q}$ $x_0 = 0$ then there are lots of sequences $a_n \to 0$ such that $\limsup_n f(a_n) = 0$. – copper.hat Dec 04 '19 at 05:28
  • In that case, how would one frame the $(b)\implies (a)$ statement since $L$ could be an arbitrary upper bound? Or is $L$ not arbitrary since there exist sequences such that $L$ is attained? – Karthik Kannan Dec 04 '19 at 05:37
  • It might help to think of $\limsup$ in the following way, which can be proved equivalent to whatever definition you're using for $\limsup.$ For each sequence that approaches $x_0,$ the corresponding sequence of $f$-images of the terms of the sequence could converge (finitely or infinitely) or it could fail to converge. Collect together into one set all possible extended-real-number limits you get from these various corresponding sequences of $f$-images when they converge (finitely or infinitely), and denote this by $C(f,x_0)$ (usually called (continued) – Dave L. Renfro Dec 04 '19 at 10:31
  • the cluster set of $f$ at $x_0$). Then $C(f,x_0)$ is always a nonempty compact set in the extended real line, $C(f,x_0)$ is a singleton set if and only if $\lim \limits_{x \rightarrow \infty} f(x)$ exists (finitely or infinitely), the least element of $C(f,x_0)$ is $\liminf \limits_{x \rightarrow \infty} f(x),$ and the greatest element of $C(f,x_0)$ is $\limsup \limits_{x \rightarrow \infty} f(x).$ (continued) – Dave L. Renfro Dec 04 '19 at 10:32
  • If desired, this can be rephrased so as to regard "limit exists" as excluding infinite limits, but the wording of these four properties will be a little more clunky. One can also define the notion of a cluster set of $f$ at $x_0$ relative to a given sequence converging to $x_0$ (the set of subsequence limits of the sequence), and this is always a subset (not necessarily proper) of $C(f,x_0)$ and the obvious analogs of the four properties I listed earlier for $C(f,x_0)$ hold for the set of subsequence limits. – Dave L. Renfro Dec 04 '19 at 10:32
  • Thank you for the useful comments and for pointing out the relevant literature. – Karthik Kannan Dec 04 '19 at 16:42
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    Regarding literature, see my answer to the Mathematics StackExchange question Name for multi-valued analogue of a limit and my answer to the Mathoverflow question A search for theorems which appear to have very few, if any hypotheses. – Dave L. Renfro Dec 04 '19 at 17:08

1 Answers1

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Here is a partial answer based on the comment by @copper.hat. One can show that if $\limsup_{x\rightarrow x_{0};x\in E}f(x) = L$ then for every sequence of elements $(a_{n})_{n = 0}^{\infty}$ in $E$ that converges to $x_{0}$ we have $\limsup_{n\rightarrow\infty}f(a_{n})\leq L$ and there exists a sequence such that $\lim_{n\rightarrow\infty}f(a_{n}) = L$.

$L = \inf_{\varepsilon > 0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$. Suppose there exists $\varepsilon > 0$ such that $\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x) = L$. Then for every $\varepsilon'\leq\varepsilon$ we have $\sup_{|x-x_{0}| < \varepsilon'; x\in E}f(x) = L$. Then one may construct a sequence $(\varepsilon_{n})_{n = 1}^{\infty}$ going to $0$ such that $\varepsilon_{n}\leq \varepsilon$, and a sequence $(x_{n})_{n = 1}^{\infty}$ such that $x_{n}\in\{x\in E:|x -x_{0}| < \varepsilon_{n}\}$ and $L-1/n< f(x_{n})\leq L$ (by the axiom of choice) for each $n\in\mathbb{N}\setminus\{0\}$. Then $\lim_{n\rightarrow\infty}x_{n} = x_{0}$ and $\lim_{n\rightarrow\infty}f(x_{n}) = L$.

On the other hand, if there is no $\varepsilon > 0$ such that the infimum is attained then for each $n\in\mathbb{N}\setminus\{0\}$ we may pick $\varepsilon_{n}> 0$ (by the axiom of choice) such that $L < \sup_{|x-x_{0}| < \varepsilon_{n}; x\in E}f(x) < L+1/n$. We claim that $\lim_{n\rightarrow\infty}\varepsilon_{n} = 0$. Suppose this is false for the sake of contradiction, then there exists $\varepsilon > 0$ such that for every $N\in\mathbb{N}\setminus\{0\}$ there exists $n\geq N$ such that $\varepsilon_{n} > \varepsilon$. For such $n$ we would then have $L+1/n>\sup_{|x-x_{0}|<\varepsilon; x\in E}f(x)$. However, by the Archimedian property there exists $M\in\mathbb{N}\setminus\{0\}$ such that $M\big(\sup_{|x-x_{0}|<\varepsilon; x\in E}f(x)-L\big) > 1$ which contradicts the previous statement. Then one may pick $(x_{n})_{n = 0}^{\infty}$ as before such that $L < f(x_{n})<L+1/n$ which is the desired sequence.

Karthik Kannan
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