# Sequence that converges to the limit inferior of a function

Let $f:\mathbb R^n\to \mathbb R\cup\{+\infty,-\infty\}$. The limit inferior of $f$ at $x_0 \in \mathbb R^n$ is defined as $$\liminf_{x\to x_0}f(x) = \lim_{\varepsilon \downarrow 0}\left(\inf\{f(x)\mid x \in (x_0 + \varepsilon B_n)\setminus\{x_0\}\}\right)$$ where $B_n$ is the closed unit ball. I am trying to prove or disprove the following conjecture:

There exists a sequence $\{x_k\}$ of $\mathbb R^n$ such that $x_k \to x_0$ and $f(x_k) \to \liminf_{x\to x_0} f(x)$ as $k \to \infty$.

My first trial was as follows:

Define $U_k = (x_0 + \varepsilon_k B_n)\setminus\{x_0\}$ and $m_k = \inf\{f(x)\mid x \in U_k\}$ with $\varepsilon_1 > \varepsilon_2 > \cdots$ and $\varepsilon_k \to 0$. Then, $m_1 \le m_2 \le \cdots$ and $m_k \to \liminf_{x\to x_0}f(x)$. If we pick a point $x_k$ from a set $V_k = \{x\in U_k\mid m_k \le f(x) \le m_{k+1}\}$, then $x_k \to x_0$ and $f(x_k) \to \liminf_{x\to x_0}f(x)$.

However, if $\liminf_{x\to x_0}f(x) = \inf\{f(x)\mid x \in U_1\}$ and $\liminf_{x\to x_0}f(x) \notin f(U_1)$, then $V_k = \emptyset$ and the proof does not work. Would you give me any hint or reference?

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I have corrected the proof thanks to Martin Argerami's answer.

Define $U(\varepsilon) = (x_0 + \varepsilon B_n)\setminus\{x_0\}$ and $m(\varepsilon) = \inf\{f(x)\mid x \in U(\varepsilon)\}$. Then, $m(\varepsilon) \to m_0 = \liminf_{x \to x_0}f(x)$ as $\varepsilon \to 0$.

1) If there exists $\varepsilon > 0$ such that $m(\varepsilon) = +\infty$, then $f(x) = +\infty$ for every $x \in U(\varepsilon)$, so $f(x_k) \to m_0 = +\infty$ for every sequence $\{x_k\}$ that converges to $x_0$.

Suppose $m(\varepsilon) < +\infty$ for all $\varepsilon >0$.

2) If there exists $\varepsilon_1 >0$ such that $m(\varepsilon) = -\infty$ for every $0 < \varepsilon \le \varepsilon_1$, then let $\varepsilon_1 > \varepsilon_2 > \cdots$ such that $\varepsilon_k \to 0$.

2-1) If $-\infty \in f(U(\varepsilon_k))$, let $x_k \in U(\varepsilon_k)$ such that $f(x_k) = -\infty$.

2-2) If $-\infty \notin f(U(\varepsilon_k))$, let $x_k \in U(\varepsilon_k)$ such that $f(x_k) < -k$; such $x_k$ exists by the definition of the infimum.

Then $x_k \to x_0$ and $f(x_k) \to m_0 = -\infty$ as $k \to \infty$ because $f(x_k) < -k$.

3) If there exists $\varepsilon_1 >0$ such that $-\infty < m(\varepsilon_1) < +\infty$, then $-\infty < m(\varepsilon) < +\infty$ for every $0 <\varepsilon \le \varepsilon_1$. Let $\varepsilon_1 > \varepsilon_2 > \cdots$ such that $\varepsilon_k \to 0$. By the definition of the infimum, there exists $x_k \in U(\varepsilon_k)$ with $|f(x_k) - m(\varepsilon_k)| < 1/k$. Then $x_k \to x_0$ and $$\lim_{k\to\infty}f(x_k) = \lim_{k\to\infty}m(\varepsilon_k) = m_0.$$

• In the $-\infty$ argument, you cannot say that $f(x_k)=-\infty$; you can say that $f(x_k)$ is very negative, but it doesn't have to be infinite. – Martin Argerami Aug 31 '16 at 15:20
• @Martin Argerami Would you explain a little bit more why $f(x_k) = -\infty$ is not possible when $f:\mathbb{R}^n\to\mathbb{R}\cup\{-\infty,+\infty\}$? I guess that if both $+\infty$ and $-\infty$ are in the range of $f$, then $\infty - \infty$ could happen but I am not sure if it is the reason. – flyingwith Aug 31 '16 at 16:12
• It is not that it is not possible; it is not certain. For instance consider $f(t)=1/t$ and $f(0)=0$. With the definition you gave, $\liminf_{t\to0} f(t)=-\infty$, but there is no $t$ with $f(t)=-\infty$. – Martin Argerami Sep 1 '16 at 5:19
• @MartinArgerami It will be handled by the case 2-2). – flyingwith Sep 1 '16 at 7:08

By definition of inf, for each $k$ there exists $x_k\in U_k$ with $|m_k-f (x_k)|<1/k$. Then $$\liminf_{x\to x_0}f (x)=\lim_{k\to\infty}m_k=\lim_{k\to\infty}f (x_k).$$
You need to deal with the $\pm\infty$ case on its own, but the spirit is the same.