Difference between lim and lim sup of a function with a so called "blow up"?

Question

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a function with a "blow up" in finite time i.e. $$\limsup\limits_{t\uparrow T_{max}}|f(t)|=\infty.$$ I don't unterstand the difference between lim and lim sup in this case. How would the definition look like written in quantifiers?

Answer 1

Here's the general definition:

Let $X$ be a topological space, $E\subset X$, $a\in\overline{E}$ (the topological closure) and $f:E\to [-\infty,\infty]$ a function. We define the limit superior as \begin{align} \limsup\limits_{x\to a}f(x):= \inf\limits_{\text{$a\in U$ open}} \sup\limits_{x\in (E\cap U)\setminus\{a\}}f(x). \end{align}

The thing to notice here is that this ALWAYS exists as an element of $[-\infty,\infty]$. Contrast this with the ordinary limit which need not always exist. So, the point is that the limit superior captures the information about how large a function eventually gets as you get closer to the point $a$. Note that we can define $\liminf$ similarly in such generality (just interchange $\inf$ and $\sup$ above), and then one can show that $\lim$ exists if and only if $\limsup$ and $\liminf$ are equal, in which case $\lim=\liminf=\limsup$.

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Example 1: $X=E=\Bbb{R}$ and $a\in\Bbb{R}$

In this case, the definition above reduces to the following: \begin{align} \limsup_{x\to a}f(x) = \inf_{\text{$a\in U$ open}} \sup_{x\in U\setminus \{a\}}f(x) \end{align} Now, notice that in $\Bbb{R}$, if $U$ is open and contains $a$ then there is an $\epsilon>0$ such that the $\epsilon$-ball is contained in $U$, i.e $B_{a,\epsilon}:= \{x\in\Bbb{R}\,:\, |x-a|<\epsilon\}\subset U$. Conversely, for any $\epsilon>0$, there is an open set $U$ containing $a$ and contained in $B_{a,\epsilon}$ (this is trivial since $B_{a,\epsilon}$ works). Therefore, we can simplify the definition further: \begin{align} \limsup_{x\to a}f(x)&= \inf_{\epsilon>0} \sup_{0<|x-a|<\epsilon}f(x). \end{align} Finally, observe that the function $\epsilon\mapsto \sup\limits_{0<|x-a|<\epsilon}f(x)$ is weakly decreasing, because "taking the supremum over a smaller set gives a smaller value". Hence, we can replace $\inf\limits_{\epsilon>0}$ with $\lim\limits_{\epsilon\to 0^+}$ to get \begin{align} \limsup_{x\to a}f(x)&=\lim_{\epsilon\to 0^+}\sup_{0<|x-a|<\epsilon}f(x). \tag{$*$} \end{align}

As you can see once again from this equality, this just reiterates what I said above: $\limsup$ captures the information about "how large" (this is the supremum part) the function becomes as you get closer and closer to $a$ (this is the $\lim\limits_{\epsilon\to 0^+}$ part). By the way, if you haven't taken some introductory topology, the previous stuff may be too abstract, so you can just take $(*)$ as your starting definition. Once again, the point here is that $\limsup\limits_{x\to a}f(x)$ is a quantity which for ANY function $f$, ALWAYS exists in $[-\infty,\infty]$.

At this point, we can also try to define things like $\limsup\limits_{x\to a, x\in S}$, where $S$ is a certain subset of the domain of your function; this will mean we take the $\limsup$ of the function while restricting $x$ to only come from the set $S$. So, in the case of the real line, we can define $\limsup\limits_{x\uparrow a}$ accordingly: \begin{align} \limsup_{x\uparrow a}f(x)&:= \lim_{\epsilon\to 0^+}\sup_{a-\epsilon<x<a}f(x). \end{align}

With this, we can answer your question of what \begin{align} \limsup\limits_{t\uparrow T}|f(t)|:= \lim_{\epsilon\to 0^+}\sup_{T-\epsilon<t<T}|f(t)|=\infty \end{align} means in terms of quantifiers:

Explicitly, this means for every $R>0$, there is a $\delta>0$ such that for all $\epsilon>0$, if $0<\epsilon<\delta$, then $\sup\limits_{T-\epsilon<t<T}|f(t)|>R$.

As I've mentioned above, the limit exists if and only if $\limsup$ and $\liminf$ are equal, in which case all three are equal. So, the thing is that in your case, $\lim\limits_{t\uparrow T}|f(t)|$ need not exist, though the $\limsup$ always exists. But, in the case that the $\lim$ exists, then it will have to also equal the $\limsup$ which is $\infty$, which means:

For every $R>0$ there is a $\delta>0$ such that for all $t\in\Bbb{R}$, if $T-\delta<t<T$ then $|f(t)|>R$.

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Example 2: $X=[-\infty,\infty]$, $E=\Bbb{R}$, $a=\infty$.

Now we shall unwind our definition to understand what $\limsup\limits_{x\to\infty}f(x)$ means for a function function $f:\Bbb{R}\to [-\infty,\infty]$. First of all, by definition if $U\subset [-\infty,\infty]$ is an open neighborhood of $\infty$, then there is some $R\in\Bbb{R}$ such that $(R,\infty]\subset U$. Conversely, for any $R>0$, $(R,\infty]$ itself is an open neighborhood of $\infty$. Thus, \begin{align} \limsup_{x\to\infty}f(x)&:=\inf_{\text{$U$ open neighborhood of $\infty$}}\sup_{x\in (\Bbb{R}\cap U)\setminus \{\infty\}}f(x)\\ &=\inf_{R\in (-\infty,\infty)}\sup_{x\in (\Bbb{R}\cap (R,\infty])\setminus\{\infty\}}f(x)\\ &=\inf_{R\in\Bbb{R}}\sup_{x\in (R,\infty)}f(x) \end{align} Now, we note that the function $R\mapsto \sup\limits_{x\in (R,\infty)}f(x)$ is weakly decreasing because ‘taking supremum over smaller set gives a smaller value’. Hence, taking the infimum over all $R\in (-\infty,\infty)$ is the same thing as taking the limit $R\to\infty$. Hence, \begin{align} \limsup_{x\to\infty}f(x)=\lim_{R\to\infty}\sup_{x\in (R,\infty)}f(x).\tag{$**$} \end{align} So, if all the above was too much topological background, then you can take $(**)$ as your definition instead.

Answer 2

The sequence may have more limit points. In that case, the limit doesn't exist and limes superior is the supremum of those limit points. For example take sequence:$$ 0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,\ldots $$ This sequence has two limit points: $0$ and $\infty$. So the limit doesn't exist but limes superior does and it's $\infty$.

In your case, you are talking about a limit of a function but I gave you an example of limit of a sequence. Well, you can create a function which is analogous to this sequence, for example $f(x)=\frac{x}2\left(\sin x + 1\right)$. In this case, the limit points are all points in $[0,\infty]$ and limes superior is $\infty$.