What are the meanings of sup$\{f,g\}$ and max$\{f,g\}$? I am reading Rudin's real and complex analysis. On page 14 we see $\underset{n} {\text{sup }} f_n$, and on page 15 we see max$\{f,g\}$. These two are obviously different. I searched some answers and found an example: 
sup $\{f,g\} = f(x)$ wehn $f(x)\geq g(x)$ and $g(x)$ when $f(x) < g(x)$ 
But then I saw max$\{f,g\}$ which makes me confused. Are they the same? What are the differences?
Can someone also give some explanation of $(\underset{n\to \infty}{\text{lim sup }} f_n)(x)$? How to visualize/understand it?
 A: "max" stands for "maximum" the greatest value in a set.  "sup" stands for "supremum" or "least upper bound" of the values in a set.  When the set is finite, these mean the same thing, and which one you use is a matter of state.  When the set is infinite, it may not have a maximum.  For example, what is $\max \{ x | 0 < x < 1\}?$  Obviously, there is no maximum, but the supremum is $1$.  So for infinite sets, you shouldn't use "max" unless somehow you know that the set has a maximum. For example, we know that a continuous function attains its maximum on a closed interval, so it would be okay to use "max" in that case, though "sup" would be just as good.
A: If $x = \max A$ for some set $A \subset \mathbb R$, then $x \ge y$ for any $y \in A$, and $x \in A$.
But the $\sup A$ does not need to be in $A$. By definition if $x = \sup A$, for some set $A \subset \mathbb R$, then $ x \ge y$ for any $y \in A$, but $x \le y$ if $y$ is an upper bound of $A$.
A: Supremum and Maximum
Let $A$ be a set of real numbers. 
If $x$ is a real number such that $a \leq x$ for all $a \in A$, then $x$ is called an upper bound for $A$. Example: $1$ is an upper bound for the interval $(0,1)$. So is $2$. 
If $x$ is an upper bound for $A$ and $x$ belongs to $A$, then $x$ is called the maximum of $A$ and we write $x = \max A$. In other words, the maximum of $A$ is the element of $A$ that is larger than every other element of $A$. Not every set has a maximum. Examples: The interval $(0,1)$ has no maximum; The finite set $\{1,2,3,4\}$ has $4$ as its maximum. If the maximum of $A$ exists, then it is unique. This justifies the use of "the maximum" rather than "a maximum."
If $x$ is an upper bound for $A$ and $x$ is smaller than every other maximum for $A$, then $x$ is called the supremum of $A$ (or the least upper bound for $A$) and we write $x = \sup A$. In symbols, $\sup A$ is the number such that (1) $a \leq \sup A$ for all $a \in A$ and (2) if $x'$ is a real number such that $a \leq x'$ for all $a \in A$, then $\sup A \leq x'$. 
The supremum of $A$ is unique if it exists. The Dedekind Completeness Property of the Real Numbers is: If $A$ has at least one upper bound, then the supremum of $A$ exists.
If $\max A$ exists, then $\sup A$ exists and $\max A = \sup A$. Example: $A=(0,1]$ has $\max A = \sup A = 1$. 
The supremum may exist even when the maximum does not. Example: $A=(0,1)$ has no maximum, but $\sup A = 1$. 
Sometimes neither the maximum nor the supremum exist. 
Example: The interval $(0,+\infty)$ has no maximum and no supremum. 
If $\sup A$ exists and belongs to $A$, then $\max A = \sup A$. Example: $A=(0,1]$ has $\max A = \sup A = 1$.
To put things another way, a real number is called...
(i) an upper bound for $A$ if it is greater than or equal to every element of $A$. 
(ii) the supremum of $A$ if it is an upper bound for $A$ and it is less than or equal to every upper bound of $A$.
(iii) the maximum if $A$ if it is the supremum of $A$ and it belongs to $A$.
Limit Superior ($\limsup$)
If $(a_n)$ is a sequence of real numbers that is bounded, we can define a new sequence $b_n = \sup \{ a_k : k \geq n\}$. Note that $(b_n)$ is a decreasing sequence. We define 
$$\limsup a_n = \lim_{n \to \infty} b_n = \lim_{n \to \infty} \sup \{ a_k : k \geq n\}.$$
A useful fact to help understand the meaning of $\limsup a_n$ is that
$$
\limsup a_n = \sup S
$$ 
where $S$ is the set of all real numbers $x$ such that $\lim_{k \to \infty} a_{n_k} = x$ for some convergent subsequence $(a_{n_k})$ of the sequence $(a_n)$.  In fact, $\sup S$ belongs to $S$, so 
$$
\limsup a_n = \max S
$$
Thus, you can think of $\limsup a_n$ as the largest limit for all convergent subsequences of $(a_n)$.
Example: The sequence $(1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4},1,\ldots)$ does not converge. It has a subsequence $(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots)$ that converges to $0$. It has another subsequence $(1,1,1,1,\ldots)$ that converges to $1$. It has infinitely many other convergent subsequences, but all of these either converge to $0$ or converge to $1$. Therefore $S = \{0,1\}$. Hence $\limsup a_n = \sup S = \max S = 1$.
For a sequence of real numbers that is bounded, the $\limsup$ always exists as a finite real number, even if the limit of the sequence does not. 
If $(a_n)$ is a sequence of real numbers that is not bounded above, then we define $\limsup a_n = +\infty$.  If $(a_n)$ is a sequence of real numbers such that $\lim_n a_n = -\infty$, then we define $\limsup a_n = -\infty$.
Given a sequence of functions $f_n:E \to \mathbb{R}$, for each fixed $x \in E$, the sequence $(f_n(x))$ is a sequence of real numbers, so we can talk about $\limsup f_n(x)$.  We define the function $\limsup f_n: E \to [-\infty,\infty]$ by setting
$$
(\limsup f_n)(x) = \limsup f_n(x). 
$$
for each fixed $x \in E$. 
Note that since we have not specified that the sequence of functions $f_n$ is bounded, it could be that $\limsup f_n(x) = \pm \infty$. That is why the range of the function $\limsup f_n$ is given as $[-\infty,\infty]$. 
