Why is the absolute sign needed in the definition of a bounded function A function $f$ is bounded if there exists a real number $M$ such that $|f(x)| \le M$ for all $x \in \operatorname{dom}(f)$. Why is the absolute sign needed?
 A: If $M\geq0$ then $|y|\leq M$ is equivalent to $-M\leq y\leq M$, so the definition is just saying there is $M\geq0$ so that $-M\leq f(x)\leq M$ for all $x$. This is just a standard economy of notation; it might just as well have said there exist $M_1,M_2$ such that $M_1\leq f(x)\leq M_2$ for all $x$. On the other hand, if the values of $f$ are complex numbers, then saying $f(x)\leq M$ would not even make sense, and this this becomes more that just a shorthand (even though $|f(x)|\leq M$ is still equivalent to saying that the real and imaginary parts of $f(x)$ are bounded).
A: A more general definition of bounded is that a (real) function $f(x)$ is bounded if there exist real numbers $m, M$ such that $m \le f(x) \le M $ for all $x$. Now there is no modulus involved and it shows that the upper and lower bounds need not be negations of each other.
It follows that if $f$ is bounded by this definition and $L = max( |m|, |M|)$ then $|f(x)| \le L$, and conversely that if $|f(x)| \le L$ then $-L \le f(x) \le L$.
A: Case 1:
$f$ is a $\mathbb{R} \rightarrow \mathbb{R}$ function:
The definition can have multiple equivalent forms:
$\exists\ m, M \in \mathbb{R} : \forall\ x \in dom(f) : m \leq f(x) \leq M$
$\exists\ M \in \mathbb{R} : \forall\ x \in dom(f) : - M \leq f(x) \leq M$
$\exists\ M \in \mathbb{R} : \forall\ x \in dom(f) : | f(x) | \leq M$
When teaching calculus it is usually done this way:
$f$ is upper bounded $\leftrightarrow \exists\ M \in \mathbb{R} : \forall\ x \in dom(f) : f(x) \leq M$
$f$ is lower bounded $\leftrightarrow \exists\ m \in \mathbb{R} : \forall\ x \in dom(f) : m \leq f(x)$
$f$ is (generally) bounded $\leftrightarrow$ $f$ is upper and lower bounded
Case 2:
$f$ is a $\mathbb{C} \rightarrow \mathbb{C}$ function:
Since complex numbers can not be compared the only option from the three definitions abowe is the third. That is also correct for real and complex and even more abstract functions if the $|.|$ is correctly defined. For example any euclidean space. So it also works for $\mathbb{R}^n \rightarrow \mathbb{R}^m$ functions.
Note that $|.|$ can be defined in multiple ways. For example:
$\sqrt{(Im z) ^ 2 + (Re z) ^ 2}$
$|Im z| + |Re z|$
A: Because if it's just "$f(x)\leq M$" then $f$ could "blow down" to $-\infty$.
A: I think the definition should be
\begin{equation}
|f(x)| \leq M.
\end{equation}
We need the absolute value so that $f$ is also bounded from below. For example, if we only had 
\begin{equation}
f(x) \leq M,
\end{equation}
$f$ can be unbounded below. That is, we can have $f$ approaching $-\infty$. 
A: Here is why: 
Suppose we let $f(x) = \begin{cases} x & x < 0 \\ 0 & x \geq 0 \end{cases}$.
Should such a function be considered "bounded"?  It is certainly bounded from above (by $1$ for example), so $f(x) \leq 1$ is true for all $x$...but it isn't bounded from below.  $\lim \limits_{x \to - \infty} f(x) = - \infty$, which means our function is not bounded from below.
This is remedied by saying $|f(x)| \leq M$, because then by definition, $|f(x)| \leq M$ implies $-M \leq f(x) \leq M$, so that $f$ is bounded from above and below.
A: An equivalent definition of bounded would be "$f$ is bounded if there exists an $M \ge 0$ such that $-M \le f(x) \le M$". But this requires $f$ to be a real function. On the other hand $|f(x)|$ is real by definition of the absolute value function $|.|$
A: One interpretation of bounded function is that a function $f:X \to \mathbb R$ is bounded when its image $f(X)$ is bounded. Now, $Y\subseteq\mathbb R$ is bounded exactly when it is contained in an interval. For simplicity, we can take this interval as being symmetric around the origin: $Y\subseteq [-M,M]$. Applying this to $Y=f(X)$, we get $|f(x)|\le M$, for all $x\in X$.
Another interpretation is that a function $f:X \to \mathbb R$ is bounded  when the magnitude of its values is bounded. Magnitude in $\mathbb R$ is measured by the absolute value.
