Math notation for location of the maximum My question is about notation. I have maximum of the function $f(x)$. This can be expressed as $\max(f)$ How can I express in compact form that $x_0$ is the location of that maximum.
 A: You can use
$$
x_0  = \mathop {\arg \max }\limits_x f(x).
$$
EDIT 1: You can (and preferably) use the above notation only if the maximum is reached at a single value. I'll elaborate later on.
EDIT 2: From Wikipedia,
$$
\mathop {\arg \max }\limits_x f(x) = \{ x|\forall y:f(y) \le f(x)\} ,
$$
that is, $\mathop {\arg \max }\limits_x f(x)$ is the set of values of $x$ for which $f$ attains its maximum*.
However, if the maximum is reached at a single value, then we define the arg max as a point (rather than a singleton set), so that
$$
x_0  = \mathop {\arg \max }\limits_x f(x) \Leftrightarrow f(x_0 ) = \mathop {\max }\limits_{x \in D} f(x),
$$
where $D$ is the domain of $f$. Thus, for example, if $f$ is defined by 
$$
f(x)=\cos (x),\;\; x \in [0,2 \pi],
$$
then
$$
\mathop {\arg \max }\limits_x f(x) = \{ 0,2\pi \} ,
$$
whereas if $f$ is defined by
$$
f(x)=\sin (x),\;\; x \in [0,2 \pi],
$$
we can (and preferably) write
$$
\mathop {\arg \max }\limits_x f(x) = \frac{\pi }{2} .
$$
*However, note that a function $f$ might not attain a maximum value over its domain; in this case,
$$
\mathop {\arg \max }\limits_x f(x) = \emptyset .
$$
As an example, define $f$ on $[0,1]$ by $f(x)=x$ if $0 \leq x < 1$, and $f(1)=0$. If, on the other hand, $f$ is continuous on a closed bounded interval $[a,b]$, then, by the Extreme value theorem, it must attain its maximum value (at least once); so, in this case, $\mathop {\arg \max }\limits_x f(x) \neq \emptyset$.
A: I think that "the max is at $x_0$" works just fine. You don't? Do you want something like $ \mathrm{max} f(x) = f( x_0 )$?
EDIT: cool - I learned something new! This is completely from Jineon Baek. But $\mathrm{arg \, max}f(x)$ is another way - it refers to the set of points $X$ that maximize the function.
