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.
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$\begingroup$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, $\mathop {\arg \max }\limits_x f(x) \neq \emptyset$.
$\endgroup$ 8 $\begingroup$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.
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