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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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    $\begingroup$ en.wikipedia.org/wiki/Arg_max $\endgroup$ Commented Aug 1, 2011 at 6:16
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    $\begingroup$ This has nothing to do with set theory. $\endgroup$
    – Asaf Karagila
    Commented Aug 1, 2011 at 6:53
  • $\begingroup$ @Mike, you can refer not only to the "maximum" of $f$, but also to a "maximizer." So $x$ is a maximizer of $f$ iff $f$ has a maximum, and $f(x)$ equals that maximum. $\endgroup$ Commented Apr 15, 2013 at 9:56

2 Answers 2

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

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  • $\begingroup$ wouldn't "$\in$" be preferable to "$=$"? $\endgroup$
    – t.b.
    Commented Aug 1, 2011 at 6:23
  • $\begingroup$ @Theo: That might be a matter of convention. I'll check different sources. $\endgroup$
    – Shai Covo
    Commented Aug 1, 2011 at 6:25
  • $\begingroup$ @Theo: For example, see de.wikipedia.org/wiki/Arg_max#Definition $\endgroup$
    – Shai Covo
    Commented Aug 1, 2011 at 6:28
  • $\begingroup$ Specifically, according to the first definition in that link $$x_0 = \arg \max f(x) \Leftrightarrow f(x_0 ) = \mathop {\max }\limits_{x \in D} f(x).$$ $\endgroup$
    – Shai Covo
    Commented Aug 1, 2011 at 6:39
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    $\begingroup$ Don't waste too much energy on this matter :) it was a nitpick, but I think one should try and avoid abuse of notation in such simple matters. $\endgroup$
    – t.b.
    Commented Aug 1, 2011 at 7:01
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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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  • $\begingroup$ Thanks to all! That is what i am looking for $\endgroup$
    – Mike
    Commented Aug 1, 2011 at 6:58

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