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Math notation for location of the maximum

Given a function $f(x)$, we can normally find $\max_i f(i)$. This expression evaluates to the maximum value of $f(x)$. Sometimes, however, what is interesting isn't as much the maximum value itself, but the value at which the function reaches its maximum, whatever that might be:

$$i:f(i)=\max_t f(t)$$

In plain English, I want to know who has eaten the most $f$ruit, rather than how much he's eaten.

This is kind of cumbersome, and perhaps needlessly requires a new symbol t. Besides, this isn't entirely correct, as $f(x)$ could very well have more than one maximum:


Is there a nicer way to express this concept?

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marked as duplicate by t.b., Srivatsan, J. M., Ilya, Asaf Karagila Dec 23 '11 at 13:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

(In Italian this concept is called "punto estremante", the English Wikipedia article suggests no equivalent phrasing.) – badp Dec 23 '11 at 0:15
Supremum and maximum are not quite the same thing - the supremum can exist when a maximum does not. – Thomas Andrews Dec 23 '11 at 0:41
@ThomasAndrews I am very well aware. Did I make that confusion anywhere in the page? – badp Dec 23 '11 at 0:43
up vote 6 down vote accepted

Often one writes $\underset{x}{\operatorname{argmax}} f(x)$ for the value of $x$ that maximizes $f(x)$.

(Despite the fact that I have frequently seen that notation for many years, I find that it's not a standard TeX operator name; one cannot type \argmax and have it understood by TeX.)

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\arg\max seems to work. – badp Dec 23 '11 at 0:18
@badp I believe your edit suggestion was made with good intentions, but I rejected it because it changed the answer too heavily. I suggest you write it as a comment instead, so that Michael can introduce that change if necessary. – Srivatsan Dec 23 '11 at 0:26
@badp : I see that you suggested \arg\max. Let's try that: $\displaystyle\arg\max_x f(x)$. The subscript $x$ ends up directly below $\max$ rather than symmetrically under $\operatorname{argmax}$. – Michael Hardy Dec 23 '11 at 0:31
@Srivatsan May I point you to the comment I have added? His post could've been improved with a comment in Michael's inbox and no further action on his part, now he's going to get this frankly uninteresting discussion in his inbox instead. Nevermind edit suggestions typically are rejected for being too minor... – badp Dec 23 '11 at 0:32 – Henry Dec 23 '11 at 0:33

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