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Let $F\subseteq (\mathbb R \to\mathbb R)$ be some space of functions, and let $G:F\to \mathbb R$ be a functional. I have a statement of the following form: $$\begin{align}\text{Let } &f^*(x):=x^2. \quad\quad\quad \text{Then }\\ &f^*\in \arg\max_{f\in F} G(f) \end{align}$$

Rather than first defining a function and then referencing it, I'd like to compress this into one equation for brevity's sake. Something like:

$$(x\mapsto x^2)\in \arg\max_{f\in F} G(f)$$

Is there a generally accepted notation like this? I'd prefer not to invent something new and unknown.

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  • $\begingroup$ Now that I understand the question - there is no typical notation for this, because it is not common to do two things at once. If $f^*$ is defined to be the arg max, then there is a claim that $f^*$ is also $x^2$. If $f^*$ is defined to be $x^2$, there is a claim that it is the arg max. Either way, it is more common to separate the definition from the claim. More generally, the $\lambda x.x^2$ notation from computer science and the related $x \mapsto x^2$ notation are not especially common in math, with the possible exception of some specific subfields. $\endgroup$
    – Carl Mummert
    Aug 31, 2018 at 14:16
  • $\begingroup$ @CarlMummert, well, let's say that we DON'T want to be able to keep the function symbol $f^*$ around, then it's not really doing two things: essentially we would be saying in one single statement: "the function that maximizes $G(f)$ is the function that maps $x$ to $x^2$". $\endgroup$
    – user56834
    Aug 31, 2018 at 14:18
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    $\begingroup$ !Programmer2134: so why not use that sentence? It is exceptionally clear. $\endgroup$
    – Carl Mummert
    Aug 31, 2018 at 14:19
  • $\begingroup$ @CarlMummert, because I want to be able to write it very often, with brevity $\endgroup$
    – user56834
    Aug 31, 2018 at 14:20
  • $\begingroup$ I think there is no standard notation for that which would be immediately clear to everyone, since as I said the $\mapsto$ notation is not very common except possibly in specialized areas. $\endgroup$
    – Carl Mummert
    Aug 31, 2018 at 14:22

2 Answers 2

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I don't understand either of your statements, so both of them are too concise to be readable.

Do you mean that $f^*$, which is $\underset{f \in F}{\operatorname{argmax}} G(f)$, turns out to be the function defined by $f^*(x) = x^2$? If so, for the sake of comprehensibility rather than brevity, you should write this out in words in a complete sentence: for example,

Let $f^* = \underset{f \in F}{\operatorname{argmax}} G(f)$. Then it turns out for mysterious reasons that $f^*(x) = x^2$.

(Possibly with an explanation why this is the function that maximizes $G(f)$.)

Or possibly (after the recent edits, this seems closer to the sort of emphasis you want):

Let $f^* \in F$ be given by $f^*(x) = x^2$. Then $f^* \in \underset{f \in F}{\operatorname{argmax}} G(f)$.

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  • $\begingroup$ $\underset{f \in F}{\operatorname{argmax}} G(f)$ is not a set: it means "the element of $F$ which maximizes the value of the functional $G$", and elements of $F$ are functions. It equals the function that maps $x$ to $x^2$. $\endgroup$
    – Misha Lavrov
    Aug 31, 2018 at 14:12
  • $\begingroup$ !Misha - that is a reasonable interpretation here, but what if the maximum happens are multiple points? I was also looking at en.wikipedia.org/wiki/Arg_max . . But now I see - if the OP had written $f^* \in \text{arg max} G(f)$, where $f^*(x) = x^2$, it would be clearer to me. $\endgroup$
    – Carl Mummert
    Aug 31, 2018 at 14:13
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    $\begingroup$ @CarlMummert Quoth wikipedia: "If the maximum is reached at a single point then this point is often referred to as the arg max, and arg max is considered a point, not a set of points." This is the standard use case, in my opinion. (Though, of course, the fact that the maximum is reached at a single point - and that it is reached at all - must be justified.) $\endgroup$
    – Misha Lavrov
    Aug 31, 2018 at 14:14
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    $\begingroup$ +1 for use words. $\endgroup$
    – Ethan Bolker
    Aug 31, 2018 at 14:16
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If you don't want to invent something new you could write $$ \big(\arg\max_{f\in F} G(f) \big)(x) = f^\star(x) = x^2. $$ However, in my opinion it would be preferable to write it as $$ f^*=\arg\max_{f\in F} G(f),\; f^*(x)=x^2, $$ which has approximately the same length.

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