What's the term for a value x that satisfies the constraint $f(x) = f$ for a function f?

Question

I know that $x$ is called the fixed point of a function $f$ if it satisfies the constraint $f(x) = x$.

However, for a function $f$ if there exists some value $x$ such that $f(x) = f$ then what is the term for the value $x$ with respect to $f$.

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Consider the following function in JavaScript:

var bind = Function.prototype.bind;
var bindable = bind.bind(bind);

Now bindable is the function $f$ that I'm talking about. It satisfies the constraint $f(x) = f$ for the value bind:

bindable(bind) = bindable;

I know that I shouldn't express mathematics in terms of programming but I didn't know any other way to put it.

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Consider that bind has the following type definition:

(a b -> c) a -> (b -> c)

It takes a function of type a b -> c and zero or more arguments which grouped together have the type a and returns another function of type b -> c where b has the type of the rest of the arguments grouped together.

Hence bind(bind) has the following type definition:

(a b -> c) -> (a -> (b -> c))

Let bindable = bind(bind), thenbindable(bind) also has the same type definition:

(a b -> c) -> (a -> (b -> c))

i.e. bindable(bind) = bindable.

Answer 1

I doubt that there is a name for this. If there is, it would be something from lambda calculus.

Unlike fixed points, where $f(x)=x$, I'm not sure what the value would be in such a concept.

There is a sense in which $f(x)=f$ is also a fixed point. Namely, if $D=\lambda y.\lambda g. g(y)$, then $f$ is a fixed point of $Dx$. But I'm not sure if that really gives you anything.

You might consider asking this question at the StackExchange site dedicate to computer science questions.

Answer 2

This situation never occurs in ordinary mathematics, because in order to specify a function you need to specify the codomain $C$ (set of possible values the function could take) first, and that set cannot contain any functions that have $C$ as codomain. Or in terms of set theory (though that point of view does not seem the most important to me) the set of functions $X\to Y$ is a subset of $X\times Y$; if some $x \in X$ and $f\in Y$ satisfied $f(x)=f$ then one would have $(x,f)\in f$ and since also $f \in^+ (x,f)$ where $\in^+$ is the "transitive closure" of the relation $\in$ (just how depends on the definition of ordered pairs) one would have $f\in^+ f$ violating the axiom of regularity.

This being said one can imagine "function application" is just some externally defined operator that takes $f$ and $x$ to produce a value written $f(x)$, and in this case $f(x)=f$ is possible. To give it a name I would say $f$ is a fixed point for $x$ under reverse function application (being-applied-to).