What do we call a function when it's applied to a modified argument?

Question

This is an embarrassingly simple question, but I realized that I haven't formalized the following idea in my mind.

Suppose we have a function $f: \Bbb{R} \to \Bbb{R}; x \mapsto x$. What am I to make of the notation $f(x-1)$? Is it "the same function" applied to a different argument? Is it a "different function?"

Is it a function at all? If $f(x-1)$ is the name of a function (in particular, a name for $x \mapsto f(x-1)$, where the latter $f(x-1)$ should not be confused with the former), then it ought to take an argument: $f(x-1)(y) = y-1$. It is not the same kind of creature as $f(x)$ (which is a function application), but is comparable to just $f$.

And yet I sometimes read things like "suppose we transform the function $f(x)$ into $f(x-1)$" in physics. It's not hard to understand what's being said, but the lack of formality in my thought is troubling me for some reason.

Answer 1

I think the most straightforward way to regard it is just as a function composition, which is formally defined as $(f \circ g)(x) = f(g(x))$ (taking note of well-definedness in the respective domains). So in your specific example, let $g : \mathbb{R} \to \mathbb{R} : x \mapsto x - 1$, then $f(x - 1) = (f \circ g)(x)$.

And yes, to add to your additional questions, the composition of two functions is again a function, assuming the composition is well-defined. If you had a function from apples to oranges, and a function from oranges to pears, then you could have a composition from apples to pears which is well-defined, but not a function from oranges to oranges.

Answer 2

A common error committed by novice students is to confuse the name of a function with the values taken by that function. When we write $$ f : \mathbb{R}\to \mathbb{R} : x \mapsto x^2, $$ then:

• $f$ is the function itself. At a very basic level, $f$ is a set; specifically, a subset of the Cartesian product $\mathbb{R}\times\mathbb{R}$. Even more specifically, in the example above, we could write $$ f = \{ (x,y) : y = x^2 \}. $$
• $f(x)$ is the value of the function $f$ when evaluated at the point $x$. Stated another way, if $(x,y_x) \in f$, then $f(x) = y_x$. In the example above, we have $$ f(x) = x^2, \quad f(-4) = 16, \quad f() =^2, \quad\text{and}\quad f(x-1) = x^2 - 2x + 1. $$

So, to clearly answer your question, $f(x-1)$ is not a function; $f(x-1)$ is a value in the codomain of $f$. Specifically, it is the value obtained when $f$ is evaluated at $x-1$. If you want to view the action $x \mapsto f(x-1)$ as a new function, you will have to introduced new notation:

• For example, define a new function $$ g : \mathbb{R} \to \mathbb{R} : x \mapsto f(x-1). $$

• Alternatively, in this case, it is common to think of the new function as composition with a translation. That is, for each $a\in\mathbb{R}$, define $$ T_a : \mathbb{R} \to \mathbb{R} : x \mapsto x+a. $$ Then we have $$ (f\circ T_{-1})(x) = f(x-1). $$

• Another possibility is to leave a dot as placeholder for the variable. For example, the notation $f(\cdot -1)$ indicates a function defined by the mapping $x \mapsto f(x-1)$.

• Finally, since these kind of translations show up a lot, some authors will introduce notation which specifically indicates translation. For example, Folland (in the text Real Analysis: Modern Techniques and Their Applications) uses the notation $\tau_a$ to indicate a translation operator acting on a function. In that text, the function $\tau_a f$ is defined by the mapping $\tau_a f(x) = f(x+a)$.