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Suppose I have a function $f:\mathbb{R}_+ \to [0,1]$ and another similar function $g$.
I consider $g$ to be a fixed function, and I want to know how close $f$ is to $g$. For this purpose, I define a `residual operator' $R$ which calculates the difference between the two functions, but as a function of $f(x)$. Can I write:

$R: f \mapsto (f(x) \mapsto (f(x) - g(x)\;)$?

In other words, the output of $Rf$ is a function whose output is the difference evaluated at $f(x)$.

P.s. I suppose to be more rigorous I could write this as

$R: (f,g) \mapsto (f(x) \mapsto (f(x) - g(x)\;)$.


I'm getting a lot of answers that I should write $R: (f,g) \mapsto f - g$. However, please notice that when I plot $Rf$ I want the horizontal axis to be the value of $f(x)$. I know this is possible in practice, because I have plotted this many times, but I'm having difficulty representing what I am calculating in function notation.


With the help of the very nice responses, I think I know how to write this in rigorous function notation. The object I'm interested in is a plot, and a plot is really just a collection of points in two dimensions. How do I write "a collection of points in two dimensions"? One such collection might be written:

$\{(x(t),y(t)) \;|\; x,y: \mathbb{R}_+ \to \mathbb{R} , t\ge 0\} \subset \mathbb{R}^2$.

My operator becomes

$R: (f,g) \mapsto \{\,(f(t),f(t)-g(t)) \;|\; f,g: \mathbb{R}_+ \to \mathbb{R} , t\ge 0\}$.

In summary, we can think of $R(f,g)$ as a subset of $\mathbb{R}^2$.

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    $\begingroup$ I don't see why $R(f) = f-g$ wouldn't work. $\endgroup$
    – whpowell96
    Commented Mar 22 at 16:04
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    $\begingroup$ More like $R: f \mapsto ( x \mapsto f(x) - g(x) )$, but why not simply $R(f) = f - g$? $\endgroup$ Commented Mar 22 at 16:04
  • $\begingroup$ @whpowell96 I'd like it to be a function of $f(x)$. When I plot it, I'd like the horizontal axis to be $f(x)$. $\endgroup$
    – Chris
    Commented Mar 22 at 16:09
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    $\begingroup$ Regarding the edit, the thing you want is not a function in general. What it sounds like you want is a parametric graph of the form $x\mapsto (f(x), f(x)-g(x))$ $\endgroup$
    – whpowell96
    Commented Mar 22 at 17:08
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    $\begingroup$ I agree. My operator is giving me a set of point in the 2D plane. Which is not the same as a function. I now see this with the help of these comments. Please see my edited question. $\endgroup$
    – Chris
    Commented Mar 22 at 18:01

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