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Is the set $\{g:\mathbb{R}\rightarrow\mathbb{R}\mid g\circ f=f\circ g\}$ infinite?

Why?

Let's suppose $f(x)=x^2$

$g(x)$ is an inverse function of $f(x)$.

$g(f(x))=(x^2)^{\frac12}=x$

Therefore, since composition of functions are associative,

$g(f(x))=f(g(x))=x$

The range of $f(x)=x$ is $[0,\infty)$

Since the range is $\infty$, the set is infinite.

My question is: What is the point of ${g:\mathbb{R}\rightarrow\mathbb{R}}$ in $g\circ f=f\circ g$?

Bernard
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Nay Sie
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    Your "dot" notation seemingly means composition of functions, usually denoted by a $\circ$ sign ($f \circ g = g \circ f$)? – Jean Marie Jul 10 '20 at 14:32
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    It's important to realize that the set being described is not a set of real numbers, it's a set of functions. To be precise it's the set of all functions commuting with $f$ (As @Dietrich Burde mentions above). There may be many such functions. If $f$ is invertible (which isn't true for your examples of $x \mapsto x^2$) then the inverse of $f$ does indeed satisfy the condition of commuting with $f$. – Jamie Radcliffe Jul 10 '20 at 14:32
  • @DietrichBurde It's not quite as simple as you suggest: $g(x)=2x$ doesn't commute with $f(x)=x^2$. – Jamie Radcliffe Jul 10 '20 at 14:44
  • Composition of functions is associative, i.e., $h \circ (g \circ f) = (h \circ g) \circ f;$ however, it is not necessarily commutative, i.e., it is not true that $g \circ f = f \circ g$ for all functions $f, g : \mathbb R \to \mathbb R.$ But you are correct that a function commutes with its inverse by definition. – Dylan C. Beck Jul 10 '20 at 14:48

1 Answers1

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If $f,g : \mathbb{R} \to \mathbb{R}$ are two functions, one can consider the two other functions $f\circ g$ and $g\circ f$ defined by $f\circ g (x) = f\left(g(x)\right)$ and $g\circ f (x) = g\left(f(x) \right)$. One can also ask whenever are these two functions the same.

If $f$ is a fixed function, the set $C(f)=\left\{g : \mathbb{R} \to \mathbb{R} | f\circ g = g \circ f \right\}$ is the set of all functions $g$ for which the two functions $f\circ g$ and $g\circ f$ are equal.

For example, if $f = \mathrm{id}$ is the identity function, then $C(f)$ is the whole set of functions, because all function commutes with the identity map.

But if $f(x)=x^2$ for example, then the function $g : x \mapsto -x$ is not in $C(f)$, as $f\circ g = f$ and $g\circ f = -f$. The set $C(f)$ depends on $f$.

Note that for every $n$, $f^{\circ n}=\underbrace{f\circ f \circ \cdots \circ f}_{n \text{ times}}$ is in $C(f)$, because $f^{\circ n} \circ f = f \circ f^{\circ n} = f^{\circ n+1}$, so for general $f$ you know an infinite family of functions in $C(f)$.

Edit It has been noted that it would be relevant to discuss the case whenever $\{f^{\circ n} | n \in \mathbb{N} \}$ is finite (with convention $f^{\circ 0} = \mathrm{id}$). As it is non-empty ($f \in C(f)$) one can ask if it can be any positive integer. The answer is yes. Suppose $n \geqslant 1$ and consider the following function : \begin{align} f(x) =\begin{cases} x+1 & \text{if} & x \in \{1,2\ldots,n-1\} \\ 1 & \text{if} & x = n \\ x & \text{if} & x \neq 1,\ldots, n \end{cases} \end{align} It is just a cylcic permutation on $\{1,\ldots,n\}$, and the identity on the complementary. It follows that $f^{\circ n}$ is the identity, and that for exery $1\leqslant k,k' \leqslant n-1$, if $k \neq k'$, then $f^{\circ k} \neq f^{\circ k'}$. Consequently, the set $\{ f^{\circ k} | k \in \mathbb{N} \}$ has exactly $n$ elements.

Didier
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  • Can the question be answered finite or infinite? Can you explain on that? – Nay Sie Jul 10 '20 at 15:04
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    @Dldier_ It would be helpful to discuss what happens when the set of composition powers is finite, though $f$ is not the identity. One example is $f(x)=-x$, but there are discontinuous examples of all finite orders. – Jamie Radcliffe Jul 10 '20 at 15:08
  • @JamieRadcliffe Thank you for the suggestion. I added a discuss on the fact that the set of powers of $f$ can be any finite non-empty set. – Didier Jul 10 '20 at 15:29