How many continuous functions $f$ are there which satisfy $(f(x))^{2} = x^{2}$ - How to approach?

Question

This is related directly to another question about this: Continuous Functions

My question though revolves around how to approach the question in a "thinking" way. So mechanically looking at the expression one would take the square root of both sides and end up with:

$$f(x) = \pm x$$

But this doesn't really say anything about how to interpret the result. So in the other post the solutions revolved around looking at the behaviour of $x$ between $(-\infty,0)$ and $(0,\infty)$ and applying the intermediate value theorem. But what is accomplished by showing that $f(x)$ does not change sign? and also how do you arrive at the second set of solutions? Specifically I'm aware of the relationship $\sqrt{x^{2}} = |x|$, but knowing the relationship doesn't mean I know how to apply it correctly.

Answer 1

There are $4$ such functions. Namely $x\mapsto\pm x$ and $x\mapsto\pm|x|$. This follows from the method suggested by @lulu in the comments. We know that $f(0)=0$ from the defining equation. Then consider some $\epsilon\gt0$. We know that $(f(\epsilon))^2=\epsilon^2$ and hence either $f(\epsilon)=\epsilon$ or $f(\epsilon)=-\epsilon$. Using the continuity of $f$ and applying the same reasoning for any other $x\gt0$ gives $f(x)=x$ or $f(x)=-x$ for all $x\gt0$ in each case respectively. Similarly we have either $f(x)=x$ or $f(x)=-x$ for all $x\lt0$. Combining these $4$ cases gives the $4$ functions originally stated.

Answer 2

This answer might be a bit pedantic and not in the spirit of the question:

It depends on the domain of the function $f$. Let us say that the domain of $f$ is a subset of $\mathbb{C}$ called $D$, i.e., $f$ is a continuous function $D \to \mathbb{C}$ with the property that $f(x) \in \{x, -x\}$ for all $x$. We can now define the function $g$ on $D \setminus \{0\}$ by setting $g(x) = f(x)/x$. Note that $g$ is continuous and $g(x) \in \{1,-1\}$ for all $x \in D \setminus \{0\}$. Because $g$ is continuous and $\{1,-1\}$ is discrete, $g$ must be constant on every connected component of $D \setminus \{0\}$. This gives two possibilities per connected component of $D \setminus \{0\}$ (check that any combination yields a valid and distinct function $f$).

---

Examples:

• If $D = \mathbb{R}$, then $D \setminus \{0\} = (-\infty,0) \cup (0,\infty)$ and you indeed get $2 \cdot 2 = 4$ possibilities.
• If $D = \mathbb{C}$, then $D \setminus \{0\} = \mathbb{C} \setminus \{0\}$, which is connected, so in that case there are only $2$ possibilities.
• If $D = (-1,1) \cup (3,5) \cup (7,10) \cup (12,100)$, you can check that you get $32$ possible functions $f$.

Answer 3

The difference between $x$ and $-x$ is $2x$. For any nonzero $x$, this is nonzero. Hence a change of sign is not possible, except at the origin.

There are four possible combinations $(\pm\to\pm$).