3
$\begingroup$

Find all functions $f$ which are continuous on $\mathbb R$ and which satisfy the equation $f(x)^2=x^2$ for all $x \in \mathbb R$.

Clearly $f(x)=x, -x, |x|, -|x|$ all satisfy the condition. However, how can I show that these must be the only possible choices? The condition guarantees that $|f(x)|=|x|$, for all $x$ so I think it's quite obvious that these four choices are the only possibilities. But I don't see why continuity is necessary. If $f$ does not need to be continuous, are there other possibilities? Then how can I use continuity to guarantee that these are the only choices?

$\endgroup$
7
$\begingroup$

From $f(x)^2 = x^2$ follows "only" that $f(x) = x$ or $f(x) = -x$ for each $x \in \mathbb R$. Without the continuity requirement, you could choose from both possibilities for each $x$ independently. For example, $f(x)=x$ if $x$ is rational and $f(x)=−x$ if $x$ is irrational.

So what you need to show is that either $$ f(x) = x \text{ for all } x > 0 $$ or $$ f(x) = -x \text{ for all } x > 0 \, . $$ and this follows from the continuity: If there were $x_1, x_2 > 0$ such that $f(x_1) = x_1 > 0$ and $f(x_2) = -x_2 < 0$ then because of the Intermediate Value Theorem there would be a $x_3 > 0$ such that $f(x_3) = 0$, contradicting the requirement $f(x_3)^2 = x_3^2$.

For the same reason you have either $$ f(x) = x \text{ for all } x < 0 $$ or $$ f(x) = -x \text{ for all } x < 0 \, . $$

This gives the four possible functions $f(x)=x, -x, |x|, -|x|$.

$\endgroup$
1
$\begingroup$

As you mention, for any $x\in \mathbb{R}$, $f(x) = \pm x$. Let $S$ denote the set on which $f(x) = x$, and $T$ be the set where $f(x) = -x$, then $S$ and $T$ are both closed and $S\cap T = \{0\}$.

Suppose $0 < x\in S$, then $f(x) > 0$, so by continuity, $f$ must be positive on a neighbourhood of $x$. Since $f(y) = \pm y$ on that neighbourhood, it follows that $f(y) = y$. Hence, $S\setminus \{0\}$ is an open set. Similarly, $T\setminus\{0\}$ is an open set.

By connectedness of $(-\infty,0)$, it follows that either $(-\infty,0) \subset S$ or $(-\infty,0) \subset T$. Similarly, for $(0,\infty)$, which gives the four different possibilities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.