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Question

Let $g:\mathbb{R}\to\mathbb{R}$ be an odd function and $h:\mathbb{R}\to\mathbb{R}$ be an even function.

Find $g(x)$ and $h(x)$ such that $g(x)h(x)=x$ for all $x\in\mathbb{R}$.

Try

I can only figure out the trivial cases where $g$ and $h$ are both polynomials.

From $g(x)h(x)=x$, the degrees of $g$ and $h$ are both at most $1$.

Since $g$ is odd, $g(x)=ax$ where $a$ is a constant.

Since $h$ is even, $h(x)=b$ where $b$ is a constant.

From $x=g(x)h(x)=(ax)(b)=abx$, $ab=1$.

Therefore, the solution would be $g(x)=Cx$ and $h(x)=\dfrac{1}{C}$ where $C$ is a non-zero constant.

Does anyone have some ideas for the non-polynomial cases?

Greg Martin
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Yuta
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    One can generalize the solution $g(x)=Cx$, $h(x)=\frac1C$ to $g(x) = C(x)x$, $h(x) = \frac1{C(x)}$ for any nonvanishing even function $C(x)$. And there are a whole lot of those, especially since we're not assuming continuity. – Greg Martin May 01 '21 at 02:11

1 Answers1

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Take your favourite even funcion $h$ such that $h(x)\neq 0$ for $x\neq 0$. (For example, $h(x)=\cos(x)+|x|$. Or $h(x)=|x|e^{|\cos(x)|+32}$.)

Let $g$ be defined as $$g(x)=\begin{cases}x/h(x)&\text{if }x\neq 0\\0&\text{if }x=0\end{cases}$$

Then $g,h$ give a solution to your problem.

In fact, all solutions are like this, as you can easily verify.

(You can also start with odd $g$ and construct $h$ similarly to obtain all solutions to your problem).

Luiz Cordeiro
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