# Functions that take out the inverse operation

Are there are there any real-values functions $f(t)$ other than $f=t,1/t$ and $\pm 1$ such that $f(t)f(t^{-1})=1$ for $t>0$?

• To be pedantic: $f = \pm 1$ isn't even a function. – Bladewood Apr 21 '18 at 3:45
• $f(t)=t^c$ for some constant $c$? – Lord Shark the Unknown Apr 21 '18 at 3:48
• @LordSharktheUnknown Yes! That's actually a pretty good example. Thanks. – Andrew Yuan Apr 21 '18 at 3:56
• I'll remark that the two existing answers have been invalidated by the latest revision. Try not to do that. – Bladewood Apr 21 '18 at 4:23
• @Bladewood Sorry about that. I forgot to mention that I would prefer nontrivial examples of existing example, i.e., if $f(t)$ is an example, then $f(t^{-1})$ is trivially an example. – Andrew Yuan Apr 21 '18 at 22:56

$f(t) = \frac{1}{t}$.

Lots of others if you don't require continuity.

Xifei Auto mentions another class of answers but there are even more.

Pick any function defined on $[1, \infty)$ with $f(1) = 1$ and then extend the definition to $(0, 1)$ with your required relationship.

If the original function is continuous then the extended one will be continuous as well.

Here is a non-continuous example:

$$f(t) = \begin{cases} \frac{1}{2} & \text{if t \in (0,1)} \\ 1 & \text{if t = 1} \\ 2 & \text{if t \in (1,\infty)} \\ \end{cases}$$

Of course, you can put any crazy stuff you liked in the last case and matching reciprocal stuff in the first.

Another crazier example:

$$f(t) = \begin{cases} 1 & \text{if t = 1} \\ \frac{1}{2} & \text{if t \in (0,1) \land t \in \mathbb{Q}} \\ \frac{1}{3} & \text{if t \in (0,1) \land t \notin \mathbb{Q}} \\ 2 & \text{if t \in (1,\infty) \land t \in \mathbb{Q}} \\ 3 & \text{if t \in (1,\infty) \land t \notin \mathbb{Q}} \\ \end{cases}$$

If the original function on $[1, \infty)$ is differentiable then the extended function is also differentiable.

Certainly:

$$f(t) = \frac1t.$$

$f(t)f(\frac1t) = \frac1t \frac1{\frac1t} = \frac1t t = 1$ for all $t \neq 0$.

It is easy to see if $f(t)=\pm t^a, a\in \Re$, $f$ satisfies your identity.