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 $t>0$?

  • $\begingroup$ To be pedantic: $f = \pm 1$ isn't even a function. $\endgroup$ – Bladewood Apr 21 '18 at 3:45
  • $\begingroup$ $f(t)=t^c$ for some constant $c$? $\endgroup$ – Lord Shark the Unknown Apr 21 '18 at 3:48
  • $\begingroup$ @LordSharktheUnknown Yes! That's actually a pretty good example. Thanks. $\endgroup$ – Andrew Yuan Apr 21 '18 at 3:56
  • $\begingroup$ I'll remark that the two existing answers have been invalidated by the latest revision. Try not to do that. $\endgroup$ – Bladewood Apr 21 '18 at 4:23
  • $\begingroup$ @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. $\endgroup$ – Andrew Yuan Apr 21 '18 at 22:56

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

Lots of others if you don't require continuity.

Expanded answer.

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.



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

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


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


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