Let $f$ be a smooth (infinitely differentiable) function satisfying $f^{-1}(x)=f(x+1/2)-1/2$ for all real $x$ (where $f^{-1}$ is an inverse function).

I have strong feeling that it also must satisfy $f(x+1)=f(x)+1$ for all real $x$, but cannot prove or disprove it.

I know that $f$ is monotone because it has inverse everywhere. Also if $f$ has fixed point $p$ (i.e. $f^{-1}(p)=f(p)$), then $f(p+k/2)=p+k/2$ where $k$ is any integer. But I don't know if $f$ even always has fixed point. In some cases $f$ can be calculated for all positive $x$ if values for $0\le x <1/2$ are given. I don't know how to invert that equation to get negative values.

Some examples which satisfy both relations:

  • $f(x)=x$
  • Define $g(x)=x-\frac{\sin{2\pi x}}{4\pi}$, then $f(x)=2g^{-1}(x)-x$, (basically a rotated and scaled sine wave)


  1. Does $f$ always satisfy $f(x+1)=f(x)+1$ for all real $x$?
  2. Will $f$ be always increasing?
  3. What else can we say about $f$?

1 Answer 1


Let $g(x) = x+1/2$. The equation can be written as $$f^{-1} = g^{-1} f g$$ Now operate with $f$, $g$ like elements of a group. We get from the above taking inverses $$g^{-1} f^{-1} g = f$$ and so $$g^{-2} f g^{2} = f$$ that is $$f(x+1) = f(x)+1$$

$\bf{Added:}$ If $f$ is continuous, then $f$ is monotone, and from the last equality we conclude $f$ strictly increasing.

$\bf{Added:}$ It's useful to look at graphs of the function $f_1(x)\colon = f^{-1}(x)$ and $f_2(x) \colon = f(x+1/2)-1/2$. We have $\Gamma_{f_1} = S(\Gamma_{f})$, and $\Gamma_{f_2} = T(\Gamma_{f})$, where $S$ is the symmetry w.r. the first bisector, and $T$ is the translation by the vector $(-1/2, -1/2)$. In other, words, the graph of $f$ is invariant under the transformation $S T^{-1} = T^{-1} S$, which is a glide reflection. Now we see there are many such functions $f$. $f$ will necessary have fixed points, and the set of fixed points is invariant under the translation $x\mapsto x+1/2$.

$\bf{Added:}$ Using the idea with the graph of functions, we see that the graph of $f$ can be parametrized by $t\mapsto (t- h(t), t+ h(t))$, where $h$ is a function satisfying $h(t+1/2) = -h(t)$. In general such functions $h$ are of the form $$h(t) = \sum_{k \ge 0} (a_k \cos 2(2k+1) \pi t + b_k \sin 2 (2k+1) \pi t)$$ that is a Fourier expansion with only terms of odd indexes. We want to make sure that $t-h(t)$ is strictly increasing, so require $h'(t) <1$ for all $t$.

  • 1
    $\begingroup$ Thanks! Much easier than I thought. Basically we don't even need smoothness condition? $\endgroup$
    – Somnium
    Aug 6 at 0:15
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    $\begingroup$ @Somnium: Yes indeed, works for all bijective functions $f$. Now it would be interesting to characterize such functions, perhaps satisfying other conditions (smooth, etc). Interesting question :-) $\endgroup$
    – orangeskid
    Aug 6 at 0:17
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    $\begingroup$ Thanks for additions. I already knew about symmetry, that's how I found second example. $\endgroup$
    – Somnium
    Aug 6 at 9:42

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