Smooth function satisfying $f^{-1}(x)=f(x+1/2)-1/2$

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)

Questions:

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 $$f$$?

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 $$t$$.
• @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 :-) Aug 6 at 0:17