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"The graph of a function $f: \mathbb{R}\to\mathbb{R}$ has two has at least two centres of symmetry. Prove that $f$ can be represented as sum of a linear and periodic function." Source: sum of a linear and periodic function

I came across this while searching for an unrelated question, and it looked interesting. I have seen some claims that this was easy, but I cannot see where even to begin with this, and do not really understand the solution given in the link (namely, why does it involve looking at rotations?).

How is this proven?

umm
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  • I have updated my answer for both cases. Still, you should specify the kind of symmetry you are talking about. – M. Winter Dec 23 '17 at 00:27
  • Unfortunately, I have no idea the type of symmetry in the problem statement (I just assumed it was some sort of not-necessarily-parallel-to-coordinate-axes line that the graph could be reflected across). This problem is from a 2003 German math competition. – umm Dec 23 '17 at 00:31

1 Answers1

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Rotational symmetry

Definition. A center of (rotational) symmetry of $f:\Bbb R\to\Bbb R$ is a pair $c_x,c_y\in\Bbb R$ with $$f(x)=2c_y-f(2c_x-x)\quad\text{for all $x\in\Bbb R$}.$$

If there are two such centers of symmetry, namely $(c_x^1,c_y^1)$ and $(c_x^2, c_y^2)$, then we find

$$f(x)=2c_y^1-f(c_x^1-x)=\underbrace{2(c_y^2-c_y^1)}_a+f(\underbrace{2(c_x^2-c_x^1)}_b+x)=a+f(b+x).$$

Define $g(x)=f(x)-a/b \cdot x$. We then have that

$$g(x+b)=f(x+b)-\frac ab(x+b)=f(x)-a+\frac abx+\frac abb=f(x)+\frac ab x=g(x).$$

This means $g$ is periodic. And since $f(x)=g(x)+a/b\cdot x$, we have that $f$ is the sum of a periodic function and a linear function.

Mirror symmetry

Definition. A center of (mirror) symmetry of $f:\Bbb R\to\Bbb R$ is a number $c\in\Bbb R$ with $$f(x)=f(2c-x)\quad\text{for all $x\in\Bbb R$}.$$

So if there are two such numbers $c_1$ and $c_2$ (assume $c_2>c_1$), then we have that

$$f(x)=f(2c_1-x)=f(2c_2-(2c_1-x))=f(2(c_2-c_1)+x).$$

This means that $f$ is periodic with period $2(c_2-c_1)$ and we can choose the linear part to be zero.

M. Winter
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