# Prove a function is constant function with the given conditions.

If $$f(4-x)=f (4+x)$$ and $$f(2-x)=f(2+x)$$, prove that $$f$$ is constant function.

I tried to solve by assuming x and y are two variable then if we get $$f (x)=f (y)$$ then $$f$$ is constant function, but can't complete the problem. Someone please help me.

$$f$$ is not necessarily constant. Here is a counter-example: $$f(x) = \begin{cases} 0, & \text{if x\in\Bbb Q} \\ 1, & \text{if x\notin\Bbb Q} \end{cases}$$

And another one, continuous this time:

$$f(x)=\cos\pi x$$

The two equations $$\,f(4-x)=f(4+x)\,$$ and $$\,f(2-x)=f(2+x)\,$$ require that the function is invariant with respect to two reflections and hence a translation by $$4$$. The simple calculation is $$\, x \mapsto 8-x \,$$ in the first reflection and then $$8-x \mapsto x-4 \,$$ after the second reflection yielding a translation by $$4.$$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $$\,[0,2].\,$$ The function must be an even function with a period of $$4$$. That is, $$\,f(x)=f(-x)\,$$ and $$\,f(x)=f(x+4)\,$$ for all real $$\,x.\,$$ This determines the function for all real numbers and that it satisfies $$\,f(2 n-x)=f(2 n+x)=f(x)\,$$ and $$\, f(4 n + x) = f(x) \,$$ for all integer $$\,n\,$$ and all real $$\,x.\,$$

For a simple example, consider $$\, f(4n+x) := |x|\,$$ for all $$\,|x|\le 2\,$$ and integer $$\,n.\,$$ This is a sawtooth function with fundamental period $$\,4.\,$$

• That should read, "and hence a translation by 4." For instance, $f(x)=\cos\frac{\pi}{2}x$ satisfies the conditions. – TonyK Oct 27 '18 at 10:37
• A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$. – TonyK Oct 27 '18 at 10:53
• @TonyK Thanks for that correction. I should be more careful. – Somos Oct 27 '18 at 11:01

Without tellings what are the domain and the codomain of $$f$$, nobody can answer that. However, if the domain is, say, $$\mathbb Q$$, then the statement is false. Consider, for instance,$$\begin{array}{rccc}f\colon&\mathbb Q&\longrightarrow&\mathbb Z\\&q&\mapsto&\begin{cases}1&\text{ if }q\in\mathbb Z\\0&\text{ otherwise.}\end{cases}\end{array}$$