1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$f$ 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:

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

$\endgroup$
2
$\begingroup$

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 $\,4.\,$

$\endgroup$
  • $\begingroup$ That should read, "and hence a translation by 4." For instance, $f(x)=\cos\frac{\pi}{2}x$ satisfies the conditions. $\endgroup$ – TonyK Oct 27 '18 at 10:37
  • $\begingroup$ A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$. $\endgroup$ – TonyK Oct 27 '18 at 10:53
  • $\begingroup$ @TonyK Thanks for that correction. I should be more careful. $\endgroup$ – Somos Oct 27 '18 at 11:01
0
$\begingroup$

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,$$\begin{array}{rccc}f\colon&\mathbb Q&\longrightarrow&\mathbb Z\\&q&\mapsto&\begin{cases}1&\text{ if }q\in\mathbb Z\\0&\text{ otherwise.}\end{cases}\end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.