What type of functions can satisfy f(x)=f(a-x)?

[McLaren Rulez]

Homework Statement

Given a function such that $f(x)=f(a-x)$ where a is some constant, what can we say about this function? This isn't actually homework; so please forgive the fact that I'm being a little vague

Homework Equations

None.

The Attempt at a Solution

I think that a periodic function (with period a) which is also even satisfies this but is there something else that can also satisfy this condition? Not really sure how to proceed in proving either that this is the only answer or to show an alternative.

 

[lurflurf]

All you can say about the function is it is symmetric about x=a/2.
What are the domain and range?
If domain were real numbers for example we could define f to be an arbitrary function when x<=a/2 then define f when x>a using the given functional equation.
or just take an arbitrary function g and define
f=(1/2)g(x)+(1/2)g(a-x)=(1/2)g(a/2+(x-a/2))+(1/2)g(a/2-(x-a/2))

 

[McLaren Rulez]

The domain and range are the set of real numbers.

Thank you for the reply lurflurf. But a function which is symmetric about a should be of the form f(a+x) = f(a-x). Not quite the same as mine, right?

The second part looks fine but what is that function g(x)+g(a-x) for an arbitrary g?

 

[lurflurf]

That should have been symmetric about x=a/2

If
f(x)=(1/2)g(x)+(1/2)g(a-x)
then
f(a-x)=(1/2)g(a-x)+(1/2)g(x)=f(x)
in fact f=g for functions of the type desired

or write any function f
f(x)=(1/2)[f(x)+f(a-x)]+(1/2)[f(x)-f(a-x)]
if
(1/2)[f(x)-f(a-x)]=0
then f is already a function of your type
regardless
(1/2)[f(x)+f(a-x)] is a function of your type and in some sense the function of your type most like f

 

[Ray Vickson]

All you can say about the function is it is symmetric about x=a/2.
What are the domain and range?
If domain were real numbers for example we could define f to be an arbitrary function when x<=a/2 then define f when x>a using the given functional equation.
or just take an arbitrary function g and define
f=(1/2)g(x)+(1/2)g(a-x)=(1/2)g(a/2+(x-a/2))+(1/2)g(a/2-(x-a/2))

There are many functions that are NOT symmetric about x = a/2, but satisfy the OPs equation. All you can say is that the function is periodic, with period a.

RGV

 

[McLaren Rulez]

Ray, could you post an example? Thank you both for replying

 

[Ray Vickson]

Ray, could you post an example? Thank you both for replying

Sorry: I mis-read the original as f(x-a) instead of your f(a-x). So, the statement regarding symmetry about x = a/2 is correct: the function _is_ symmetric. And: it is not (necessarily) periodic. Mea culpa.

RGV