What functions satisfy such equation?

Question

Let $f:(-a,a)\rightarrow \mathbb R$ be a continuous function such that $$ f(0)=\frac{f(-x)+f(x)}{2} \textrm{ for } |x|<a. $$

What about $f$? Is it necessarilly an odd function?

...what about $f$? What do you want to know about $f$? (Other than whether or not it's odd.) — Gyu Eun Lee, Feb 20 '15 at 15:49
I want to know a general form of solution of that equation with unknow function. — user172903, Feb 20 '15 at 15:50

score 17 · Accepted Answer · answered Feb 20 '15 at 15:53

17

One thing is for sure, for $x\in (-a,a)$, $g(x):=f(x)-f(0)$ is an odd function. Apart from that nothing else seems to be evident.

answered Feb 20 '15 at 15:53

Samrat Mukhopadhyay

16,277

8

And more to the point, any function of the form $f(x)=f(0)+g(x)$ with $g$ odd is a solution, so this completely classifies the set of solutions. – Mario Carneiro Feb 20 '15 at 16:09
Yes, that's true Mario. – Samrat Mukhopadhyay Feb 20 '15 at 16:10
I would add that if it's not clear how you found these solutions, the OP should try a simple case: $f(0)=0$. Then draw pictures for $f(0)\neq0$. – MichaelChirico Feb 20 '15 at 18:35
@MichaelChirico Probably found it this way: $f(0)=\frac{f(-x)+f(x)}{2}\iff \frac{(f(-x)-f(0))+(f(x)-f(0))}{2}=0\iff f(-x)-f(0)=-(f(x)-f(0))$. – user26486 Feb 20 '15 at 23:48

score 8 · Answer 2 · answered Feb 20 '15 at 15:51

8

It doesn't have to be odd. Consider: $f(x)=x+1$

answered Feb 20 '15 at 15:51

rafalpw

600

score 2 · Answer 3 · edited Feb 20 '15 at 18:15

2

f is not necessarily an odd function. Consider any constant function: f(x) = c. Any constant function satisfies the property you stated, but does not satisfy the properties of odd functions: f(-x) = - f(x) and f(x) + f(-x) = 0.

edited Feb 20 '15 at 18:15

Community

1

answered Feb 20 '15 at 16:59

Lang

36

What functions satisfy such equation?

3 Answers3