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?

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

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.

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    $\begingroup$ 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. $\endgroup$ – Mario Carneiro Feb 20 '15 at 16:09
  • $\begingroup$ Yes, that's true Mario. $\endgroup$ – Samrat Mukhopadhyay Feb 20 '15 at 16:10
  • $\begingroup$ 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 $f(0)\neq0$. $\endgroup$ – MichaelChirico Feb 20 '15 at 18:35
  • $\begingroup$ @MichaelChirico Probably found it this way: $f(0)=\frac{f(-x)+f(x)}{2}\iff \frac{(f(-x)-f(0))+(f(x)-f(0))}{2}=0\iff f(-x)-f(0)=-(f(x)-f(0))$. $\endgroup$ – user26486 Feb 20 '15 at 23:48

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


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.


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