# if a function is bounded then both its even part and odd part are bounded?

Assuming that bounded function $f(x)$ is defined by $|f(x)|<\infty$ for all $x$. We know the function can be defined as sum of even and odd function, therefore I was wondering if it's true to say that if a function is bounded then both its even part and odd part are bounded.

My thought is that it's true. if we assume that the odd part is not bounded in a point, and also assume that somehow the sum with the even part can "bounded" the sum (in general the sum of two unbounded numbers is undefined), then it can be true only for one side of the X axis. and my thought was the same for all other possibilities.

If I'm wrong, please correct me. Moreover, if it's true, then it will be nice to have a real proof.

• Do you know about the following decomposition: $$f(x) = \frac{f(x) - f(-x)}{2} + \frac{f(x) + f(-x)}{2}$$ ? – Onil90 Apr 17 '18 at 8:21
• I knew from long past, didn't thought about using it, thank you :) – Mr.OY Apr 17 '18 at 8:24

$\left|\dfrac{f(x)+f(-x)}{2}\right|\leq\dfrac{|f(x)|+|f(-x)|}{2}\leq\dfrac{M+M}{2}=M$, where $\sup|f|:=M<\infty$.
Assuming that $m\le f(x)\le M$ we have
• $a(x)=\frac{f(x)+f(-x)}{2}\implies m'\le a(x)\le M'$
• $b(x)=\frac{f(x)-f(-x)}{2}\implies m''\le b(x)\le M''$