If $f$ is odd and periodic then a translation of $f$ is even?

Question

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a odd and periodic function, with period $L>0$. If we define $$g(x):=f\left(x-\frac{L}{2}\right), \; \forall \; x \in \mathbb{R},$$ then $g$ is even?

I tried to prove it, as follows: let $x \in\mathbb{R}$ arbitrary. Thus, $$g(-x)=f\left(-x-\frac{L}{2}\right)=f\left(-\left(x+\frac{L}{2}\right)\right)=-f\left(x+\frac{L}{2}\right).$$

But I couldn't conclude that $g(-x)=g(x)$.

Is this true in general? What did I do is right?

Answer 1

From your last line, $$g(-x)=-f(x+L/2)$$ $$=-f(x-L/2+L)$$ $$=-f(x-L/2)=-g(x)$$ $g$ is in fact odd.

Answer 2

I think $g(x)$ is odd.

$g(-x)=f(-x-L/2)=f(-x+L/2)=-f(x-L/2)=-g(x)$

Answer 3

If you want to get even function, then for $\left( -\frac{L}{2}, \frac{L}{2} \right)$ let's define: $$ g(x)=\begin{cases} f(x), & x>0 \\ -f(x), & x<0 \end{cases} $$