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

I think $g(x)$ is odd.

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