Check proof that if a function is even with respect to 2 distinct values of $x$ then it is periodic

Question

I'm trying to prove that:

If a function $f(x): \mathbb R \to \mathbb R$ is even with respect to $x = a,b \in \mathbb R;\; a\ne b$ then it is periodic with period $T_{min} = 2|a-b|$

Here is what i made:

Consider some $x_1 = a$, $f(x)$ is symmetric with respect to $a$ therefore $f(x) = f(2a - x)$, but at the same time the function is odd about some $x_2=b$ and hence $f(x) = f(2b-x)$. So:

$$ f(x) = f(2a-x)\\ f(x) = f(2b-x)\\ $$

So we may write:

$$ f(2a-x) = f(2b-x) $$ On one hand:

$$ f(2a - x)= f(2b-x + 2a - 2b) \iff T=2(a-b) $$

On the other:

$$ f(2b-x) = f(2a-x+2b-2a) \iff T=2(b-a) $$

Since both periods satisfy the equations that means the only solution is $T=2|a-b|$.

update:

To check that $T$ is indeed period of$f(x)$ let $b > a$, using initial conditions that $f(2a-x) = f(2b-x)$: $$ f(x+T) = f(2b - (x-T)) = f(2b-x-2b+2a)=f(2a-x)=f(x) \; \Box $$

Answer 1

$$f(x)=f(2a-x)$$therefore by substituting $x\to 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$\Large T_{\min}=2|a-b|$$

Answer 2

Once you have $f(2a-x) = f(2b-x)$, put $y = 2a-x$. Then $x = 2a-y$ so that $2b-x =2b-(2a-y) =y+2(b-a) $ so $f(y) = f(y+2(b-a))$.