Error in the proof for scaling the period of a function that leads to $f(x + p) = f(x) \Rightarrow f(x + p / k) = f(x)$

Question

The question is about scaling the period of a periodic function. This question has been asked and answered. What bothers me is something else:

For $k \ne 0$

$f(x + p) = f(x) \Rightarrow f(kx + p) = f(kx)$ (1)

$f(kx) = f(k (x + p / k))$ (2)

$f(x) = f(kx / k) = f(x + p / k)$ (3)

I must have misunderstood something, because (3) is definitely not valid for every $k\ne0$. Counter example for (3): $cos(0) \ne cos(0 + 2\pi/100)$.

Answer 1

You're mixing up $f$ and $g$ from that answer. It's true that $g(0) = g(0 + 2\pi/100)$ does not hold for $g = \cos$, but it holds for $g(x) = \cos(100x)$.

Answer 2

$(3)$ is not valid of course, because it claims that $f$ is $\frac{p}{k}$-periodic $\forall k \ne 0$, which is cleary an absurd condition.

On the contrary the new function $g(x) := f(kx)$ has period $\frac{p}{k}$ because $g$ is simply a dilation/contraction of $f$: $$g\left(x+\frac{p}{k}\right) = f\left(k\left( x+\frac{p}{k}\right)\right) = f\left(k x+p\right) = f\left(k x \right) = g(x)$$