Prove if infinite product of $f(x)$ is $0$ then so is infinite product of $f(xvarphi)$

Let is say we want to disprove it. Then we need a $varphi$ for which this product is not zero. One way to obtain that is for $f$ to be $1$ at all integer multiples of $varphi$. So let's decide that $varphi = 2$ and $f$ is $1$ for all even integers. Then we need the values of $f$ on odd integers to give $0$ in the first product. One way is to have $f(text{odd}) = 1/text{odd}$. Sharpening that up, we obtain...

This is false as written. Let $$ f(x) = begin{cases}1 ,& text{$x$ is an even integer } \ 1/x ,& text{otherwise} end{cases} $$

Then $prod_{x=2}^infty f(x) = 0$, but $prod_{x=2}^infty f(2x) = 1$. In fact, for any $varphi geq 2$ an even integer, $prod_{x=2}^infty f(varphi x) = 1$.

Not really: you can consider your favourite function $f$ such that $f(n)=1-frac1n$ for all odd $nin Bbb N$ and $f(n)=1-2^{-n-1}$ for all even $ninBbb N$. Then clearly $prod_n f(n)$ diverges to $0$, whereas $prod_n f(2n)$ converges to a positive number.

This is not true. Consider any continuous function defined over $mathbb{N}$ by $$f(n)=cases{frac12&$n$ odd\1&$n$ even}$$ An explicit example is given by $$f(x)=frac{3+cos{(pi x)}}4$$ Then we have $$prod_{x=2}^infty f(x)=0$$ But $$prod_{x=2}^infty f(2x)=1$$