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

Question

Prove or disprove that if $$\prod\limits_{x=2}^{\infty} f(x)=0$$ and $f(x)\neq0$ for any $x\geq0$ then $$\prod\limits_{x=2}^{\infty} f(x\varphi)=0$$ for any constant $\varphi\geq2$

This seems true but I'm not quite sure how to prove it since the constant is inside a function f.

Answer 1

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 $$\prod_{x=2}^\infty f(2x)=1$$

Answer 2

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

Answer 3

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, $\prod_{x=2}^\infty f(\varphi x) = 1$.