Do $\sum_{x,y} f(x)g(x,y) = 1$ and $\sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $\sum_{y} g(x,y) = 1$?

Question

Do $\sum_{x,y} f(x)g(x,y) = 1$ and $\sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $\sum_{y} g(x,y) = 1$? If so, how can I proof that?

Note: I am trying derive that $g(x,y)$ is a conditional probability without assuming it.

Previous mistake in the question removed.

Answer 1

Of course not.

If $f(x)=1$ then you're saying that $\sum_{x,y} g(x,y) = 1$ implies that $\sum_y g(x,y)=1$. But this is only possible if there is only one $x$.

Edit: This is an answer for the edited version. Still the answer is no.

Suppose there are two $x$'s and two $y$'s then:

Let $f(x_1)=\frac{1}{2}$ and $f(x_2)=\frac{1}{2}$ then $\sum_x f(x) = 1$.

Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$

Then $\sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.

However $$\sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= \frac{1}{2}\cdot 0.2 + \frac{1}{2}\cdot 1.8 = 1$$

Answer 2

There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:

Let $f(n)=\frac{1}{n}$ and $g(n,m)=n\cdot2^{-n}\cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.

EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.

Let $f(n)=\frac{6}{\pi^2n^2}$ and $g(n,m)=\frac{\pi^2n^2}{6}\cdot2^{-n}\cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.