# Does null average against every smooth function implies independence?

Are these assertions equivalent?

• $$f:\mathbb{S}^1\times \mathbb{S}^1\to\mathbb{C}$$ is such that $$\int_0^{2\pi}\int_0^{2\pi}f(x,y)\psi(y)dydx=0$$ for all $$\psi\in C^{\infty}(\mathbb{S}^1).$$

• $$f:\mathbb{S}^1\times \mathbb{S}^1\to\mathbb{C}$$ is such that that $$f$$ does not depend on $$y$$, that is, $$f(x,y) = f(x,0),\, \forall y\in\mathbb{S}^1$$, and $$\int_0^{2\pi}f(x,y)dx = 0.$$

It's clear that the second implies the first, though I'm having some difficulty to prove that the first implies the second. My idea was trying to prove that $$f(x,y) - f(x,0)$$ is the null function, by contradiction: suppose it isn't, then taking an appropriate $$\psi$$ to arrive at a contradiction, though I ran into some problems trying to fit $$f(x,0)$$ inside the integral...

No the two assertions are not equivalent. Take $$f(x,y) = ye^{ix}$$. Then for every $$\def\S{\mathbb{S}^1}\psi \in C^\infty(\S)$$, we have $$\int_{[0,2\pi]^2} f(x,y)\psi(y)\,dxdy = \int_0^{2\pi} \psi(y)\left(\int_0^{2\pi} f(x,y)\, dx\right) \,dy = 0$$ since $$\int_0^{2\pi} f(x,y) \, dx = 0$$ for every $$y$$. But obviously $$f$$ depends on $$y$$.
Assertion $$1$$ should be equivalent to:
• $$f \colon \S \times \S \to \mathbb{C}$$ satisfies $$\int_0^{2\pi} f(x,y)\, dx = 0$$ for every $$y \in \S$$.
• Use the fact that if $g \in L^1(\mathbb{S}^1)$ is such that $\int_0^{2\pi} g(y)\psi(y) \, dy = 0$ for every $y \in \mathbb{S}^1$ then $g = 0$ a.e. The idea is to apply this to $g(y) = \int_0^{2\pi} f(x,y)\, dx$ but you need some regularity on $f$ to ensure that $g$ is integrable, for example $f \in L^1(\mathbb{S}^1 \times \mathbb{S}^1)$. @MathNewbie – Michh Jun 3 at 22:38