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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...

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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 $y \in \S$.
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    $\begingroup$ 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 $\endgroup$ – Michh Jun 3 at 22:38

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