upper bound on the integral of a product of non-negative functions

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i have two functions $G_1(\theta,\phi)$, $G_2(\theta,\phi)$,

$G_{1,2}: [0,\pi] \times [0,2\pi] \to \mathbb{R}^+_0$

i know that both functions satisfy that: \begin{equation*} \int_0^\pi\int_0^{2\pi}G_{1,2}(\theta,\phi)\,d\phi\, d\theta = 4\pi \end{equation*} and i'm trying to find an upper bound B, for the integral of the product of both functions over their domain: \begin{equation*} \int_0^\pi\int_0^{2\pi}G_1(\theta,\phi)G_2(\theta,\phi) \, d\phi \, d\theta \leq B \end{equation*}

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There is no such upper bound.

Let $h\in (0,\pi]$ and $$ G_1(\theta,\phi) = G_2(\theta,\phi) = \begin{cases} 0 & \text{if} & (\theta,\phi)\not\in (0,h)\times (0,2\pi) \\ \frac{2}{h} & \text{if} & (\theta,\phi)\in (0,h)\times (0,2\pi) \\ \end{cases} $$ Then $$ \int_0^{\pi}\int_0^{2\pi} G_1(\theta,\phi) G_2(\theta,\phi) \, d\phi \, d\theta = \int_0^h\int_0^{2\pi} \left(\frac{2}{h}\right)^2 \, d\phi \,d\theta \\ = h\cdot 2\pi \cdot \left(\frac{2}{h}\right)^2 = \frac{8\pi}{h} $$ which becomes arbitrarily large as $h$ approaches $0$.