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

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 $0$.
