Consider
$f : [0, 1] \to [0, \infty)$ and$h : [0, 1] \to \mathbb{R}$,i.e., $h$ can be both positive and negative.Define $A = [0,1] \times [0,1]$.Consider the following integrals:\begin{align*}I_{1}& = \int_{A}f(x) f(y) \, k(x,y) \, dx dy, \\I_{2}& =\int_{A} f(x) f(y) h(y) k(x,y) dx dy, \\I_{3} & =\int_{A} f(x) f(y) h(x) h(y)k(x,y) dx dy,\end{align*}where $k : A \to \mathbb{R}$ isa positive semidefinite functionwith $k \ge 0$.Then how can we prove that$$I_{2}^{2}\le I_{1} I_{3}?$$
That is, consider$$g_{1}(x,y) = \sqrt{f(x) f(y) k(x,y)}\quad \mbox{and}\quad g_{2}(x,y) = \sqrt{f(x) f(y) k(x,y)}h(y).$$
Then from Cauchy-Schwarz Inequality,we have\begin{align*}\left( \int_{A} g_{1}(x,y) g_{2} (x,y) dx dy \right)^{2}& \le \left( \int_{A} g_{1}^{2}(x,y) dxdy \right) \left( \int_{A} g_{2}^{2} (x,y) dx dy \right) \\\mbox{i.e., }I_{2}^{2}& \le I_{1} \left( \underbrace{\int_{A} f(x) f(y) k(x,y) h^{2}(y) dxdy}_{=:B} \right) . \end{align*}Now, I do not see how $B = I_{3}$,or $B \le I_{3}$.