Looking for an inequality relating $\int_Efg$ to the integrals $\int_Ef$ and $\int_Eg$

Question

Let $f,g:E \to [0, \infty]$ be nonnegative, integrable functions. (I mean, Lebesgue integrable. $E \subseteq \mathbb{R}$.) Assume that $fg$ is also integrable.

I'm trying to look for an inequality relating $\int_Efg$ to the integrals $\int_Ef$ and $\int_Eg$. I'm thinking perhaps $\int_Efg \leq (\int_Ef)(\int_Eg)$, but this seems "too good to be true." If so, then what's the next best thing?

Just to give some motivation, I'm trying to prove a lemma for a bigger proof. I'm assuming $\int_Ef, \int_Eg<\epsilon$ and trying to prove $\int_Efg< \epsilon$. (Or $\int_Efg< \epsilon^2$, or something nice like this. Doing a change of variables, it won't really matter for my purpose.)

I did some research and came across the Cauchy-Schwarz inequality for integrals (before, I only knew the versions for sums and dot products) but I tried manipulating it and I don't think it gives what I'm looking for.

Answer 1

Instead of looking $fg$, if you do convolution $f\ast g$, you have \begin{align*} \int_{\mathbb{R}}f\ast g=\left(\int_{\mathbb{R}}f\right)\left(\int_{\mathbb{R}}g\right). \end{align*} This is the cloest form regarding product of two integrals, I think.

Answer 2

There is no such bound, in general. For example, when $E=[0,1]$ you can consider functions $f,g$ both equal to $n^2$ on the interval $[0,1/n^3]$ and equal to $0$ everywhere else. Then $\int f=\int g=1/n$ whereas $\int fg=n$. Taking $n\to\infty$ shows that we can keep $\int f$ and $\int g$ arbitrarily small while at the same time making $\int fg$ arbitrarily large.