# inequality for two integral expressions

Given bounded positive functions $$f, g, h \in L^1$$, with $$g, h$$ finitely supported, I would like to compare the following two expressions: \begin{aligned} a_1 &= \int_{0}^{\infty} \! dx \, f(x) \int_{-\infty}^{\infty} \! dy \, g(y) \int_{-\infty}^{\infty} \! dz \, h(z) \, \frac{\sin[x \, (y-z)]}{\pi \, (y-z)} \\ a_0 &= \int_{0}^{\infty} \! dx \, f(x) \int_{-\infty}^{\infty} \! dy \, g(y) \, h(y) \end{aligned} In other words, $$a_0$$ is $$a_1$$ with the sinc kernel replaced by a delta distribution, so an alternative formulation is to look at the behaviour of $$a_\epsilon = \int_{0}^{\infty} \! dx \, f(x) \int_{-\infty}^{\infty} \! dy \, g(y) \int_{-\infty}^{\infty} \! dz \, h(z) \, \frac{\sin[\tfrac{x}{\epsilon} \, (y-z)]}{\pi \, (y-z)} \\$$ as $$\epsilon$$ goes to zero.

I am interested in conditions for when something can be said about the relative values of $$a_0$$ and $$a_1$$. Ideally, there would be some bounds for $$a_0$$ in terms of $$a_1$$, but even just conditions for when there is a simple inequality would be helpful.

• $a_0$ need not to be finite under $f,g,h \in L^1$, only. Could you clarify the question? Thanks. – Giorgio Metafune Aug 11 at 13:51
• @GiorgioMetafune My mistake, both $g$ and $h$ are bounded and with finite support as well. Their support overlaps so that $a_0$ is not trivially zero. – student Aug 11 at 16:36