Bounding $\int_{\mathbb{R}_{+}}W(x,\lambda)W(y,\lambda)d\lambda$

Question

Let $$W:\mathbb{R}_{+}^{2}\rightarrow(0,1]$$ be a symmetric, integrable function. Let $$f(x)=\int_{\mathbb{R}_{+}}W(x,\lambda)d\lambda$$ and assume this function is monotonically non-increasing. Is there a constant C (depending on W but independent of x and y) such that $$\int_{\mathbb{R}_{+}}W(x,\lambda)W(y,\lambda)d\lambda\le Cf(x)f(y) \hspace{5mm} \mathrm{?}$$

It's relatively easy to see that this is really a question about the tail behaviour of the functions on the right and left hand sides. Constructing examples where the left hand side dies off much faster than the right hand side seems straightforward, but I haven't had any luck with the converse.

Thanks!

Answer 1

Counterexample:

$$ W(x,\lambda)= \begin{cases} 1 & x+\lambda \leq 1\\ 0 & x+\lambda >1 \end{cases} $$

Now your inequality for $x=y$ takes the form $(1-y)\leq C(1-y)^{2}$ which does not hold as $y \to 1$.

Of course you can modify this example to achieve the condition $W: \mathbb{R}^{2}_{+} \to (0,1].$