Proving some sup over integrals is finite (could we exchange $\sup$ and $\int$?). Could someone help me to prove that $$I=\sup_x \int_{|x-y|<1}|q(y)|^2 \omega(x-y) \, dy <\infty$$ where $x,y\in \mathbb{R}^n$, $q(y)$ is bounded for all $y$ and $\omega(x-y)$ is a function given by: $\mathbf{i})$ negative powers of $|x-y|$, e.g., $|x-y|^{-1}$ or $\mathbf{ii})$ $\omega(x-y)=\log |x-y|$ or $\mathbf{iii}$ $\omega(x-y)=1$?
What about to exchange supremum and integral?
 A: It's definitely less than infinity, its producing a real number. So therefore its definitely true. However, if it was something like $$I=\sup_x \int_{|x-y|<1}{|q(y)|^2 \omega(x-y) \, dy >\infty}$$ it'd be false.
Also, $\lim_{n\rightarrow \infty} f_n = \lim_{n\rightarrow \infty} \inf (f_n)  = \lim_{n\rightarrow \infty} \sup (f_n) = f$
Its LIMITING to infinity, therefore it's not necessarily infinity. :)
A: i) is false:  On $\mathbb R^1,$ take $q \equiv 1,\omega (x-y) = 1/|x-y|.$ For $x=0$ we get $\int_{-1}^1 1/|y|\, dy = \infty.$
ii) is true for obvious reasons: The integrand is $\le 0$ on the region of integration.
If you want to know the absolute value of the integral is bounded independent of $x,$ you can argue this way: Suppose $|q|\le M.$ Then the absolute value of the integral is bounded above by
$$\int_{|x-y|<1} q^2(y) |\ln |x-y|| \, dy \le M^2\int_{|x-y|<1} |\ln |x-y|| \, dy = M^2\int_{|y|<1} |\ln |y|| \, dy <\infty.$$
Ask if you have questions on why the last integral is finite.
iii) True, and it's very simple.
