I would like to verify my solution for the following problem:
Let $g:(0,\infty)\to\mathbb{R}$ be a nonzero, continuous function and define $G:[0,1]\times[1,\infty)\to\mathbb{R}$ by
$$G(x,y)=g(xy).$$ Show that $G$ is not in $L^1([0,1]\times[1,\infty))$.
Solution:
$$\int_1^\infty \int_0^1|G(x,y)|dx dy = \int_1^\infty \int_0^1|g(xy)|dx dy = \int_1^\infty \int_0^y\cfrac{|g(t)|}{y}dt dy \geq \int_1^\infty \cfrac{1}{y} \inf_{t\in(0,y]} |g(t)|\; dy \geq \inf_{t\in(0,\infty)} |g(t)|\int_1^\infty \cfrac{1}{y} dy = \infty$$
I'm a little concerned about taking the infimum over the non compact set $(0,y]$ ($g$ is not defined at $0$).
Is this correct? And if so, how do I address my concern above? Thanks for your time.
Choose $T$ such that $\int_0^{T}|g(t)|dt >0$. Note that $\int_1^{\infty}\int_0^{y} \frac {|g(t)|} y dt dy \geq \int_T^{\infty}\int_0^{y} \frac {|g(t)|} y dt dy$. Now use the fact that $\int_0^{y}|g(t)|dt \geq \int_0^{T}|g(t)|dt$ for $y >T$. Can you finish?