# Pointwise multiplication by unbounded function throws us out of $L^2(\mu)$

Let $$(X,\mathcal{A},\mu)$$ be a $$\sigma$$-finite measure space and let $$\phi$$ be a measurable function that is not an element of $$L^\infty(\mu)$$, i.e. $$\phi\not\in L^\infty(\mu)$$. I am trying to construct a function $$g\in L^2(\mu)$$ such that $$\phi\cdot g\not\in L^2(\mu)$$.

I tried partitioning $$X$$ in the sets $$A_k=\{k\leq|\phi| so i could somehow control the behavior of $$g$$ on those sets relative to $$\phi$$, but I needed something more. I considered a partition of $$X$$ in disjoint sets of finite measure say $$(X_n)$$ and then I tried to take their co-partition. The problem arising is that I cannot determine which of the sets $$A_k\cap X_n$$ are of non-zero measure. I am really stuck here. Any ideas?

Replacing $$g$$ with $$\frac{g\phi}{|\phi|}$$ on $$X$$, which doesn't change the $$L^2$$ norm of $$g$$ we can first treat the case $$\phi=|\phi| \ge 0$$ and then use the above to complete the general case; writing $$a_k=\mu(A_k)$$, the hypothesis gives $$a_k$$ non zero for infinitely many $$k$$; for any $$a_k= \infty$$, we replace $$A_k$$ with a subset of finite positive measure $$b_k$$ which exists by sigma finitness, and make $$g$$ zero outside that on $$A_k$$, so we can actually assume $$a_k$$ finite to start with.
Then for all $$k>0$$ with $$a_k \ne 0$$, let $$g= \frac{1}{k\sqrt{a_k}}$$ on the corresponding $$A_k$$ and zero everywhere else.
It is obvious that the integral of $$g^2$$ is finite, being dominated by $$\zeta(2)$$, while the integral of $$(\phi g)^2$$ is at least an infinite sum of $$1$$'s, so infinite.
• Oh, wow. I didnt think of replacing $A_k$'s, which created a whole lot of trouble for me and I let that idea pass. Anyway, thank you very much! – JustDroppedIn May 27 at 14:15