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 $A_k=\{k\leq|\phi|<k+1\}$ 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.

  • $\begingroup$ 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! $\endgroup$ – JustDroppedIn May 27 at 14:15
  • $\begingroup$ You are welcome $\endgroup$ – Conrad May 27 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.