When is this estimate about L2-norms true?

Question

Let $f$ be a function in $L^2(U)$ where $U$ is some (not necessarily bounded) domain. For example, if $f$ is bounded, then for all $g \in L^2(U)$,

$$\|fg\|_{L^2(U)} \leq \|f\|_{\infty}\|g\|_{L^2(U)} = C\|g\|_{L^2(U)}. $$

Are there any other cases when this is true (i.e $f$ not necessarily bounded) or is this a necessary assumption?

Answer 1

Suppose $f \not \in L^\infty(U)$. Take $n \ge 1$, and let $E_n = \{x \in U : |f(x)| \ge n\}$. Then $\mu(E_n) > 0$, so if we let $g = \frac{1}{\sqrt{\mu(E_n)}}1_{E_n}$, then $||g||_2 = 1$ and $||fg||_2 \ge n$. So there is no $C$ with $||fg||_2 \le C||g||_2$.

Answer 2

No, there are no other examples.

There exists $c$ such that $||fg||_2\le c||g||_2$ for all $g\in L^2$ if and only if $f\in L^\infty$.

Proof: One direction is obvious. For the other direction, suppose $||fg||_2\le c||g||_2$ for every $g\in L^2$. Fix $h\in L^1$ for a minute. There exists $g\in L^2$ with $h=g^2$. Then $$||f^2h||_1=||fg||_2^2\le c^2||g||_2^2=c^2||h||_1.$$So $h\mapsto\int f^2h$ defines a bounded linear functional on $L^1$.

Since we're in a $\sigma$-finite measure space, $(L^1)^*=L^\infty$. So $f^2\in L^\infty$, hence $f\in L^\infty$.