Show a function is in $L_\infty$ Let's assume we're working on a measure space $(X,\Sigma,\mu)$, where $\mu$ is a $\sigma$-finite measure.  Suppose that $g$ is a measurable function such that  $\forall f\in L^2$, $||fg||_2\leq ||f||_2$.  Now, show that $g\in L^\infty$ and find $||g||_\infty$. 
First thing I tried was to do it by contradiction (i.e. assume that $\forall M\in\mathbb{R}^+$, $\mu\Big(\{x\in X: g(x)>M\}\Big)>0$), and tried to play with some inequalities, but didn't get anywhere. Also tried setting $M>||f||_2$, and dealing with the two sets $E_M:=\{x\in X: g(x)>M\}$ and $X\setminus E_M$ in that case, but couldn't do it.  I was aiming to show that that would lead to $||fg||_2>||f||_2$. 
It seems like a question that would use one of the common inequalities (likely Hölder's,  but Minkowski's or Young's too), I just couldn't put it together.
Any ideas?

You want to show that $g$ is essentially bounded, so you want to integrate something positive over the set where $|g| > M$ and see that this has to be $0$ if $M$ is large. This might give you the idea to choose for $f$ the characteristic function of the set where $|g|>M$. This will already give you the result on a finite measure space (where characteristic functions are always in $L^2$.) The last step is to conclude that this will also hold on $\sigma$-finite measure spaces, and that one is standard.
