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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?

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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.

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I understand what you're saying, but not how it helps me, or how to put the proof together. I defined $E_\alpha:=\{x\in X: |g(x)|>\alpha\}$ and $f_\alpha:=\chi_{E_\alpha}$. So yes, if $\mu(X)<\infty$, $f_\alpha\in L^2$. But now what? How do I use the inequality stated in the problem? –  FPP Nov 2 '12 at 0:43
    
You know $\|f\|_2^2 \ge \|fg\|_2^2 = \int |fg|^2 \, d\mu \ge \alpha^2 \int |f|^2 \, d\mu = \alpha^2 \|f\|_2^2$. Does this help? –  Lukas Geyer Nov 2 '12 at 0:51
    
Not really, no, I'm not seeing it (in fact it took me a while to see why that second inequality is true, but anyway). What should I look at? $\int_{E_\alpha} |f|\ d\mu$ for an arbitrary $f\in L^2$? Or $\int_{E_\alpha} |f_\alpha|\ d\mu$? Or what? I get the idea of integrating something positive over $E_\alpha$ and having to see that that is zero for large enough $\alpha$, but don't see how to incorporate the inequality above, or anything really. –  FPP Nov 2 '12 at 1:09
    
In fact, isn't $||f||_2^2\geq \alpha^2||f||_2^2$ impossible when $\alpha>1$? Does that imply that $|\alpha|\leq 1$? Is that relevant at all? –  FPP Nov 2 '12 at 2:46
1  
Exactly, you get that $\alpha \le 1$ from that inequality, and that implies $\|g\|_\infty \le 1$. The inequality is true, because $f$ is the characteristic function of the set where $|g| \ge \alpha$, so on that set $f=1$ and $|g| \ge \alpha$, that is how you get the inequality. And it is in fact only true for the special case where you choose $f$ to be the characteristic function $f_\alpha$. –  Lukas Geyer Nov 2 '12 at 3:55

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