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My last question hasn't got any replies so I'll try another..

Is there a way to split the following integral ($g$ is arbitrary) $\int{f^2g}$ so that I instead have an expression involving the $L^2$ norm on $f$ and either $L^2$ or $L^\infty$ norm on $g$? In particular, I want something like $A\lVert f \rVert_{L^2}^2 + B\lVert g \rVert^c_{L^p} \leq \int{f^2g}$ where $A$ and $B$ and $c$ are constants. I tried to use Holder's inequality but that gives me an upper bound instead.

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    @AD. This question comes from my previous thread (http://math.stackexchange.com/questions/153263/a-priori-estimate-help-something-to-do-with-gronwall-lemma). I'm not even sure what I want now. In the linked thead, I want to get rid of the third integral in the first equation. Basically I can bound $g$ in the $L^\infty$ norm and want to use that fact.2012-06-04

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As you have noticed, it's not possible for $A$ or $B$ to be positive (just set $f=0$ or $g=0$). So we're looking for nonnegative $a,b$ such that:

$\forall g,-a\|f\|^2_{L^2} - b\|g\|^c_{L^p} \leq \int f^2 g$

Since this has to hold for all $g$, we can take $-g$, and therefore

$\forall g, \int f^2 g \leq a\|f\|^2_{L^2} + b\|g\|^c_{L^p}$

is what we're looking for. Since $f^2 \geq 0$ and $g \leq |g|$, we have that $f^2g \leq f^2|g|$, and therefore:

$\int f^2 g \leq \int f^2 |g| \leq \|g\|_\infty \int f^2 \leq \|g\|_\infty \|f\|^2_{L^2}$

And using the famous trick $ab \leq {a^2+b^2 \over 2}$, we get:

$\int f^2g \leq \frac12\|g\|_\infty^2 + \frac12\|f\|^4_{L^2}$

And I don't think you'll do better than that, for scaling reasons as Giuseppe mentioned. And unfortunately here the constant is $0.5$. But if you really want $a < 0.5$, you can use the just-as-famous trick $ab \leq \frac12 \left(\frac1Pa^2+Pb^2 \right)$ for nonzero $P$.