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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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    You will never find such positive constants, for starters. Take $g = 0$, then $\int f^2g = 0$ but $\|f\|_{L^2}^2$ can be arbitrarily large. Same if you take $f = 0$.2012-06-04
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    Negative constants are ok as long as $A > -0.5$.2012-06-04
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    Why "as long as $A<0.5$"?2012-06-04
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    Which $L^p$ spaces are you referring to? $L^p(\mathbb{R}^n)$? If this is the case then there is some scaling problem, I'm afraid that this inequality can never hold. Indeed, if you substitute $f(x)\to f(\lambda x)$, $g(x)\to g(\lambda x)$, then the right-hand side scales as $\lambda^n$ and the left-hand-side (which is homogeneous only if $c=p/2$) scales as $\lambda^{n/2}$.2012-06-04
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    @GiuseppeNegro Thanks for the comment. The domain is a closed surface in $\mathbb{R}^n$.2012-06-04
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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 $P$.