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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 $f = 0$. – Najib Idrissi Jun 4 '12 at 18:34
Negative constants are ok as long as $A > -0.5$. – soup Jun 4 '12 at 18:41
Why "as long as $A<0.5$"? – AD. Jun 4 '12 at 18:45
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 $\lambda^{n/2}$. – Giuseppe Negro Jun 4 '12 at 18:47
@GiuseppeNegro Thanks for the comment. The domain is a closed surface in $\mathbb{R}^n$. – soup Jun 4 '12 at 18:58
up vote 2 down vote accepted

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 $P$.

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