Is it true that $ \int_Sf(x)g(x)dx \leq \int_Sf(x)dx\int_Sg(x)dx \ \forall f(x),g(x)\geq0 $ ?
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3In general, no. You could try some easy examples. Why this question? Are you trying to do something else? – Jan 17 '22 at 17:04
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Look up "Cauchy-Schwarz Inequality" for a correct version... – David C. Ullrich Jan 17 '22 at 17:09
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Please consider adding some more details around your claim. A similar inequality , which goes the opposite way (but can be made to go your way with a simple tweak) is Harris' inequality. It is definitely not true for merely positive $f,g$. – Sarvesh Ravichandran Iyer Jan 17 '22 at 17:19
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Left hand side doesn't have to be finite if we merely assume $g,f \in L^1$ are positive. – Presage Jan 17 '22 at 18:00
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Take $S$ to be the interval $[0,\epsilon]$, Take $f,g$ to be the functions that send $x$ to $x$. Then $$\int^{\epsilon}_0f(x)dx=\int^{\epsilon}_0xdx={\epsilon}^2/2$$
$$\int^{\epsilon}_0g(x)dx=\int^{\epsilon}_0xdx={\epsilon}^2/2$$
$$\int^{\epsilon}_0f(x)g(x)dx=\int^{\epsilon}_0x^2dx=\frac{{\epsilon}^3}{3}$$
However $\frac{{\epsilon}^3}{3}\leq {\epsilon}^2/2\times {\epsilon}^2/2$ is false for small ${\epsilon}$. It is false if $\epsilon$ equals $10^{-6}$ for example.
