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For square integrable functions $f, g$, the Cauchy-Schwarz inequality says,

$$ \left(\int_{\mathbb{R}^n}f(x)g(x)dx\right)^2 \leq \left(\int_{\mathbb{R}^n}f^2(x)dx\right)\left(\int_{\mathbb{R}^n}g^2(x)dx\right) $$

I was wondering if there was some surface integral equivalent to this theorem. For instance, suppose $B \subset \mathbb{R}^n$ is some ball in $\mathbb{R}^n$. Then, would the following "Cauchy-Schwarz" inequality still work?

$$ \left(\int_{\partial B}fgdS\right)^2 \leq \left(\int_{\partial B}f^2dS\right)\left(\int_{\partial B}g^2dS\right) $$

where $\partial B$ represents the surface of $B$. I haven't had luck finding this in literature anywhere.

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For general measure $\mu$ (positive measure) one has $\|fg\|_{L^{1}(\mu)}\leq\|f\|_{L^{2}(\mu)}\|g\|_{L^{2}(\mu)}$.

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  • $\begingroup$ Search Holder's inequality in Wikipedia for details. $\endgroup$ – user284331 Dec 8 '17 at 3:48
  • $\begingroup$ In this case, is our measure the surface measure, dS? $\endgroup$ – Flowsnake Dec 8 '17 at 3:52
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    $\begingroup$ Yes, the surface measure is a positive measure. $\endgroup$ – user284331 Dec 8 '17 at 3:53

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