# Cauchy Schwarz inequality for surface integrals?

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.

## 1 Answer

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

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