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Let $\Omega $ a bounded smooth surface with boundary of $\mathbb R^n$. What does $$f|_{\partial \Omega }\in L^p(\partial \Omega )\ \ ?$$ Is it $$\int_{\partial \Omega }|f|_{\partial \Omega }|^p<\infty \ \ ?$$ To be honnest, I don't really understand the meaning of $$\int_{\partial \Omega }|f|_{\partial \Omega }|^p,$$ because $f:\mathbb R^n\to \mathbb R$ and here we integrate on "$n-1$" value.

Henri
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  • See http://mathworld.wolfram.com/Lp-Space.html – Winther Sep 03 '18 at 17:08
  • @Winther: I know what are $L^p$ space, it's not my question. Sorry for the typo, I corrected it :) – Henri Sep 03 '18 at 17:18
  • Yes that is the definition of being in $L^p$. That the integral exists – Winther Sep 03 '18 at 17:19
  • what mean $\int_{\partial \Omega }|f|\Omega |^p$ ? Because $\partial \Omega $ has measure $0$, so $f|{\partial \Omega }$ could be anything (and in particular $0$)... Moreover, $f$ has $n$ variable, si $\int_{\partial \Omega }f$ looks strange since $\partial \Omega $ has dimension $n-1$... – Henri Sep 03 '18 at 17:20
  • @Winther: If $f\in {L^1}(\mathbb R^2)$ then on a set of measure $0$ the function $f$ is not well defined (it can be everything). Indeed, let $g=0$ in $\partial \Omega $ and $g=f$ in $\mathbb R^2\backslash \partial \Omega $, then $f=g$ in $L^2$, and thus $\int_{\partial \Omega }f=\int_{\partial \Omega }g=0$. – Henri Sep 03 '18 at 17:24
  • @Winther: Henri is correct, a function in $L^p$ is defined almost every only (it may no sense to specify the values of a function on a set of measure $0$). But here $f|_{\partial \Omega }$ is the trace of $f$ and is defined as a limit (see my answer). – Surb Sep 03 '18 at 17:33
  • @Surb Ah my bad, I did not see the tag. I read this as a PDE question where $f|_{\partial \Omega}$ is the boundary condition. – Winther Sep 03 '18 at 17:35

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Here thing are different. Do you know trace operator ? $\mathcal C^0(\bar \Omega )$ is dense in $L^p(\Omega )$. Let $(f_n)$ a sequence of $\mathcal C^0(\bar \Omega )$ functions that converge to $f$ in $L^p$. Then, by definition $$f|_{\partial \Omega }=\lim_{n\to \infty }f_n|_{\partial \Omega },$$ in $L^p(\partial \Omega )$ sense, that mean $$\lim_{n\to \infty }\int_{\partial \Omega }|f-f_n|=0.$$ For your second question, notice that $\partial \Omega $ is a subset of $\mathbb R^n$, and thus $\int_{\partial \Omega }f|_{\partial \Omega }$ make sense.

Surb
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  • The set of continuous functions with compact support is also dense in $L^p(\Omega)$. That definition would imply that $f|_{\partial\Omega}=0$ for all $f\in L^p$. – Julián Aguirre Sep 03 '18 at 17:55
  • @JuliánAguirre: Thank you for your comment, rigorously you are totally right. I just wanted to give a motivation on how to define $f|_{\partial \Omega }$ (or an illustration on how it could work). Indeed, it's a little more subtly and require Sobolev spaces to define the trace properly (because $\mathcal C^1(\bar \Omega )$ is dense in $W^{1,p}(\Omega )$ whereas $\mathcal C^{1}(\Omega )$ with compact support is not). – Surb Sep 03 '18 at 20:33