My Math Forum Equivalence of integrals

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 September 9th, 2013, 11:50 AM #1 Member   Joined: Aug 2012 Posts: 32 Thanks: 0 Equivalence of integrals The following question is simpler than it initially looks, its about the equivalence of two integrals which are slightly different, but I thought it best to give some detail about where they come from in case I am missing something that might influence the question. Let me know if any detail is required. By the way this is based on notes from the book Functional Anlaysis, Sobolev Spaces by Brezis, section 'Extension Operators'. Given $x \in \mathbb{R^{N}}$, write $x= (x#39;, x_{N})$ with $x' \in \mathbb{R^{N-1}},\text{~~~} x' = (x_{1},x_{2}, ..., x_{N-1})$ and set $|x'| = (\sum_{i=1}^{N-1}x_{i}^{2})^{1/2}$. We define $\mathbb{R_{+}^{N}}= \lbrace x = (x#39;, x_{N});\text{~~}x_{N} > 0 \rbrace$ $\text{~~~~~~~~~~~~~}\mathbb{R_{-}^{N}}= \lbrace x = (x#39;, x_{N});\text{~~}x_{N} < 0 \rbrace$ $\text{~~~~~~~~~~~~~}Q= \lbrace x = (x',x_{N});\text{~~} |x#39;| < 1 \rbrace$ $\text{~~~~~~~~~~~~~}Q_{+}= Q \cap \mathbb{R_{+}^{N}}$ $\text{~~~~~~~~~~~~~}Q_{-}= Q \cap \mathbb{R_{-}^{N}}$ Q can be thought of as a cylinder in way ("in a way" since its N dimensional). Let $u \in L^{P}(Q_{+})$ and $\phi \in C_{c}^{\infty}(Q)$. Consider the following extension of $u$ to all of $Q$: $u*(x)= u(x', x_{N}) \text{ if } x_{N} > 0 \text{ and } u(x=#39;,-x_{N}) \text{ if } x_{N} < 0$ Consider the following integral: $\int_{Q_{-}}u(x', -x_{N})\frac{\partial \phi}{\partial x_{N}}(x', x_{N})dx$ change of variable $y= (x#39;,-x_{N})$ gives us: $\int_{Q_{+}}u(y', y_{N})\frac{\partial \phi}{\partial x_{N}}(y', -y_{N})dy$ can we then state that: $\int_{Q_{+}}u(y', y_{N})\frac{\partial \phi}{\partial x_{N}}(y', -y_{N})dy =$$\int_{Q_{+}}u(y', y_{N})\frac{\partial \phi}{\partial y_{N}}(y', -y_{N})dy$? If so why would this be the case? Thanks a lot, let me know if any extra detail is required.
 September 16th, 2013, 12:36 PM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Equivalence of integrals The summary of what u're saying is that $x_N= -y_N$, in which case you would have to include a minus sign on your last equation line. Now I have a question for you - how much do you know or have acquired interest for partial differential equations with Neumann boundary condition?
 September 18th, 2013, 09:15 AM #3 Member   Joined: Dec 2006 Posts: 65 Thanks: 0 Re: Equivalence of integrals I have a question. ? ? Qc? (Qv) so ? and it’s partial derivatives are continuous then they are Borl measurable .In this case the integrand is measurable but is that integrable ?
 September 18th, 2013, 10:31 AM #4 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Equivalence of integrals Yes, because $\phi$ is in $C_c^{\infty}(Q)$ so the function is both $C^{\infty}$ and compactly supported.
 September 19th, 2013, 02:45 AM #5 Newbie   Joined: Sep 2013 Posts: 10 Thanks: 0 Re: Equivalence of integrals This might be a silly question, but is your reason why the integrals in the first post should be equivalent if $y= (x#39;, -x_{N})$ simply because $dx_{N}= -dy_{N}$ and therefore $\frac{\partial \phi}{\partial x_{N}}= -\frac{\partial \phi}{\partial y_{N}}$ ?

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