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Let

  • $\lambda$ denote the Lebesgue measure on $\mathbb R^2$
  • $\tilde\Delta,\Delta\subseteq\mathbb R^2$ be nonempty and compact
  • $F:\tilde\Delta\to\Delta$ with $$F(\tilde x)=A\tilde x+b\;\;\;\text{for all }\tilde x\in\tilde\Delta\tag1$$ for some invertible $A\in\mathbb R^{d\times d}$ and $b\in\mathbb R^d$

If $v\in L^1(\Delta)$, we obtain $$\int_\Delta v\:{\rm d}\lambda=\int_{\tilde\Delta}\left(v\circ F'\right)\left|\det F'\right|\:{\rm d}\lambda\tag2$$ (where $F'=A$, but that's not important in the sequel) via integration by substitution. Now, if $v$ admits a weak gradient $\nabla v\in L^1(\Delta,\mathbb R^d)$, I want to express $$\int_\Delta\nabla v\:{\rm d}\lambda\tag3$$ in the same way in terms of an integral over $\tilde\Delta$.

How do we need to apply integration by substitution for $(3)$?

I was quite sure that the solution is $$\int_\Delta\nabla v\:{\rm d}\lambda=\int_{\tilde\Delta}\left(\nabla v\circ F\right)\left|\det F'\right|\:{\rm d}\lambda\tag4\;.$$ However, I've read (for example, in Remark 5.45 of a lecture note) that we have $$\int_\Delta\nabla v\:{\rm d}\lambda=\int_{\tilde\Delta}\nabla\left(v\circ F\right)\left|\det F'\right|\:{\rm d}\lambda\tag5$$ instead.

The crucial point is that in $(4)$ we take the weak gradient of $v$ and evaluate it at $F(\tilde x)$ whereas we take the weak gradient of $v\circ F$ (which has to be calculated by the chain rule) and evaluate it at $\tilde x$ in $(5)$.

Actually, $(5)$ doesn't make sense to me. The function we're considering in the integration by substitution rule should be $x\mapsto\nabla v(x)$. I guess there is something I'm missing which prevents us from using the conventional integration by substitution rule and hence $(5)$ is the result of some modified rule for integrands involving the gradient.

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