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Let $\Omega,\hat{\Omega}\subset\mathbb{R}^n$ be two bounded open connected sets with Lipschitz boundary related by the following relation.

There is an affine transformation $$ T:\hat{\Omega}\to\Omega\\ \qquad\;\;\:\: \hat{x}\mapsto B\hat{x}+b, $$ where $B$ is an invertible matrix.

Let now consider a function $u:\Omega\to\mathbb{R}$ and its pull-back $\hat{u}:=u\circ T:\hat{\Omega}\to \mathbb{R}$.

Suppose that both $u,\hat{u}$ are smooth enough functions to consider derivatives up to order $k$.

I read that:

For every $p\in \Omega$, $D^ku(p)\in {(\mathbb{R}^n)^\ast}^{\otimes k}$ (space of $k$-linear maps from $\mathbb{R}^n\times\dots\times\mathbb{R}^n$ to $\mathbb{R}$) and moreover it's symmetric.

Now, what I don't understand is the following relation.

For every $p\in\hat{\Omega}$,

$$ D^k\hat{u}(\hat{p})(\xi_1,\dots,\xi_k)=D^k u(T\hat{p})(B\xi_1,\dots,B\xi_k), $$ for any $\xi_i\in\mathbb{R^n}$.

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    What don't you understand about it? – 2017-01-12
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    @Hurkyl how to recover the last relation – 2017-01-12
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    @Hurkyl yes, thanks – 2017-01-12

1 Answers 1

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To begin with, for each $p \in \hat \Omega$, one has:

$$D\hat u(p) = D(u \circ T)(p) = Du(Tp)\circ DT(p) = Du(Tp) \circ B$$

where $B$ is identified with the linear mapping associated to it relative to the canonical basis. Hence: $D \hat u = \Phi \circ Du \circ T$, where $\Phi: L \in L(\Bbb R^n, \Bbb R^n) \mapsto \Phi L = L \circ B$. Evidently $\Phi$ is linear and continuous ($\|\phi\| \le \|B\|$), so for each $L$, $D\Phi(L) = \Phi$. Thus:

$$D^2 \hat u(p) = D\Phi(Du(Tp)) \circ D^2 u(Tp) \circ DT(p) = \Phi \circ D^2 u(Tp) \circ B$$

In other words,

$$[D^2\hat u(p)](\xi_1,\xi_2) \equiv [D^2 \hat u(p) (\xi_1)](\xi_2) = [\Phi(D^2u(Tp)(B(\xi_1))](\xi_2) \\= [D^2 u(Tp)(B\xi_1)](B\xi_2) \equiv D^2 u(Tp)(B\xi_1,B\xi_2)$$

The pattern is now clear. Try to do it by induction.

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    If you define $\|D^k v(x)\|:=sup \{|D^k v(x)(\xi_1,\dots,\xi_k)|: \|\xi_i\|=1 \}$, is it possible to write the usual seminorm as $|u|_{H^1}=\sqrt{\int \|D u(x)\|dx}$? – 2017-01-12
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    @avati91 that doesn't look right. Why do you think that such a writing is possible? – 2017-01-12
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    I can't edit now. But I meant $\|Du(x) \|^2$ under the integral sign – 2017-01-12
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    @avati91 it's still not true. Do you know what the $|\cdot|_{H^1}$ norm is? – 2017-01-12
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    Yes, it's $\sqrt{\sum_{i=1}^n\int | \partial_{x_i} u|^2 }$, where $\partial_{x_i} u$ is the weak derivative, $\partial_{x_i} u (x) = D u (x)(e_i)$ – 2017-01-12
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    @avati91 no, it's $\sqrt{ \int u^2 + \int \partial u^2}$ – 2017-01-12
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    No, that's the Sobolev norm which is usually denoted as $\|\cdot \|_{H^1}$. I wrote "seminorm" – 2017-01-12
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    @avati91 oh, I didn't pay close attention. Sorry about that. But I still don't see the equality. I'll think about it, but don't expect much as I'm not an expert. – 2017-01-12
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    No, problem! Okay, thank you very much! – 2017-01-12
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    @avati91 you're welcome! – 2017-01-12