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We define the affine transformation on distributions by $\langle A^{*}u, \phi \rangle=\frac{1}{\det(A)}\langle u,\phi(A^{-1}x)\rangle$

Assume this we should have $\langle \partial_{i}(A^{*}u), \phi\rangle=\langle A^{*}u,-\partial_{i}\phi\rangle=\frac{1}{\det(A)}\langle u,-\partial_{i}\phi(A^{-1}x)\rangle$

To make it equal to $A^{*}(\sum^{n}_{j=1}a_{ji}\partial_{j} u)$ we would expect $\langle A^{*}(\sum^{n}_{j=1}a_{ji}\partial_{j}u),\phi\rangle=\frac{1}{\det(A)}\langle \sum^{n}_{j=1}a_{ji}\partial_{j} u,\phi(A^{-1}x)\rangle=\frac{1}{\det(A)}\langle u,-\sum_{j=1}^{n}a_{ji}\partial_{j}\phi(A^{-1}x)\rangle$

Therefore we should have $\partial_{i}\phi(A^{-1}x)=\sum_{j=1}^{n}a_{ji}\partial_{j}\phi(A^{-1}x)$

Now we have $A^{-1}=\frac{1}{\det(A)}B$, where $B_{ij}=A^{*}_{ji}$(the adjoint matrix). Thus we should have $(A^{-1}x)_{j}=\frac{1}{\det(A)}\sum A^{*}_{ji}x_{i}$

And by chain rule we should have $\partial_{i}\phi(A^{-1}x)=\sum \partial_{j}\phi(A^{-1}x)\partial_{i}((A^{-1}x)_{j})$

But this does not match our result given since $A^{*}_{ji}\not=a_{ji}$; so I want to ask what is wrong in my derivation.

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    Just in case someone cannot solve this, I solved it.2012-07-19

0 Answers 0