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I know that the Sobolev space $H^s$ with the inner product given by $$ \langle u,v \rangle_{H^s} := \sum_{| \alpha| \leqslant s} \int_{\Bbb R^n} \nabla^\alpha u \cdot \nabla^\alpha v $$ is Hilbert space. ($u = (u_1, \cdots , u_N), v = (v_1, \cdots ,v_N)$)

Then if we set the bracket as $$ \langle u,v \rangle_{A,H^{s}} := \sum_{| \alpha| \leqslant s} \int_{\Bbb R^n} (A \nabla^\alpha u) \cdot \nabla^\alpha v $$

with $N \times N$ positive definite matrix $A$, then is $(H^s, \langle \cdot , \cdot \rangle_{A, H^{s}} ) $ also Hilbert space?

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up vote 1 down vote accepted

Yes, this is again a Hilbert space, since the norms induced by the standard inner product and the one induced by $A$ are equivalent.

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