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Talk:Friedrichs extension - Wikipedia, the free encyclopedia

Talk:Friedrichs extension

From Wikipedia, the free encyclopedia

[edit] Typo

The article gives an example

[T \phi](x) = -\sum_{i,j} \partial_{x_i} a_{i j}(x) \partial_{x_j} \phi(x) \quad x \in U, \phi \in  \operatorname{C}_0^\infty(U),

and then asks that the matrix a_ij be positive semidefinite. The seems to be in error; either an expression requiring the derivatives of a_ij should be pos semi-def, or the example should be

[T \phi](x) = -\sum_{i,j}  a_{i j}(x) \partial_{x_i}\partial_{x_j} \phi(x) \quad x \in U, \phi \in  \operatorname{C}_0^\infty(U),

and I believe the latter was intended, i.e. differentiation acting to the right, and not acting on a_ij. linas 14:27, 21 March 2006 (UTC)

No I think it was right as stated; maybe another paranthesis would have been better
[T \phi](x) = -\sum_{i,j} \partial_{x_i} \bigg\{a_{i j}(x) \partial_{x_j} \phi(x)\bigg\} \quad x \in U, \phi \in  \operatorname{C}_0^\infty(U),
Use integration by parts, and then apply the definition of non-negativity for operators on L2 you get the result.--CSTAR 15:10, 21 March 2006 (UTC)
Right, of course, silly me. Sometimes, I completely fail to engage my brain before engaging the keyboard. linas 16:38, 21 March 2006 (UTC)

[edit] Clarification

The article states:

Let H1 be the completion of dom T with respect to Q.

Let ξ_n be a Cauchy sequence in H that converge to ξ in H under the usual norm. Then, given various different Cauchy sequences ξ_n, it seems to me that these sequences can converge to different places under the Q norm; that is, different sequences converge to different values of Q(ξ_n,ξ_n) if/when Q is not bounded. So there may be many distinct elements in H_1 that are identified with a single element in H. Right?

There is a mapping from the completion of dom T into H. It is not immediately clear that it is injective. That requires a short (one or two line) argument not in the article.--CSTAR 16:06, 21 March 2006 (UTC)

Then, later in the article, it states:

Define an operator A ... Q(ξ, η) is bounded linear

From what I can tell, the requirement that Q(ξ, η) is bounded linear is exactly what it takes to single out just one point in the space of Cauchy sequences in H_1 (since boundedness implies continuity, then all sequences Q(ξ_n,ξ_n), by continuity, converge to the same value, and all sequences Q(ξ_n,η) converge to the same value. Thus the point is unique, and serves as the extension point.).

I think what I describe above is the intent of the definition of A, but perhaps this point could be belabored a bit. If you agree that my interpretation is correct, then perhaps I'd like to modify the article to belabor this point. linas 14:57, 21 March 2006 (UTC)

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