Operator theory and Sobolev theory

[pabryan]

• We use "strong" solutions in the sense that $u \in C^1([0, T) \to \dom(\Delta))$ such that the derivative $\partial_t u$ is equal to the function $\Delta u \in L^2$. Such solutions are then necessarily weak solutions as in Lemma 5.2 and Lemma 5.4.
• We should move some version of this statement to section 3.
• What about the converse? Need a weak solution have $\partial_t u \in \dom(\Delta))$? Even still, if $f$ has a weak derivative, need it be in the domain of $\nabla$?
• The remark that without closability, we don't get a space of functions deserves further comment. Does this imply that our Sobolev spaces are indeed the closure of the appropriate smooth functions in the Sobolev norms?
• Where we define the sobolev spaces and norms, there is a remark about closability gives densely defined, closed adjoints. This is standard, but give a ref. Preferably not Dumford and Schwartz unless a precise ref is given!

These are natural questions that surely people will ask so should be addressed. In particular, it seems our notion of solution is weaker than one would generally seek. See issue #2 Euclidean space where our method does not pick up some solutions. This is perhaps related also to the question of uniqueness/minimality of the heat kernel. We obtain perhaps stronger conclusions because of a more restrictive notion of solution. This could well be important in that perhaps our conclusions are false in a broader context. For instance, they are certainly false for classical solutions on all of Euclidean space. This is no problem - it's a well known standard fact coming from Tychonoff's example but deserves mention.

[lashputin]

• Our heat kernels are such strong solutions. So, this is the notion we need.
• Which statement?
• I'm not sure about this again?
• By construction, our sobolev spaces are exactly the closure of such smooth functions. That's why we need to know that the operator is closable. Without the closability of the operator, there is nothing to say that we'll even be inside L^2 when taking the completion.
• Yes, we can quote Kato here. Theorem 5.28 in Chapter III.

[lashputin]

P.S., with regards to issue 2, we can only get this kind of behaviour if solutions do not have finite energy. That's why we only look in L^2. BUt that's the only place we can look at, the reason you can do stuff in the smooth setting beyond L^2 is because you can define the Laplacian as -\tr\nabla^2. We don't have a hessian. The only Laplacian we have is this operator theoretic kind, which agrees with the objects in the smooth world when you look in L^2.