Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I wonder if someone can help me on what is, probably, a simple question but is baffling me at the moment!

In standard texts on functional analysis, something like the following is written Let $L\colon X \to Y$ where $X$ and $Y$ are both Hilbert spaces. Then the adjoint operator $L^\dagger$ exists and is unique, where $L^\dagger \colon Y \to X$ and $ (Lf,g)= (f,L^\dagger g)$ for all $f \in X$ and $g \in Y$.

However, I am having some trouble with this definition in practice. When dealing with differential operators, boundary conditions have to be taken into account and invariably instead of (e.g.) $L\colon L^2 \to L^2$, we have something like $L\colon A \to L^2$ where $A$ is the subset of $L^2$ that satisfies a certain linear boundary condition like $f(1) = 0$. Since $A$ is not a closed subset, it is not a Hilbert space in its own right and I can't see that it is immediate that we can apply the standard theorem as given above. Perhaps we can extend $A$ to $L^2$ somehow - or have I completely missed the point?

share|improve this question
    
By "differential operator" do you mean: the operator given by differentiation? If so, then this is very far from been defined for all $L^2$ functions-- you need to restrict to those which can be differentiated! So you're really studying an unbounded, densely defined operator. There is a lot of literature about such things-- in particular, taking adjoints does make sense, but you need some technology, and you have to be careful. –  Matthew Daws Feb 22 '11 at 8:47
1  
I think you're getting mixed up between "bounded" and "unbounded" operators. The standard theory works well for bounded operators, but most differential operators and others used in quantum mechanics, etc. are unbounded - they cannot be defined on the whole space in any nice way. The theory of unbounded operators is a lot more complicated and subtle than bounded ones. Yosida, Functional Analysis is a good reference, I believe. –  Zen Harper Feb 22 '11 at 8:49
4  
No you haven't missed the point. The work-around is to restrict the domain of definition of $A^{\dagger}$ to the subspace consisting of $g$'s such that $g \mapsto (Lf,g)$ is continuous. However, this is not a research-level question. You should ask this on math.stackexchange.com instead where you'll get a detailed answer. –  Theo Buehler Feb 22 '11 at 8:51
    
I apologize for the two typos in my previous comment. It should have been $L^{\dagger}$ and $f \mapsto (Lf,g)$. You might also want to have a look at en.wikipedia.org/wiki/Unbounded_operator –  Theo Buehler Feb 22 '11 at 10:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.