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Trial And Error.

A portrait of a famous Mathematician who was often criticized for not being rigorous enough in his new inventions.

"Anyone who has never made a mistake has never tried anything new." - Albert Einstein


2d
comment The resolvent of a differentation operator on $C[a,b]$
For $\lambda\in\rho(A)$, the function $g=R_{A}(\lambda)f$ is the unique solution of $(A-\lambda I)g=f$, which means $g'-\lambda g= f$ and $g(a)=g(b)$. For a fixed $\lambda$: If you cannot solve for $g$ given a general $f$, then $\lambda\in\sigma(A)$.
Dec
14
comment Self-adjointness
@TobiasHurth : The Schrodinger operators on infinite intervals with periodic potentials are not particularly well-known to me, even though I do know some things. Hopefully someone here will be able to help. I've wanted to learn about this subject myself, for exactly the application that you cite. Good stuff. Hopefully we can learn from someone here.
Dec
14
comment Self-adjointness
@TobiasHurth : Yes, that little result comes in very handy in showing that domains are dense and operators are selfadjoint. I think you can why very it is very generally true that multiplication by real functions on $L^{2}$ spaces are densely-defined and selfadjoint. Concerning the example you reference in the previous problem: $\frac{1}{1-x^{2}}T\frac{1}{1-x^{2}}f$ is symmetric on its domain, whatever that domain may be, but no guarantees of selfadjointness for a general selfadjoint $T$.
Dec
14
answered Difference between an eigenvalue and a spectral value
Dec
14
answered Self-adjointness
Dec
13
comment Bump function on set
@TobiasHurth : I think you mean $K\subset V$ instead of the other way around?
Dec
13
answered The orthogonal operator onto $ran(T)$
Dec
13
comment Square root of unbounded operator
@TobiasHurth : Thanks for this problem; your problems are always interesting. I'm a little shocked that such a clean treatment is possible for the problem.
Dec
13
comment Square root of unbounded operator
@TobiasHurth : $\frac{1}{i}\partial$ is selfadjoint on absolutely continuous periodic functions on $[a,b]$. So $(\partial)^{\star}=-\partial$ on this same domain. It is easy to check that adding a bounded operator $W$ leaves $\pm \partial +W$ closed on the same domain with adjoints $\mp\partial + W$. The domain of $(-\partial+W)(\partial+W)$ consists of periodic absolutely continuous $f$ for which $f'+Wf$ is periodic and absolutely continuous. If $Wf$ leaves $f$ periodic, then $f'+Wf$ is absolutely continuous and periodic iff $f$ is twice a.c. with $f$, $f'$ periodic.
Dec
13
answered Holomorphic Functional Calculus vs Borel Functional Calculus
Dec
13
comment Square root of unbounded operator
@TobiasHurth: Imposing endpoint conditions on the unconstrained domain is how operator domains are defined.
Dec
12
comment Square root of unbounded operator
@TobiasHurth : I'm assuming $\psi'/\psi$ is smooth on $[a,b]$, which seems warranted if $\psi > 0$ on $[a,b]$ (i.e., has no nodes.) Then the boundary functionals associated with $\pm\partial+W$ are endpoint evaluations only. So everything reduces to a 2d linear algebra in that case.
Dec
12
comment Square root of unbounded operator
@TobiasHurth : Look back up a couple of comments.
Dec
12
revised Square root of unbounded operator
added 650 characters in body
Dec
12
awarded  Cleanup
Dec
12
revised Square root of unbounded operator
rolled back to a previous revision
Dec
12
comment Learning functional analysis and measure theory
@AdityaKashi : I think the Chapter on Lebesgue Integration would be enough for $\mathbb{R}^{n}$, and wouldn't take you too far off into left field.
Dec
12
comment Square root of unbounded operator
@TobiasHurth : $((\partial+W)f,g)-(f,(-\partial+W)g)=(f',g)+(f,g')=\int_{a}^{b}(fg)'\,dx = fg|_{a}^{b}$. Looks like periodic for the domain of both $\partial+W$ and $-\partial +W$ works out just fine in terms of adjoints. The condition for $(-\partial+W)(\partial+W)f$ would be that $f$ and $(f'+Wf)$ are periodic. Ichh. That would work I suppose if $W$ is periodic because that then reduces to $f$, $f'$ being periodic, which is classical. But check these details for yourself.
Dec
11
comment Square root of unbounded operator
@TobiasHurth : When you say $\mathcal{C}^{\infty}$, you should make it clear that this is over the interior of an interval. Without conditions, you can't know the domain of $T$, which means that you can know the domain of $A$. When you write $T=A^{\star}A$, that's an operator equation where $A^{\star}$ is an adjoint. Some conditions may rule out such a relation because $A^{\star}A \ge 0$, and $T \ge 0$ may not be true.
Dec
11
comment Square root of unbounded operator
@TobiasHurth : I see that you added on MathOverflow that your operator is defined on a bounded domain and $V$ is smooth on that domain. Then you need to specify endpoint conditions to make your operator selfadjoint. Part of choosing the domain of $A$ has to involve those conditions.