Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have successfully checked self-adjointness of simple and classic differential operator - 1D Laplacian $$D = \frac {d^2}{dx^2}: L_2(0,\infty) \rightarrow L_2(0,\infty)$$ defined on $$\{f(x) | f'' \in L_2(0,\infty), f(0) = 0\},$$ open an article and see the first Example that this operator is not self-adjoint but stricly simmetric (hermitian).

Can anybody point out error in reasoning below?

Find adjoint operator, i.e. it's domain. $$ (Df,g) = \int_0^\infty f''\overline g dx = ... = \left. (f'\overline g - f \overline g') \right|_0^{\infty} + (f, D^*g). $$

To satisfy adjointness we should zero out free term with fixed $g$ and for all $f \in D_D$ - domain of D $$ \left. (f'\overline g - f \overline g') \right|_0^{\infty} = 0. $$

Second term in it zero outs because of $f(0) = 0$, so $$ \left. (f'\overline g ) \right|_0^{\infty} = 0. $$

Because $f$ arbitrary from domain of $D$ and hence can have not zero first derivative, so this equality holds for fixed $g$ if and only if $$g(0) = 0.$$

Thus domain of direct and adjoint operator the same, which means that it is self-adjoint.

What I see in article.

Boris Pavlov wrote: Example. Symplectic extension procedure for the differential operator Consider the second order differential operator $$ L_0u = - \frac {d^2u}{dx^2} $$ defined on all square integrable functions, $u \in L_2(0, \infty)$, with square- integrable derivatives of the first and second order and vanishing near the origin. This operator is symmetric and it’s adjoint $L^+_0$ is defined by the same differential expression on all square integrable functions with square integrable derivatives of the first and second order and no additional boundary condition at the origin.

Where is error? It certainly is, because operator $D$ with domain $f'(0)=\alpha f(0)$ is symmetric, It's domain is superset of regarded in my message domain hence my operator is not maximal symmetric and hence can not be self-adjoint.

share|cite|improve this question
Why do you conclude "Thus domain of direct and adjoint operator the same" after merely checking $g(0)=0$? The other requirement in the definition of $D$ was $f''\in L^2$. – user31373 Sep 23 '12 at 4:50
It should be twice-differentiable because of scalar product on the right side – theambient Sep 23 '12 at 4:56
up vote 2 down vote accepted

The operator in Pavlov's article is not the same as yours. His has a domain of functions "vanishing near the origin", i.e. on a neighborhood of 0. For your operator, functions in the domain need only vanish at the origin. So there is no error; your operator is self-adjoint and his is not.

Regarding your last paragraph, the operator $D$ with domain $f'(0) = \alpha f(0)$ (Robin boundary conditions) is not an extension of yours. Consider for instance $f(x) = e^{-x} \sin x$; it has $f(0)= 0$ but $f'(0) \ne 0$, so it is in the domain of your original operator but not the latter.

share|cite|improve this answer
Do I understand correctly that vanishing near origin impose first derivative vanishing at the origin? – theambient Sep 23 '12 at 5:02
I how this "vanishing near origin" is formally defined? – theambient Sep 23 '12 at 5:03
$f$ vanishes near the origin if there is an open neighborhood $U$ of 0 such that $f = 0$ (almost) everywhere on $U$. In particular, all derivatives of $f$ vanish at $0$. – Nate Eldredge Sep 23 '12 at 5:04
Thank you very much Nate! I didn't recognize difference between "near" and "at". I promise I will ) – theambient Sep 23 '12 at 5:15

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.