# Error in proof of self-adjointness of 1D Laplacian

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.

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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

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.
$f$ 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