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I have known that if $A$ is a densely defined (unbounded) operator with domain $D(A)$ such that $\langle Ax,x\rangle=0$ for all $x\in D(A)$, then this does imply that $Ax=0$ for all $x\in D(A)$. This result may be found e.g. in Schmudgen's new book on unbounded self-adjoint operators.

However the following is a purported counter-example in these notes:

Consider the differential operator $T:x\mapsto \frac{dx}{dt}$ defined on $C_c^{\infty}(\mathbb{R})$ which is a dense subset of $L^2(\mathbb{R})$. Suppose then $x\in C_c^{\infty}(\mathbb{R})$, then

$$ \int_{\mathbb{R}}\frac{dx}{dt} x\,dt = x^2\bigg|_{-\infty}^{\infty} - \int_{\mathbb{R}} x\frac{dx}{dt}\,dt. $$

Hence $\langle Tx,x\rangle = 0$ for all $x\in C_c^{\infty}(\mathbb{R})$ but $Tx\neq 0$ for some $x$.

Any help please!

Cheers...

Community
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Math
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2 Answers2

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It looks like in that pdf they're considering a real Hilbert space. The positive result holds only for complex Hilbert spaces; it's false even for finite-dimensional real Hilbert spaces.

For example define $T:\Bbb R^2\to\Bbb R^2$ by saying $T(x,y)=(-y,x)$.

David C. Ullrich
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This holds (regardless of the (un)boundedness of the relevant operators) only if the Hilbert space under consideration is complex.

Note that the scalar product used in the solution is

$$ \langle f,g \rangle =\int f(x) g(x)\, dx $$ which is only a valid scalar product if the functions are all real (note the missing conjugation on $g$).

For real vector spaces, you can even get an easier example:

$$ A =\left( \begin{matrix} 0 & -1\\ 1 & 0\end{matrix}\right). $$

PhoemueX
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  • Cheers that's what I have missed...I may also add that the result is true if $A$ is self-adjoint even on a real Hilbert space. – Math Aug 05 '15 at 21:06
  • I think guys there is another problem...Going back to the $<Ax,x>=0$ above (with $A=d/dx$). I think $<Ax,x>=0$ for all $x$ implies that $Ax=0$ for all $x$ (only). What you think? – Math Aug 06 '15 at 18:33
  • @Math: What do you mean? It is clear that there are $x$ with $Ax \neq 0$. – PhoemueX Aug 06 '15 at 18:51
  • you're right, the result I am referring to is an a.e. equality and if $x$ is only locally integrable. – Math Aug 07 '15 at 10:07