Let $X$ be a Hilbert space and let $X'$ be the dual space of $X$ with respect to the duality pairing $\langle\cdot,\cdot\rangle$.

Let $A: X \mapsto X'$ be a bounded linear operator. We assume that $A$ is self-adjoint, i.e., we have $\langle Au,v\rangle = \langle u,Av\rangle $ for all $u,v \in X$

I am confused about the meaning of being self-adjoint, since if $\langle Au,v\rangle := (Au)(v) $ then what does $\langle u,Av\rangle $ mean?


In a Hilbert space, you have the Riesz Representation Theorem, which tells you that given any $f\in X'$, there exists $y\in X$ such $$\tag{1}f(y)=\langle y,x\rangle,\ \ \ x\in X.$$ And this assignment is isometric so $X'$ is isomorphic to $X$. In practice, one thinks that $X'=X$ via the duality $(1)$.

In summary, in the case of a Hilbert $X$ space the duality $\langle\cdot,\cdot\rangle$ is precisely the inner product of $X$.

  • $\begingroup$ Just to clarify, are we allowed to treat bilinear maps $X \times X' \mapsto R$ as inner products $X \times \ X \mapsto R$ due to $X'$ being isomorphic to $X$ ? What if Bilinear map is not Positive-definite? $\endgroup$ – Sam Jul 14 '16 at 15:42
  • $\begingroup$ Yes (blinear if $X$ is real, sesquilinear if $X$ is complex). If $\varphi:X\times X'\to\mathbb R$ is bilinear and bounded, then for each $f\in X'$ you have $$x\longmapsto\varphi(x,f)$$ is in $X'$, so there exists $y\in X$ such that $$\varphi(x,f)=\langle x,y\rangle$$ for all $x\in X$. $\endgroup$ – Martin Argerami Jul 14 '16 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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