2
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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 space $X$ the duality $\langle\cdot,\cdot\rangle$ is precisely the inner product of $X$.

$\endgroup$
2
  • $\begingroup$ 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
    Commented Jul 14, 2016 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 $x\in X$. $\endgroup$ Commented Jul 14, 2016 at 16:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .