2
$\begingroup$

If we consider a bounded linear operator $A$ between two complex Hilbert spaces $X,Y$ (i.e. $A: X \to Y$) then we can define the adjoint operator $A^*: Y \to X$ via the dual operator $A_{dual}^*: Y^* \to X^*$ and the isometric isomorphisms $\iota_X, \iota_Y$ by defining $$A^*:=\iota_X^{-1} \circ A_{dual}^* \circ \iota_Y$$

So far so good. Now comes the part which I'm struggling with:

Very analogous to the dual operator the adjoint is uniquely defined via $$\langle A^*y,x\rangle_X=\langle y,Ax \rangle_Y \space \forall x\in X \space \forall y \in Y$$

I've been trying to show this by using the fact that the dual operator is uniquely defined via $$\langle A_{dual}^*y^*,x\rangle = \langle y^*,Ax\rangle$$ and the Riesz reprensation of complex Hilbert spaces but somehow I'm not getting there.

$\endgroup$
1
$\begingroup$

First, let's be careful with notation. For a linear functional $f$ on a Hilbert space $H$, we will denote its action on an element $x$ by $f(x)$. We will denote the inner product on $H$ by $\langle \cdot,\cdot\rangle_H$. This way we won't get confused about which spaces are involved and whether a bracket is the duality pairing or an inner product.

In this notation, the dual map $A_{dual}^*:Y^*\to X^*$ is defined uniquely by the equation $$ (A_{dual}^*y^*)(x) = y^*(Ax) ~\text{for all}~y^*\in Y^*, x\in X. $$ Let $y\in Y$ denote the Riesz representative of $y^*$, and let $z\in X$ denote the Riesz representative of $A_{dual}^*y^*$. Then we can write the above as $$ \langle z,x\rangle_X = \langle y,Ax\rangle_Y. $$ On the other hand since $A^*$ is claimed to be defined by the property $\langle A^*y,x\rangle_X = \langle y,Ax\rangle_Y$, to prove our claim it is necessary to show that $$ \langle z,x\rangle_X = \langle A^*y,x\rangle_X $$ So our job is really to prove that $z = A^*y$, that is, the Riesz representative vector of $A_{dual}^*y^*$ is $A^*y$. To do this, go back to the definition of $A^*$ and the Riesz representative. The Riesz representative of any $x^*\in X^*$ is defined to be $i_X^{-1}x^*$, and for any $y\in Y$ its Riesz representative functional $y^*$ is given by $y^* = i_Yy$. Therefore the Riesz representative of $A_{dual}^*y^*$ is $$ i_X^{-1}A_{dual}^*y^* = i_X^{-1}A_{dual}^*(i_Yy) = (i_X^{-1}A_{dual}^*i_Y)y = A^*y. $$ By the identification of $X$ with $X^*$ by $i_X$, the claim is proved.

$\endgroup$
  • $\begingroup$ this is perfect and exactly the answer I've been looking for, thank you very much! $\endgroup$ – noctusraid Jan 18 '16 at 22:35

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.