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.


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.

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

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