If we define the adjoint operator of linear operator $A:E\to E$, where $E$ is a complex or real Euclidean, $n$- or $\infty$-dimensional, space, as operator $A^\ast:E\to E$ such that $\forall x,y\in E\quad \langle Ax,y\rangle=\langle x,A^\ast y\rangle$, I wonder whether for any $A$ the adjoint exists. If it does, how can we prove it?
In the case that it exists, I think -please correct me if I am wrong- it is unique as in the case of the adjoint operator of linear operator $A:E\to E_1$, with $E$ a Banach space, defined as $A^\star: E_1^\star\to E^\star$, $f\mapsto f\circ A$.
Thank you very much for any answer!
Best Answer
From comments, it sounds like by "Euclidean space" you mean "Hilbert space". Let's call it $H$.
It is certainly true that for any bounded linear operator $A : H \to H$, the adjoint map exists. It's really the same as for Banach spaces: define the linear operator $A^*$ on $H^*$ via $A^* f = f \circ A$, and then use the Riesz representation theorem to identify $H^*$ with $H$.
More explicitly: for each $y \in H$, the map $x \mapsto \langle Ax, y \rangle$ is a continuous linear functional on $H$. By the Riesz representation theorem, there exists a unique $z_y \in H$ such that $\langle Ax, y \rangle = \langle x, z_y \rangle$. Define $A^*$ as the map taking $y$ to this uniquely determined $z_y$. Then it is an easy verification that $A^*$ is linear and bounded, and $\langle Ax, y \rangle = \langle x, A^* y \rangle$ holds by construction.
If $A$ is an unbounded linear operator on $H$ having dense domain $D(A)$, we can also define an adjoint $A^*$ as an unbounded operator. We let $D(A^*)$ be the set of all $y$ such that the map $x \mapsto \langle Ax, y \rangle$ is a continuous linear functional on $D(A) \subset H$, which by (uniform) continuity extends uniquely to a continuous linear functional on $H$. Then for such $y$ we may define $A^* y$ just as before. It is an exercise to show that $A^*$ is a closed operator, and is densely defined iff $(A, D(A))$ is closable.