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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!

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    $\begingroup$ If by "Euclidean space" you mean finite dimensional, then it is very easy in this case: you can identify linear operators with matrices, and then the adjoint is just the conjugate transpose matrix. $\endgroup$ – Nate Eldredge Dec 6 '14 at 20:23
  • $\begingroup$ Thank you for the comment! No, I mean even $\infty$-dimensional spaces... $\endgroup$ – Self-teaching worker Dec 6 '14 at 21:10
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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.

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  • $\begingroup$ Thank you so much! I intended a real or complex Euclidean space of finite or infinite dimension, but not necessarily complete. I know Riesz representation theorem, but can we apply it to non-complete (non-Hilbert) Euclidean spaces? Thank you very much again! $\endgroup$ – Self-teaching worker Dec 7 '14 at 9:46

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