# Adjoint and Hilbert adjoin of unbounded (linear) opeartors

Given a Banach Space $X$, a densely defined linear operator $A$, one can define an adjoint of $A$, $A':X'\to X'$ (Here $X'$ is dual of $X$) as follows:

$$\phi\in D(A'),\; A'(\phi)=\psi \iff \phi(Ax)=\psi(x) \;\; \forall x\in D(A)$$

One can check that if domain of $A$ is dense, then this uniquely defines $\psi$ on whole of $X$ (by Hahn Banach).

On the other hand, if $X$ is additionaly a Hilbert space, then one defines a Hilbert adjoint $A^{*}$ via

$$x\in D(A^{*}),\; A^{*}(x)=y \iff \langle x,Az\rangle = \langle y,z\rangle \;\; \forall z\in D(A)$$

But apparently, in the setting of Hilbert space Riesz Representation theorem identifies each element of $X$ with that of $X'$ via conjugate linear isometry $x\mapsto \langle \cdot,x\rangle$. Upon realizing this, we can reformulate the definition of $A'$ as follows:

$$\langle \cdot,x\rangle \in D(A'),\; A'(\langle \cdot,x\rangle)=\langle \cdot,y\rangle\iff \langle Az,x\rangle=\langle z,y\rangle \;\; \forall z\in D(A)$$

This gives an identification of $A'$ with an operator $B$ on $X$, defined by $Bx=y$ via above relation. That is, $\langle Az,x\rangle=\langle z,Bx\rangle$ or in other words, $\overline{\langle x,Az\rangle}=\overline{\langle Bx,z\rangle}$ whence $\langle x,Az \rangle=\langle Bx,z\rangle$. So $B=A^{*}$

So it seems that different definitions of adjoints coincide via Riesz representation lemma, but why is Hilbert adjoint defined a bit differently from that of $B$? (although they are equivalent via conjugate symmetry of inner product), is it just a convention, or am I missing something?

## 1 Answer

Short answer: The Hilbert space adjoint is defined differently to make its domain the same space that the operator acts on, not its dual.

When working with a Hilbert space, we have the opportunity to define the adjoint of an operator $A:X\to X$ so that it also acts from $X$ to $X$. This allow us to consider compositions such as $A^*A$ which turn out to be quite useful (it's a positive operator; and its kernel is the same as of $A$, and $\|A^*Ax\|=\|Ax\|^2$, and the square root $|A|=\sqrt{A^*A}$ is like the "magnitude" of $A$, allowing us to create polar factorization $A=U|A|$, ...). It would be unwise to ignore this opportunity. It comes with a modest cost of making the adjoint correspondence $A\mapsto A^*$ conjugate-linear instead of linear.

On a general Banach space, we don't have such luxury. E.g., the adjoint of an operator $A:\ell^1\to\ell^1$ has to be an operator $A^*:\ell^\infty\to\ell^\infty$, so compositions like $A^*A$ do not make any sense. So we are stuck with dual spaces ... as a small consolation prize, the correspondence $A\mapsto A^*$ is now linear.