The adjoint of a linear functional

Question

Let $l\in \mathcal{H}^*$ be a real linear functional on a hilbert space $\mathcal{H}$.

I want to compute the adjoint $l^*$ in $\mathcal{H}$. Using the Riesz-Fischer representation, the unique existence of a $h_0\in \mathcal{H}$ with $l(h)=\langle h,h_0\rangle $ for every $h\in\mathcal{H}$ is guaranteed.

Hence I obtain with the definition of the adjoint operator via inner product,

$\langle \langle h,h_0\rangle u, v\rangle=\langle h,h_0\rangle\langle u, v\rangle =\langle u,\langle h,h_0\rangle v\rangle$ for all $v,u\in \mathcal{H}$.

I have my doubts that this is correct.

Answer 1

You're not working with the correct definition of adjoint. The adjoint $l^* : \mathbb{R} \to H$ is the unique bounded linear map such that

$$l(u)\cdot v = \langle u ,l^*(v)\rangle, \text{ for } u \in H, v \in \mathbb{R}$$

Note that $\cdot$ is just multiplication in $\mathbb{R}$, and $\langle \cdot, \cdot\rangle$ is the inner product in $H$.

Riesz representation theorem gives that $\exists h_0 \in H$ such tha $l(u) = \langle u, h_0\rangle, \forall u \in H$.

For all $u \in H, v \in \mathbb{R}$ we have

$$l(u)\cdot v = \langle u, h_0\rangle \cdot v = \langle u, vh_0\rangle$$

so the adjoint is given by $l^{*}(v) = vh_0$, which it's just multiplication by $h_0$.

Answer 2

I'm afraid you may be a bit confused. The inner product characterization of adjoint ($\langle Au,v \rangle = \langle u, A^*v \rangle$) applies only to operators $A: \mathcal H \to \mathcal H$. You can define an adjoint of an operator $l : \mathcal H \to \mathbb R$ but not using this characterization.