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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.

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  • $\begingroup$ Yes this a correct (notice that by denoting $\lambda=\langle h, h_0 \rangle \in \Bbb R$) you have: $$\langle \lambda u, v\rangle=\lambda \langle u, v\rangle =\langle u,\lambda v\rangle$$ there is no need to use anything beside the bi linearity of the inner product. $\endgroup$ – Delta-u May 25 '18 at 16:47
  • $\begingroup$ You mean $\mathcal H$ is a real Hilbert space? $\endgroup$ – Berci May 25 '18 at 16:47
  • $\begingroup$ How is the adjoint of a functional defined? $\endgroup$ – Berci May 25 '18 at 16:49
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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 $h_0$.

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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.

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  • $\begingroup$ Yes, you are right. I forgot to adapt to the general case. My book only showed Riesz and corrolaries for endomorphisms. $\endgroup$ – EpsilonDelta May 25 '18 at 17:00

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