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Let $H$ be a Hilbert space, and let $z \in H$. Let $T_z \colon H \to K$, where $K$ is the field of scalars for $H$ and $K$ is either $\mathbb{R}$ or $\mathbb{C}$, be defined by $$ T_z (x) \colon= \langle x, z \rangle_H \ \mbox{ for all } \ x \in H.$$ Then $T_z$ is a bounded linear functional with $$\Vert T_z \Vert = \Vert z \Vert.$$ Am I right?

How to find the Hilbert adjoint operator $T_z^*$ of $T_z$? Is there a way of finding a simple enough formula for $T_z^*$?

By Theorem 3.9-2 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig, if $H_1$ and $H_2$ are Hilbert spaces over the same field $K$, where $K$ is either $\mathbb{R}$ or $\mathbb{C}$, and if $T \colon H_1 \to H_2$ is a bounded linear operator, then there exists a unique bounded linear operator $T^* \colon H_2 \to H_1$ such that $$ \langle T(x), y \rangle_{H_2} = \langle x, T^*(y) \rangle_{H_1} \ \mbox{ for all } \ x \in H_1, \ y \in H_2,$$ and $$\Vert T^* \Vert = \Vert T \Vert.$$

If $\dim H_1 = n < \infty$, then let $\{ e_1, \ldots, e_n \}$ be an orthonormal basis for $H_1$. Then for each $u \in H_1$, we have $$u = \langle u, e_1 \rangle_{H_1} e_1 + \cdots + \langle u, e_n \rangle_{H_1} e_n.$$ Now, for each $y \in H_2$, as $T^*(y) \in H_1$, so we have $$T^*(y) = \langle T^*(y), e_1 \rangle_{H_1} e_1 + \cdots + \langle T^*(y), e_n \rangle_{H_1} e_n = \langle y, T(e_1) \rangle_{H_2} e_1 + \cdots + \langle y, T(e_n) \rangle_{H_2} e_n.$$ So, for each $y \in K$, we can write $$T_z^*(y) = \langle y, T_z(e_1) \rangle_K e_1 + \cdots + \langle y, T_z(e_n) \rangle_K e_n.$$ Now since $K$ is either $\mathbb{R}$ or $\mathbb{C}$, there are two cases.

Case 1. If $K = \mathbb{R}$, then $$T_z^*(y) = y \sum_{j=1}^n \langle e_j, z \rangle_H e_j = y \sum_{j=1}^n \langle z, e_j \rangle e_j = yz. $$

Case 2. If $K = \mathbb{C}$, then $$T_z^*(y) = y \sum_{j=1}^n \langle z, e_j \rangle_H e_j = yz. $$ Is this reasoning correct?

What if $\dim H = \infty$?

N.B.: The last equalities in either of the above two cases have been included in the light of the comment by celtschk.

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    $\begingroup$ If I'm not mistaken, $T_z^*(y) = yz$. No answer because I didn't really think this through. $\endgroup$ – celtschk Jul 18 '16 at 10:31
  • $\begingroup$ For $x\in H$, $y\in K$ one has $\langle x,T^{*}y \rangle$=$\langle Tx,y \rangle$=$\langle \langle x,z \rangle,y\rangle$=$\langle x,z \rangle$ $y^{\bar}$=$\langle x,yz \rangle$ so $T^{*}y=yz$. $\endgroup$ – PHU CUONG LE VAN Jul 18 '16 at 13:15
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\begin{eqnarray} \langle T_z^*(\alpha),x \rangle_{H}&=&\langle \alpha , T_z(x)\rangle_\mathbb{K}\\ &=& \alpha \overline{T_z(x)}\\ &=& \alpha\langle z,x\rangle_H\\ &=& \langle \alpha z , x\rangle_H\\ \end{eqnarray} So $$T_z^*(\alpha)=\alpha z $$ It's clear now that $\|T_z\|=\|T_z^*\|=\|z\|$

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