Kreyszig's Functional analysis seems to introduce the hilbert adjoint operator by means of an explicit representation. I haven't seen this anywhere else and I would like to confirm this explicit solution for the Hilbert adjoint operator, that is $T^*:H_2\rightarrow H_1$ such that $\langle Tx,y\rangle=\langle x, T^*y\rangle, \forall x\in H_1, y\in H_2$.
The following discussion is pieced together from 2 theorems from Kreyszig's Introduction to Functional Analysis text.
First, statements of the 2 Theorems whose results piece together into the adjoint operator. I've left out some of their details that I didn't deem relevant.
3.8-1 Riesz's Theorem (Functionals on Hilber spaces): Every bounded linear functional $h$ on a Hilbert space $H$ can be represented in terms of the inner product, namely, $f\left(x\right)=\left\langle x,z\right\rangle $ where $z$ is unique and given by $z=\frac{\overline{f\left(z_{0}\right)}}{\left\langle z_{0},z_{0}\right\rangle }z_{0},\hspace{1em}z_{0}\in\mathcal{N}\left(f\right)^{\perp} $
($\mathcal{N(f)}$ is the null space of $f$. $z_0$ is chosen as such so that $\langle x, z \rangle = 0 $ for all $x : f(x)=0$)
3.8-4 Riesz representation: Let $H_{1} , H_{2} $ Hilbert spaces and $h:\ H_{1}\times H_{2}\rightarrow K $ a bounded sesquilinear form. Then $h$ has a representation $h\left(x,y\right)=\left\langle Sx,y\right\rangle $
The proof from the text considers $\overline{h\left(x,y\right)} $, keeps $x$ fixed (so as if $\overline{h\left(x,y\right)}$ is a function of 1 variable, $y$ ), and then applies Riesz's Theorem (above) such that $\overline{h\left(x,y\right)}=\left\langle y,z\right\rangle ,\hspace{1em}z=\frac{h\left(x,z_{0}\right)}{\left\langle z_{0},z_{0}\right\rangle }z_{0},\hspace{1em}z_{0}\in\mathcal{N}\left(h\left(x,y\right)\right)^{\perp} $
and taking conjugates:$ h\left(x,y\right)=\left\langle z,y\right\rangle $
and now we let $z=Sx$
and there we have the unique representation.
Now an explicit form for $T^{\star}$: Working backwards from the previous proof (replacing $S$ with $T^{*}$ , $x$ with $y$ ) I have $T^{*}y=z=\frac{h\left(y,z_{0}\right)}{\left\langle z_{0},z_{0}\right\rangle }z_0,\hspace{1em}z_{0}\in\mathcal{N}\left(h\left(y,x\right)\right)^{\perp} $
with$h\left(y,x\right)=\left\langle z,x\right\rangle =\left\langle T^{*}y,x\right\rangle $
which, by definition of the Hilbert adjoint operator $\cdots=\left\langle y,Tx\right\rangle $
so does this give us an explicit form for the adjoing operator? i.e. $T^{*}y=\frac{\left\langle y,Tx\right\rangle }{\left\langle z_{0},z_{0}\right\rangle }z_0,\hspace{1em}z_{0}\in\mathcal{N}\left(\left\langle y,Tx\right\rangle \right)^{\perp} $
Is this a correct "explicit" form of the adjoint operator? the hard part seems to just find the $z_0$. If so, why can't I find this form of the adjoint operator anywhere? It seems difficult to "solve" for $T^*$ straight from the definition $\langle Tx,y\rangle=\langle x, T^*y\rangle$ (like as they did in here: Finding the adjoint of an operator, or as I've seen $V^* = A^H$ proved for matrices)
Thanks for your help.